Ulysses’ trip

Multi-pass Gaussian blur filter

2025-07-17T01:06:08-07:00

In image processing, a Gaussian blur filter is the transformation of an image that blurs it by a Gaussian function. Mathematically, a Gaussian blur filter is a linear transformation $G$ on a measurable function $f:\bR^d\to\bR$ that does not grow too fast at infinity (or more generally, $f:\bR^d\to V$ , where $V$ is a locally convex topological vector space) which results in a function $Gf:\bR^d\to\bR$ defined by the integral $\fc{Gf}x\ceq\int\frac{\d^dy}{\sqrt{\p{2\pi}^d\det A}}\fc\exp{-\fr12\p{A^{-1}}_{jk}y_jy_k}\fc f{x+y},$ where $A$ is a symmetric positive definite matrix called the covariance matrix of the Gaussian kernel.

By looking at this formula, we can see that $\fc{Gf}x$ is exactly the expectation value of $\fc f{x+Y}$ , where $Y$ is a random vector distributed according to the multivariate normal distribution with mean $0$ and covariance matrix $A$ . Let’s say, however, that instead of a multivariate normal distribution, we have a discrete distribution on a finite set of points $\B{a_i}$ , each with probability $w_i$ , satisfying $\sum_i w_i=1,\quad\sum_i w_ia_i=0.$ Then, for $Y$ distributed according to this discrete distribution, the expectation value of $\fc f{x+Y}$ is $\bopc E{\fc f{x+Y}}=\sum_i w_i\fc f{x+a_i}.$ This distribution has the covariance matrix $A_{jk}=\sum_i w_ia_{ij}a_{ik}$ , but this expectation value is not the same as the result $\fc{Gf}x$ of the Gaussian blur filter. In order to approach the Gaussian distribution, we need to sum $N$ independent samples $\B{Y_\alpha}$ , each distributed according the same discrete distribution with covariance matrix $A/N$ . Without changing the relation between $A$ and $\B{a_i}$ , we can achieve this by taking the distribution to be among the points $\B{a_i/\sqrt N}$ , with the same probabilities $w_i$ . Then, according to the central limit theorem, we have $\fc{Gf}x=\lim_{N\to\infty}\bopc E{\fc f{x+\sum_\alpha Y_\alpha}} =\lim_{N\to\infty}\sum_{\B{i_\alpha}}\p{\prod_\alpha w_{i_\alpha}}\fc f{x+\fr{1}{\sqrt N}\sum_\alpha a_{i_\alpha}}.$ By staring at this formula, one may notice that it is actually the result of the same linear transformation $P_N$ applied $N$ times to the function $f$ , where $\fc{P_Nf}x\ceq\sum_iw_i\fc f{x+\fr{a_i}{\sqrt N}}.$ Thus, we can write $G=\lim_{N\to\infty}P_N^N.$

We can now see why we can implement a Gaussian blur filter as a multi-pass filter: each pass of the filter is equivalent to applying the linear transformation $P_N$ once, and the Gaussian blur filter is the limit of applying this transformation infinitely many times. After choosing the points $\B{a_i}$ and the weights $w_i$ , we can easily implement the filter in a fragment shader. After implementing $P_N$ , one can implement the multi-pass filter by flip-flopping between two textures $N$ times. Here is an example implementation of $P_N$ for a 2D ( $d=2$ ) texture, with $a_0=\begin{pmatrix}\sgm\\0\end{pmatrix},\quad a_1=\begin{pmatrix}-\sgm\\0\end{pmatrix},\quad w_0=w_1=\fr12,$ where $\sgm$ would be the standard deviation of the resulting Gaussian blur filter:

          
            varying vec2 vTextureCoord;

            

            uniform sampler2D uTexture;

            uniform float uStrength; // sigma / sqrt(N)

            

            void main() {

            	gl_FragColor  = texture2D(uTexture, vTextureCoord + vec2( uStrength, 0.0)) * 0.5;

            	gl_FragColor += texture2D(uTexture, vTextureCoord + vec2(-uStrength, 0.0)) * 0.5;

            }

When using the shader, the texture is input as the uniform uTexture, and $\sgm/\sqrt N$ is input as the uniform uStrength. This example is then called the 2-tap horizontal Gaussian blur filter (because to find $\fc fx$ for some $x$ one needs to evaluate $f$ twice).

To get a good Gaussian blur effect, $N$ should be large enough, which can quickly become computationally expensive. We can reduce the number of passes by rewriting $P_N^N=\p{P_N^n}^{N/n}$ so that the number of passes is now reduced from $N$ to $N/n$ without changing the result, but the cost is to implement $P_N^n$ instead of $P_N$ in the shader. With the $P_N$ example above, the $P_N^n$ filter is then a $\p{n+1}$ -tap horizontal Gaussian blur filter. The total number of times to fetch the texture is $\p{n+1}N/n$ to render the whole blurred texture, which decreases as $n$ increases.

By utilizing the linear sampling feature of the GPU, this number can be further reduced (by half) if $\sgm/\sqrt N=1$ . One can read further about this in this article.

Let us now move on to an interesting relation to the heat equation. The heat equation is a partial differential equation that reads $\partial_t\fc{f_t}x=\sum_j\partial_j\partial_j\fc{f_t}x.$ The physical meaning of $\fc{f_t}x$ is the temperature at position $x$ at time $t$ , and the solution to this equation gives the time evolution of the temperature distribution in a medium with unit thermal diffusivity. More generally, we can have some other thermal diffusivities, which may even be anisotropic, in which case the equation reads $\partial_t\fc{f_t}x=\fr12\sum_{jk}A_{jk}\partial_j\partial_k\fc{f_t}x,$ where $A$ is a symmetric positive definite matrix. Such an equation can always be reduced to the form with unit thermal diffusivity by a linear transformation of the coordinates $x\mapsto Lx$ , where $L$ satisfies $A/2=L^\mrm TL$ (the Cholesky decomposition of $A/2$ ). By using operator exponentiation, we can write the solution to the heat equation as $\fc{f_t}x=G^t\fc{f_0}x$ , where $G^t\ceq\fc\exp{\fr t2\sum_{jk}A_{jk}\partial_j\partial_k}$ is called the time evolution operator (which is not generally defined for $t<0$ , which means that $\B{G^t}$ cannot form a group but only a monoid).

The reason that I denote it as $G^t$ is that $G^1$ gives exactly the Gaussian blur filter $G$ (when acting on real analytic functions). This is not immediately obvious, and I will justify its correctness by starting from the multi-pass expression of the Gaussian blur filter. For analytic functions, we can get the translation operator by exponentiating the derivative operator, so $\fc f{x+\fr{a_i}{\sqrt N}}=\fc\exp{\sum_j\fr{a_{ij}}{\sqrt N}\partial_j}\fc fx.$ Therefore, we can write $\begin{align*} P_N&=\sum_iw_i\fc\exp{\sum_j\fr{a_{ij}}{\sqrt N}\partial_j}\\ &=\underbrace{\sum_iw_i}_1 +\fr1{\sqrt N}\sum_j\underbrace{\sum_iw_ia_{ij}}_0\partial_j +\fr1{2N}\sum_{jk}\underbrace{\sum_iw_ia_{ij}a_{ik}}_{A_{jk}}\partial_j\partial_k +\order{N^{-3/2}}. \end{align*}$ Now, in the limit of infinite $N$ , we have $G=\lim_{N\to\infty}P_N^N =\lim_{N\to\infty}\p{1+\fr1{2N}\sum_{jk}A_{jk}\partial_j\partial_k}^N =\fc\exp{\fr12\sum_{jk}A_{jk}\partial_j\partial_k}=G^1.$ This means that applying the Gaussian blur filter is equivalent to evolving one unit of time according to the heat equation. Therefore, the equation in the beginning of this article, the definition of the Gaussian blur filter, can then be used to express the unit time evolution under the heat equation in the form of a integral transformation. To get the general $G^t$ , we can just replace $A$ with $tA$ .

As a byproduct, we can then show that the heat kernel is a Gaussian kernel. The heat kernel $\fc{K_t}{x,x'}$ is defined as the solution to the heat equation with initial condition $\fc{K_0}{x,x'}=\fc\dlt{x-x'}$ . Directly applying $G^t$ to $\fc\dlt{x-x'}$ gives $\begin{align*} \fc{K_t}{x,x'}&=G^t\fc\dlt{x-x'}\\ &=\int\fr{\d^dy}{\sqrt{\p{2\pi t}^d\det A}}\fc\exp{-\fr1{2t}\p{A^{-1}}_{jk}y_jy_k}\fc\dlt{x+y-x'}\\ &=\fr{1}{\sqrt{\p{2\pi t}^d\det A}}\fc\exp{-\fr1{2t}\p{A^{-1}}_{jk}\p{x_j-x'_j}\p{x_k-x'_k}}, \end{align*}$ which is the form that you would find in textbooks.

The reason that I decided to study the Gaussian blur filter is that I spotted a flaw in the implementation of the blur filter in PixiJS, where the uniform uStrength in the example shader above is scaled by $1/N$ instead of $1/\sqrt N$ . I was very happy to find the bug because this is the first time that I found a bug without actually producing an unexpected phenomenon first but by just staring at the source codes and deducing the mathematical formulas by hand. I opened an issue for my findings.

The role of particle indistinguishability in statistical mechanics

2025-03-03T22:53:47-08:00

Classical vs. quantum statistical mechanics

Previously, I have written two blog articles (part 1 about thermal ensembles and part 2 about non-thermal ensembles) about a formalism of statistical ensembles. I will be using it as the formalism of classical statistical mechanics in this article.

In that formalism, the space of microstates of a system is a measure space $\mcal M$ , and the physical meaning of the measure is the number of microstates. A macrostate is described by the extensive quantities, which are a function of the microstate, so designating a macrostate restricts the microstates that can realize it to a subset of $\mcal M$ .

A state of the system is a probability density function $p$ on $\mcal M$ , whose physical meaning is an ensemble of microstates. The macroscopically measured extensive quantities of the system are defined to be the ensemble average, i.e., the measured value of $A:\mcal M\to\bR$ is $\int_{\mcal M}pA$ . Generally, any probability density function is a perfectly valid state, but the most important ones are those that are thermal equilibrium states, including the microcanonical ensembles, the thermal ensembles, and the non-thermal ensembles. The term about an ensemble being thermal or non-thermal is made up by me, but for most practical reasons, we only need to focus on thermal ensembles (because both canonical ensembles and grand canonical ensembles are thermal ensembles).

To avoid subtleties about measure theory and topology, in this article, we will only use counting measure and discrete spaces for the space of microstates and the space of extensive quantities.

Possible confusion of macrostate vs. state and an example

The distribution when indistinguishable balls are put into boxes

2023-05-09T12:19:26-07:00

If there are 200 typographical errors randomly distributed in a 500 page manuscript, find the probability that a given page contains exactly 3 errors.

We can abstract this type of problems as follows:

Suppose there are $n$ distinguishable boxes and $k$ indistinguishable balls. Now, we randomly put the balls into the boxes. For each of the boxes, what is the probability that it contains $m$ balls?

For example, if the first page contains 3 errors, the second page contains 197 errors, and the rest of the pages contain no errors, then the situation corresponds to the situation where the first box contains 3 balls, the second box contains 197 balls, and the rest of the boxes contain no balls. The balls are indistinguishable because we can only determine how many errors are on each page but not which errors are on the page.

To deal with the problem, we simply need to find these two numbers:

the number of ways to put $k$ indistinguishable balls into $n$ distinguishable boxes, and
the number of ways to put $k-m$ indistinguishable balls into $n-1$ distinguishable boxes.

The latter corresponds to the number of ways to put the balls into the boxes provided that we already know that the given box contains $m$ balls. After we find these two numbers, their ratio is the probability in question.

To find the number of ways to put $k$ indistinguishable balls into $n$ distinguishable boxes, we can use the stars and bars method. To see this, we write a special example. Here is an example of $n=4$ and $k=6$ : ${}|{}\star{}\star{}|{}\star{}|{}\star{}\star{}\star{},$ which corresponds to the distribution $0,2,1,3$ . We can see that there are $n-1$ bars and $k$ stars. Therefore, the number of ways to put the balls is the same as the number of ways to choose the $k$ positions of the stars among $n+k-1$ positions. Therefore, the number of ways is $N_{n,k}=\binom{n+k-1}{k}=\frac{\left(n+k-1\right)!}{k!\left(n-1\right)!}.$ Therefore, the final probability of the given box containing $m$ balls is $P_{n,k}(m)=\frac{N_{n-1,k-m}}{N_{n,k}} =\frac{\left(n-1\right)k!\left(n+k-m-2\right)!}{\left(k-m\right)!\left(n+k-1\right)!}.$

Another easy way to derive this result is by using the generating function. The number $N_{n,k}$ is just the coefficient of $x^k$ in the expansion of the generating function $\left(1+x+x^2+\cdots\right)^n$ . The generating function is just $\left(1-x\right)^{-n}$ , which can be easily expanded by using the binomial theorem.

We are now interested in the limit $n,k\to\infty$ with $\lambda\coloneqq k/n$ fixed. By Stirling’s approximation, we have $P_{n,k}(m)\sim\left(n-1\right) \frac{k^{k+1/2}\left(n+k-m-2\right)^{n+k-m-2+1/2}}{\left(k-m\right)^{k-m+1/2}\left(n+k-1\right)^{n+k-1+1/2} } \mathrm e^{k-m+n+k-1-k-n-k+m+2}.$ The $1/2$ ’s in the exponents can just be dropped because you may find that if we extract the $1/2$ ’s, the factor tends to unity. The exponential is just constant $\mathrm e$ . Therefore, we have $\begin{align*} P_{n,k}(m)&\sim\left(n-1\right) \frac{\left(\lambda n\right)^{\lambda n}\left(n+\lambda n-m-2\right)^{n+\lambda n-m-2} } {\left(\lambda n-m\right)^{\lambda n-m}\left(n+\lambda n-1\right)^{n+\lambda n-1}}\mathrm e\\ &=\left(\tfrac{n+\lambda n-m-2}{n+\lambda n-1}\right)^n \left(\tfrac{\left(n+\lambda n-m-2\right)\lambda n}{\left(\lambda n-m\right)\left(n+\lambda n-1\right)}\right)^{\lambda n} \left(\tfrac{\lambda n-m}{n+\lambda n-m-2}\right)^m \tfrac{\left(n-1\right)\left(n+\lambda n-1\right)}{\left(n+\lambda n-m-2\right)^2}\mathrm e\\ &\to\mathrm e^{-\frac{m+1}{\lambda+1}}\,\mathrm e^m\, \mathrm e^{-\frac{m+1}{\lambda+1}\lambda}\left(\tfrac\lambda{\lambda+1}\right)^m\tfrac1{\lambda+1}\mathrm e\\ &=\left(\tfrac\lambda{\lambda+1}\right)^m\tfrac1{\lambda+1}. \end{align*}$ This is just the geometric distribution with parameter $p=1/(\lambda+1)=n/(k+n)$ .

If you want to simulate the number of balls in a box, here is a simple way to do this. First, because each box is the same, we can just focus on the first box without loss of generality. Then, we just need to randomly generate the positions of the $n-1$ bars among the $n+k-1$ positions, and then return the index of the first bar (which is the number of balls in the first box).

We can then write the following Ruby code to simulate the number of balls in the first box:

          
            def simulate n, k

              (n-1).times.inject(npkm1 = n+k-1) { |bar, i| [rand(npkm1 - i), bar].min }

            end

Compare the simulated result with the theoretical result:

          
            def frequency m, n, k, trials

              trials.times.count { simulate(n, k) == m } / trials.to_f

            end

            

            def truth m, n, k

              (n-1) * (k-m+1..k).reduce(1,:*) / (n+k-m-1..n+k-1).reduce(1,:*).to_f

            end

            

            def approx m, n, k

              n*k**m / ((n+k)**(m+1)).to_f

            end

            

            srand 1108

            m, n, k = 3, 5000, 8000

            p frequency m, n, k, 10000 # => 0.0902

            p truth m, n, k # => 0.08965012972626446

            p approx m, n, k # => 0.08963271594131858

A measure-theoretic formulation of statistical ensembles (part 2)

2023-05-01T16:26:42-07:00

This article follows part 1.

Introduction

In part 2, I will focus on non-thermal ensembles.

Before I proceed, I need to clarify that almost all ensembles that we actually use in physics are thermal ensembles, including the microcanonical ensemble, the canonical ensemble, and the grand canonical ensemble (the microcanonical ensemble can be considered as a special case of thermal ensemble where $\vec W^\parallel$ is the trivial).

The theory of thermal ensembles is built by letting the system in question be in thermal contact with a bath. Similarly, if we let the system in question be in non-thermal contact with a bath, we can get the theory of non-thermal ensembles. An example of non-thermal ensembles that is actually used in physics is the isoenthalpic–isobaric ensemble, where we let the system in question be in non-thermal contact with a pressure bath.

However, we will see that it is harder to measure-theoretically develop the theory of non-thermal ensembles if we continue to use the same method as in the theory of thermal ensembles.

Introducing non-thermal contact with an example

A thermal contact is a contact between thermal system that conducts heat (while exchanging some extensive quantities). A non-thermal contact is a contact between thermal system that does not conduct heat (while exchanging some extensive quantities). For reversible processes, thermodynamically and mathematically, heat is equivalent to a form of work, where the entropy is the displacement and where the temperature is the force. However, this is not true for non-reversible processes because of the Clausius theorem. This should have something to do with the fact that entropy is different from other extensive quantities (as is illustracted in part 1).

First, I may introduce how we may cope with the reversible processes of two subsystems in non-thermal contact in thermodynamics. As an example, consider a tank of monatomic ideal gas separated into two parts by a thermally non-conductive, massless, incompressible plate in the middle that can move. The two parts can then adiabatically exchange energy ( $U$ ) and volume ( $V$ ) but not number of particles ( $N$ ). For one of the parts, we have $0=\delta Q=\mathrm dU+p\,\mathrm dV=\mathrm dU+\frac{2U}{3V}\,\mathrm dV,$ which is good and easy to deal with because it is simply a differential 1-form.

However, this convenience is not possible for non-reversible processes because then we do not have the simple relation $p=2U/3V$ . Actually, the pressure is only well-defined for equilibrium states, and it is impossible to define a pressure that makes sense during the whole non-reversible process, which involves non-equilibrium states. Therefore, although it seems that the “thermally non-conductive” condition imposes a stronger restriction on what states can the composite system reach without external sources, it actually does not because the energy exchanged by the subsystems when they exchange volume is actually arbitrary (as long as it does not violate the second law of thermodynamics) if the process is not reversible.

The possible states of the non-thermally composite system then cannot be simply described by a vector subspace of $W^{(1)}\times W^{(2)}$ . If we try to use the same approach as constructing the thermally composite system to construct the non-thermally composite system, the attempt will fail.

Continuing with our example of a tank of gas. Although the pressure is not determined in the non-reversible process, there is one thing that is certain: the pressure on the plate by the gas on one side is equal to the pressure on the plate by the gas on the other side. This is because the plate must be massless (otherwise its kinetic energy would be an external source of energy; also, remember that it is incompressible: this means that it cannot be an external source of volume). Therefore, the relation between the volume exchanged and the energy exchanged is determined as long as at least one side of the plate is undergoing a reversible process because then the reversible side has determined pressure, which determines the pressure of the other side.

This is the key idea of formulating the non-thermal ensembles without formulating the non-thermally composite system. In a thermal or non-thermal ensemble, the composite system consists of two subsystems, one of which is the system in question, and the other is the bath which we are in control of. We can let the bath have zero relaxation time (the time for it to reach thermal equilibrium) so that any process of it is reversible. Then, the pressure (or generally, any other intensive quantities that we are in control of times the temperature) is determined (and actually constant), and we can express the non-conductivity restriction as $\mathrm dU+p\,\mathrm dV=0,$ where $p$ is the pressure, which is a constant. This is a homogeneous linear equation on $\vec W^{\parallel(1)}$ (whose vectors are denoted as $(\mathrm dU,\mathrm dV)$ in our case) which defines a vector subspace of $\vec W^{\parallel(1)}$ , which we call $\vec W^{\parallel\parallel(1)}$ . The dimension of $\vec W^{\parallel\parallel(1)}$ is that of $\vec W^{\parallel(1)}$ minus one. The physical meaning of $\vec W^{\parallel\parallel(1)}$ in this example is the hyperplane of fixed enthalpy.

Note that our bath actually has the fixed intensive quantities $i=\left(1/T,p/T\right)\in\vec W^{\parallel(1)\prime}$ , we can rewrite the above equation as $\vec W^{\parallel\parallel(1)} =\left\{s_1\in\vec W^{\parallel(1)}\,\middle|\,i\!\left(s_1\right)=0\right\}.$ Wait! What does $T$ do here? It is supposed to mean the temperature of the bath, but the temperature of the bath is irrelevant since the contact is non-thermal. Actually, it is. The temperature of the bath serves as an overall constant factor of $i$ , which does not affect $\vec W^{\parallel\parallel(1)}$ as long as it is not zero or infinite. So far, this means that the temperature of the bath is not necessarily fixed, so the actual number of fixed intensive quantities is the dimension of $\vec W^{\parallel(1)\prime}$ minus one, which is the same as the dimension of $\vec W^{\parallel\parallel(1)}$ . Later we will see that anything that is relevant to the temperature of the bath will finally be irrelevant to our problem. This seems magical, but you will see the sense in that after we introduce another way of developing the non-thermal ensembles (that do not involve baths and non-thermal contact) later.

We can define a complement of $\vec W^{\parallel\parallel(1)}$ in $\vec W^{\parallel(1)}$ as $\vec W^{\parallel\perp(1)}$ . Then, we have $\vec W^{\parallel(1)}=\vec W^{\parallel\parallel(1)}+\vec W^{\parallel\perp(1)}$ . The space $\vec W^{\parallel\perp(1)}$ is a one-dimensional vector space.

For convenience, define $W^{\star\perp(1)}\coloneqq W^{\perp(1)}+\vec W^{\parallel\perp(1)}$ . The vector space $\vec W^{\star\perp(1)}$ associated with it is a complement of $\vec W^{\parallel\parallel(1)}$ in $\vec W^{(1)}$ . To make the notation look more consistent, we can use $\vec W^{\star\parallel(1)}$ as an alias of $\vec W^{\parallel\parallel(1)}$ . They are the same vector space, but $\vec W^{\star\parallel(1)}$ emphasizes that it is a subspace of $\vec W^{(1)}$ , and $\vec W^{\parallel\parallel(1)}$ emphasizes that it is a subspace of $\vec W^{\parallel(1)}$ . Then, we have $W^{(1)}=W^{\star\perp(1)}+\vec W^{\star\parallel(1)}$ . Every point in $W^{(1)}$ can be uniquely written as a sum of a point in $W^{\star\perp(1)}$ and a vector in $\vec W^{\star\parallel(1)}$ . We can describe the decomposition by a projection $\pi^{\star(1)}:W^{(1)}\to W^{\star\perp(1)}$ .

We will heavily use the “ $\star$ ” on the superscripts of symbols. Any symbol labeled with “ $\star$ ” is dependent on $i$ (but independent on an overall constant factor on $i$ ). You can regard those symbols to have an invisible “ $i$ ” in the subscript so that you can keep in mind that they are dependent on $i$ .

Example. Suppose we have a tank of gas with three extensive quantities $U,V,N$ . It is in non-thermal contact with a pressure bath with pressure $p$ so that it can exchange $U$ and $V$ with the bath. Then, the projection $\pi^{\star(1)}$ projects macrostates with the same enthalpy and number of particles into the same point. Because a complement of a vector subspace is not determined, there are multiple possible ways of constructing the projection. One possible way is $\pi^{\star(1)}\!\left(U,V,N\right)\coloneqq\left(U+pV,0,N\right).$ Here the fixed intensive quantity $p$ is involved. Note that this projection is still valid for different temperatures of the bath, so an overall constant factor of $i$ does not affect the projection.

Non-thermal contact with a bath

Now, after introducing non-thermal contact with an example, we can now formulate the non-thermal contact with a bath.

Suppose we have a system $\left(\mathcal E^{(1)},\mathcal M^{(1)}\right)$ . The main approach is constructing a composite system out of the composite system for the $\vec W^{\parallel(1)}$ -ensemble.

The composite system for the $\vec W^{\parallel(1)}$ -ensemble was introduced in part 1. We denote the bath that is in contact with our system as $\left(\mathcal E^{(2)},\mathcal M^{(2)}\right)$ .

Consider this projection $\pi^\star:W\to W^{\star\perp}$ (where $W^{\star\perp}$ is an affine subspace of $W$ and the range of $\pi^\star$ ): $\pi^\star\!\left(e_1,e_2\right) \coloneqq\left(\pi^{\star(1)}\!\left(e_1\right), \rho_{\pi(e_1,e_2)}\!\left(\pi^{\star(1)}\!\left(e_1\right)\right)\right).$ To ensure that it is well-defined, we need to guarantee that $\pi^{\star(1)}\!\left(e_1\right)\in W^{\parallel(1)}_{\pi(e_1,e_2)}$ for any $e_1,e_2$ , and this is true.

The two spaces $W^{\star\perp}$ and $W^{\perp}$ do not have any direct relation. The only relation between them is that the dimension of $W^{\star\perp}$ is one plus the dimension of $W^{\perp}$ (if they are finite-dimensional).

What is good about the projection $\pi^\star$ is that it satisfies $\vec W^{\star\parallel(1)}=\vec c^{(1)}\!\left(\vec\pi^\star(0)\right)$ . This makes our notation consistent if we construct another composite system out of $\pi^\star$ . Now, consider the composite system of $\left(\mathcal E^{(1)},\mathcal M^{(1)}\right)$ and $\left(\mathcal E^{(2)},\mathcal M^{(2)}\right)$ under the projection $\pi^\star$ . In the notation of the spaces and mappings that are involved in the newly constructed composite system, we write “ $\star$ ” in the superscript.

Just like how $\vec W^{\star\parallel(1)}$ is a subspace of $\vec W^{(1)}$ , $\vec W^{\star\parallel(2)}$ is also a subspace of $\vec W^{(2)}$ . This means that both $\vec\rho^{-1}\circ\vec\rho^\star$ and $\vec\rho\circ\vec\rho^{\star-1}$ are well-defined. The former maps $\vec W^{\star\parallel(1)}$ to another subspace of $\vec W^{(1)}$ , and the latter maps $\vec W^{\star\parallel(2)}$ to another subspace of $\vec W^{(2)}$ .

We can regard the construction of the new composite system as replacing the “plate” between the subsystems in the original composite system from a “thermally conductive plate” to a “thermally non-conductive plate”. Suppose that in the new situation, the intensive quantities “felt” by subsystem 1 is $i^\star\in\vec W^{\star\parallel(1)\prime}$ . Then, because the bath is still the same bath in the two situations, we have $-i^\star\circ\vec\rho^{\star-1}=-i\circ\vec\rho^{-1}.$ Therefore, $i^\star\coloneqq i\circ\vec\rho^{-1}\circ\vec\rho^\star$ would be a good definition of $i^\star$ . However, actually $i^\star$ is trivial: $i^\star=0.$ This is because 2 shows that $\rho\!\left(W^{\star\parallel(1)}_e\right)=W^{\star\parallel(2)}_e$ , and thus $\vec\rho^{-1}\!\left(\vec\rho^\star\!\left(\vec W^{\star\parallel(1)}\right)\right) =\vec W^{\star\parallel(1)},$ which is the kernel of $i$ by definition.

Because $i^\star$ is trivial, it is irrelevant to the temperature of the bath because it is zero no matter what temperature the bath is at.

Example. Suppose a system described by $U_1,V_1,N_1$ is in non-thermal contact with a pressure bath, and they can exchange energy and volume. The projection $\pi$ is $\pi\!\left(U_1,V_1,N_1,U_2,V_2,N_2\right) =\left(\frac{U_1+U_2}2,\frac{V_1+V_2}2,N_1,\frac{U_1+U_2}2,\frac{V_1+V_2}2,N_2\right).$ Then, the projection $\pi^\star$ can be $\pi^\star\!\left(U_1,V_1,N_1,U_2,V_2,N_2\right) =\left(U_1+pV_1,0,N_1,U_2-pV_1,V_1+V_2,N_2\right).$ By choosing a different $\pi^{\star(1)}$ or a different $\pi$ , we can get a different $\pi^\star$ . They physically mean the same composite system.

The space $W^\perp$ is four-dimensional, and the space $W^{\star\perp}$ is five-dimensional. We can denote the five degrees of freedom as $U,V,H_1,N_1,N_2$ , where $U\coloneqq U_1+U_2$ is the total energy, $V\coloneqq V_1+V_2$ is the total volume, and $H_1\coloneqq U_1+pV_1$ is the enthalpy of subsystem 1. Then, the projection $\pi^\star$ can be written as $\pi^\star\!\left(U_1,V_1,N_1,U_2,V_2,N_2\right) =\left(H_1,0,N_1,U-H_1,V,N_2\right).$ We can get $W^{\star\parallel}_e$ by finding the inverse of the projection, where $e\coloneqq\left(H_1,0,N_1,U-H_1,V,N_2\right)$ : $W^{\star\parallel}_e\coloneqq\pi^{\star-1}\!\left(e\right) =\left\{\left(H_1-pV_1,V_1,N_1,U-H_1+pV_1,V-V_1,N_2\right)\middle|\,V_1\in\mathbb R\right\}.$ Because it is parameterized by one real parameter $V_1$ , it is a one-dimensional affine subspace of $W$ . Projecting it under $c^{(1)}$ and $c^{(2)}$ will respectively give us $W^{\star\parallel(1)}_e$ and $W^{\star\parallel(2)}_e$ : $W^{\star\parallel(1)}_e \coloneqq\left\{\left(H_1-pV_1,V_1,N_1\right)\middle|\,V_1\in\mathbb R\right\},$ $W^{\star\parallel(2)}_e \coloneqq\left\{\left(U-H_1+pV_1,V-V_1,N_2\right)\middle|\,V_1\in\mathbb R\right\}.$

The affine isomorphism $\rho^\star_e$ is then naturally $\rho^\star_e\!\left(H_1-pV_1,V_1,N_1\right)=\left(U-H_1+pV_1,V-V_1,N_2\right).$ Its vectoric form is then $\vec\rho^\star\!\left(-p\,\mathrm dV_1,\mathrm dV_1,0\right) =\left(p\,\mathrm dV_1,-\mathrm dV_1,0\right).$

Our fixed intensive quantities are $i$ , which is defined as $i\!\left(\mathrm dU_1,\mathrm dV_1,0\right)=\frac1T\,\mathrm dU_1+\frac pT\,\mathrm dV_1$ . We can then get $i^\star$ by $i^\star\coloneqq i\circ\vec\rho^{-1}\circ\vec\rho^\star =\left(-p\,\mathrm dV_1,\mathrm dV_1,0\right)\mapsto0.$ This is consistent with Equation 4.

Non-thermal ensembles (bath version)

Now, we can define the non-thermal contact with a bath to be the same as the thermal contact with a bath under $\pi^\star$ . Utilizing this definition, we can define the composite system for non-thermal ensembles.

Definition. A composite system for the non-thermal $\vec W^{\parallel(1)}$ -ensemble of the system $\left(\mathcal E^{(1)},\mathcal M^{(1)}\right)$ with fixed intensive quantities $i$ is the same as the composite system for the thermal $\vec W^{\star\parallel(1)}$ -ensemble with fixed intensive quantities $i^\star=0$ (given by Equation 4), where $\vec W^{\star\parallel(1)}$ is defined by Equation 1.

This definition looks very neat. Also, just like how we define the domain of fixed intensive quantities of a thermal ensemble, we can define the domain of fixed intensive quantities of a non-thermal ensemble to consist of those values that make the integral in the definition of the partition function converge.

Because we already derived the formula of the partition function in part 1 that does not involve information about the bath anymore, we can drop the “ $(1)$ ” in the superscripts. The partition function of the non-thermal ensemble is then $Z^\star\!\left(e,i^\star\right)=\int_{s\in\vec E^{\star\parallel}_e} \Omega\!\left(e+s\right) \mathrm e^{-i^\star\left(s\right)}\,\mathrm d\lambda^{\parallel}\!\left(s\right),\quad e\in E^{\star\perp},\quad i^\star\in I^\star_e\subseteq\vec W^{\star\parallel\prime}.$ Here, the $i^\star$ is not fixed at the trivial value $0$ (I abused the notation here) but actually is an independent variable serving as one of the arguments of the partition function that takes values in $I^\star_e$ (which is not the domain of fixed intensive quantities of the non-thermal ensemble that was mentioned above).

However, the only meaningful information about this non-thermal ensemble is in the behavior of $Z^\star$ at $i^\star=0$ instead of any arbitrary $i^\star\in I^\star_e$ , but we do not know whether $0\in I^\star_e$ or not. This is then a criterion of judge whether $i$ is in the domain of fixed intensive quantities of the non-thermal ensemble or not. To be clear, we define $J\coloneqq\left\{i\in\vec W^{\parallel\prime}\,\middle|\, \exists e\in E^{\star\perp}:0\in I^\star_{e}\right\}.$ A problem about this formulation is that it is possible to have two $i$ ’s that share the same thermal equilibrium state. In that case, the non-thermal ensemble is not defined.

Because $i^\star=0$ , the observed extensive quantities in thermal equilibrium are just $\varepsilon^\circ =e+\left.\frac{\partial\ln Z^\star\!\left(e,i^\star\right)}{\partial i^\star}\right|_{i^\star=0} =e+\frac{\int_{s\in\left(E-e\right)\cap\vec W^{\star\parallel}} s\Omega\!\left(e+s\right)\mathrm d\lambda^{\parallel}\!\left(s\right)} {\int_{s\in\left(E-e\right)\cap\vec W^{\star\parallel}} \Omega\!\left(e+s\right)\mathrm d\lambda^{\parallel}\!\left(s\right)},$ and the entropy in thermal equilibrium is just $S^\circ=\ln Z^\star\!\left(e,0\right) =\ln\int_{s\in\left(E-e\right)\cap\vec W^{\star\parallel}} \Omega\!\left(e+s\right)\mathrm d\lambda^{\parallel}\!\left(s\right).$ We can cancel the parameter $e$ by Equation 5 and 6 to get $S^\circ=\ln Z^\star\!\left(\pi^\star\!\left(\varepsilon^\circ\right),0\right) =\ln\int_{s\in\left(E-\varepsilon^\circ\right)\cap\vec W^{\star\parallel}} \Omega\!\left(\varepsilon^\circ+s\right)\mathrm d\lambda^{\parallel}\!\left(s\right).$

What is interesting about Equation 7 is that it actually does not guarantee the intensive variables to be defined in $\vec W^\parallel$ . Physically this means that the temperature is not necessarily defined, unlike the case of thermal ensembles (this is because the thermal contact makes the temperature the same as the bath and thus defined). The thing that is guaranteed is that the intensive variables are defined in $\vec W^{\star\parallel}$ and they must be zero. Therefore, whenever the intensive variables are defined in $\vec W^\parallel$ , it must be parallel to $i$ (and remains the same if we scale $i$ by an arbitrary non-zero factor). Physically, this means that the system must have the same intensive variables as the bath up to different temperatures.

Non-thermal ensembles (non-bath version)

It may seem surprising that we can define non-thermal ensembles without a bath. How is it possible to fix some features about the intensive variables without a bath? The inspiration is looking at Equation 1. We can make a guess here: if we contract the system along $\vec W^{\star\parallel}$ , the contraction satisfy the equal a priori probability principle. We make this guess because of the following arguments:

Mathematically, contraction is a legal new system, so it should also satisfy the axioms that we proposed before.
Physically, because the temperature of the bath is arbitrary, the different accessible macrostates should not be too different because otherwise the temperature would matter (as appears in the expression of the partition function).

After finding the equilibrium state of the contraction, we can use the contractional pullback to find the equilibrium state of the original system.

If you do it right, you should get the same answer as Equation 7.

Summary

The only axiom that we used is the equal a priori probability principle. Then, we formulated three types of ensembles: microcanonical, thermal, and non-thermal.

A measure-theoretic formulation of statistical ensembles (part 1)

2023-03-30T21:49:51-07:00

I feel that the process of using statistical ensembles to find properties of thermal system is not rigorous enough. There are some operations that need to be defined precisely. Also, it is not generalized enough. Currently, the only generally used statistical ensembles are the microcanonical ensemble, the canonical ensemble, and the grand canonical ensemble, but there are actually other possible ensembles that are potentially useful. Therefore, I feel it necessary to try to have a mathematical formulation.

Mathematical tools and notations

Suppose $(\Omega,\sigma(\Omega),P)$ is a probability space. Suppose $W$ is an affine space. For some map $f:\Omega\to W$ , we define the $P$ -expectation of $f$ as $\mathrm E_P\!\left[f\right]\coloneqq\int_{x\in\Omega}\left(f(x)-e_0\right)\mathrm dP(x)+e_0,$ where $e_0\in W$ is arbitrary. Here the integral is Pettis integral. The expectation is defined if the Pettis integral is defined, and it is then well-defined in that it is independent of the $e_0$ we choose.

Suppose $X,Y$ are Polish spaces. Suppose $(Y,\sigma(Y),\mu),(X,\sigma(X),\nu)$ are measure spaces, where $\mu$ and $\nu$ are σ-finite Borel measures. Suppose $\pi:Y\to X$ is a measurable map so that $\forall A\in\sigma(X):\nu(A)=0\Rightarrow\mu\!\left(\pi^{-1}\!\left(A\right)\right)=0.$ Then, for each $x\in X$ , there exists a Borel measure $\mu_x$ on the measurable subspace $\left(\pi^{-1}(x),\sigma\!\left(\pi^{-1}(x)\right)\right)$ , such that for any integrable function $f$ on $Y$ , $\int_{y\in Y}f\!\left(y\right)\mathrm d\mu(y) =\int_{x\in X}\mathrm d\nu(x)\int_{y\in\pi^{-1}(x)}f\!\left(y\right)\mathrm d\mu_x(y).$

Proof

Relationship between the Gini coefficient and the variance

2023-02-06T16:38:25-08:00

This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2021-04-25 10:06 +0800.

First, define the Lorenz curve: it is the curve that consists of all points $(u,v)$ such that the poorest $u$ portion of population in the country owns $v$ portion of the total wealth.

The Gini coefficient $G/\mu$ is defined as the area between the Lorenz curve and the line $u=v$ divided by the area enclosed by the three lines $u=v$ , $v=0$ , and $u=1$ .

Now, suppose the wealth distribution in the country is $p(X)$ , where $p\!\left(x\right)\mathrm dx$ is the portion of population that has wealth in the range $[x,x+\mathrm dx]$ .

Then, the Lorenz curve is the graph of the function $g$ defined as $g(F(x))=\frac1\mu\int_{-\infty}^xtp\!\left(t\right)\mathrm dt,$ where $F\!\left(x\right)\coloneqq\int_{-\infty}^xp\!\left(t\right)\mathrm dt$ is the cumulative distribution function of $p(X)$ , and $\mu\coloneqq\int_{-\infty}^{+\infty}tp\!\left(t\right)\mathrm dt$ is the average wealth of the population, which is just $\mathrm E[\mathrm X]$ ( $X$ is a random variable such that $X\sim p(X)$ ).

Then, the Lorenz curve is $v=g(u)\coloneqq\frac1\mu\int_{-\infty}^{F^{-1}(u)}tp\!\left(t\right)\mathrm dt.$

According to the definition of the Gini coefficient, $\begin{align*} G&\coloneqq2\mu\int_0^1\left(u-g(u)\right)\mathrm du\\ &=\mu-2\mu\int_0^1g\!\left(u\right)\mathrm du\\ &=\mu-2\int_{u=0}^1\int_{t=-\infty}^{F^{-1}(u)}tp\!\left(t\right)\mathrm dt\,\mathrm du. \end{align*}$ Interchange the order of integration, and we have $\begin{align*} G&=\mu-2\int_{t=-\infty}^{+\infty}\int_{u=F(t)}^1tp\!\left(t\right)\mathrm dt\,\mathrm du\\ &=\mu-2\int_{-\infty}^{+\infty}\left(1-F(t)\right)tp\!\left(t\right)\mathrm dt. \end{align*}$ Substitute Equation 1 into the above equation, and we have $\begin{align*} G&=\int_{-\infty}^{+\infty}2tF\!\left(t\right)p\!\left(t\right)\mathrm dt-\mu\\ &=\int_{-\infty}^{+\infty}\left(2tF\!\left(t\right)-1\right)tp\!\left(t\right)\mathrm dt\\ &=\int_0^1\left(2u-1\right)F^{-1}\!\left(u\right)\mathrm du. \end{align*}$ Now here is the neat part. Separate it into two parts, and write them in double integrals: $\begin{align*} G&=\int_0^1uF^{-1}\!\left(u\right)\mathrm du-\int_0^1\left(1-u\right)F^{-1}\!\left(u\right)\mathrm du\\ &=\int_{u_2=0}^1\int_{u_1=0}^{u_2}F^{-1}\!\left(u_2\right)\mathrm du_1\,\mathrm du_2 -\int_{u_1=0}^1\int_{u_2=u_1}^1F^{-1}\!\left(u_1\right)\mathrm du_1\,\mathrm du_2. \end{align*}$ Interchange the order of integration of the second term, and we have $\begin{align*} G&=\int_{u_2=0}^1\int_{u_1=0}^{u_2}\left(F^{-1}\!\left(u_2\right)-F^{-1}\!\left(u_1\right)\right)\mathrm du_1\,\mathrm du_2\\ &=\frac12\int_{u_2=0}^1\int_{u_1=0}^1\left|F^{-1}\!\left(u_2\right)-F^{-1}\!\left(u_1\right)\right|\mathrm du_1\,\mathrm du_2\\ &=\frac12\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\left|x_2-x_1\right|p\!\left(x_1\right)p\!\left(x_2\right)\mathrm dx_1\,\mathrm dx_2\\ &=\frac12\mathrm E\!\left[\left|X_2-X_1\right|\right], \end{align*}$ where $X_1$ and $X_2$ are two independent random variables with $p$ being their respective distribution functions: $\left(X_1,X_2\right)\sim p\!\left(X_1\right)p\!\left(X_2\right)$ .

By this result, we can easily see how the Gini coefficient represents the statistical dispersion.

We can apply similar tricks to the variance $\sigma_X^2$ . $\begin{align*} \sigma_X^2&=\mathrm E\!\left[X^2\right]-\mathrm E\!\left[X\right]^2\\ &=\int_{-\infty}^{+\infty}t^2p\!\left(t\right)\mathrm dt -\left(\int_{-\infty}^{+\infty}tp\!\left(t\right)\mathrm dt\right)^2\\ &=\int_0^1F^{-1}\!\left(u\right)^2\,\mathrm du -\left(\int_0^1F^{-1}\!\left(u\right)\mathrm du\right)^2. \end{align*}$ Separate the first into two halves, and write the altogether three terms in double integrals: $\begin{align*} \sigma_X^2&=\frac12\int_0^1F^{-1}\!\left(u_2\right)^2\,\mathrm du_2\int_0^1\mathrm du_1\\ &\phantom{=~}{}-\int_0^1F^{-1}\!\left(u_1\right)\mathrm du_1\int_0^1F^{-1}\!\left(u_2\right)\mathrm du_2\\ &\phantom{=~}{}+\frac12\int_0^1F^{-1}\!\left(u_1\right)^2\,\mathrm du_1\int_0^1\mathrm du_2\\ &=\frac12\int_0^1\int_0^1 \left(F^{-1}\!\left(u_2\right)^2-2F^{-1}\!\left(u_1\right)F^{-1}\!\left(u_2\right)+F^{-1}\!\left(u_1\right)^2\right) \mathrm du_1\,\mathrm du_2\\ &=\frac12\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \left(x_2-x_1\right)^2p\!\left(x_1\right)p\!\left(x_2\right)\mathrm dx_1\,\mathrm dx_2\\ &=\frac12\mathrm E\!\left[\left(X_2-X_1\right)^2\right]. \end{align*}$ Then we can derive the relationship between the Gini coefficient and the variance: $2\sigma_X^2-4G^2=\sigma_{\left|X_2-X_2\right|}^2.$

This is what will happen after you get a haircut

2023-01-18T12:11:41-08:00

Denote the length distribution of one’s hair to be $f(l,t)$ . This means that, at time $t$ , the number of hairs within the length range from $l$ to $l+\mathrm dl$ is $Nf\!\left(l,t\right)\mathrm dl$ , where $N$ is the total number of hairs.

Each hair grows at constant speed $v$ . However, they cannot grow indefinitely because there is a probability of $\lambda$ per unit time for a hair to be lost naturally (this is the same assumption as the exponential decay). After a hair is lost, it restarts growing from zero length.

Suppose that at $t=0$ you have got a haircut so that the hair length distribution becomes $f(l,0)=f_0(l).$ Then, how does the distribution evolve with time?

Because $f$ is a distribution function, There is a normalization restriction on $f$ : $\int_0^\infty f\!\left(l,t\right)\mathrm dl=1,\quad t\ge0.$ This normalization condition also applies to the initial condition ( $t=0$ ). This means that the function $f_0$ also satisfies the normalization restriction (Equation 2) $\int_0^\infty f_0\!\left(l\right)\mathrm dl=1.$ Because of the natural loss of hair, only a $1-\lambda\,\mathrm dt$ portion of hair will survive $\mathrm dt$ . According to this, we can construct the following equation: $f\!\left(l+v\,\mathrm dt,t+\mathrm dt\right)= \left(1-\lambda\,\mathrm dt\right)f\!\left(l,t\right).$ This equation can be reduced to a first-order linear PDE: $v\frac{\partial f}{\partial l}+\frac{\partial f}{\partial t}+\lambda f=0.$

Define $\theta\coloneqq l-vt$ , and define a new function $g\!\left(\theta,t\right)\coloneqq f\!\left(\theta+vt,t\right).$ Then, $\frac{\partial g}{\partial t}=v\frac{\partial f}{\partial l}+\frac{\partial f}{\partial t}.$ Then, Equation 4 can be reduced to $\frac{\partial g}{\partial t}+\lambda g=0.$ The solution is $g\!\left(\theta,t\right)=\Phi\!\left(\theta\right)\mathrm e^{-\lambda t},$ where $\Phi$ is an arbitrary function defined on $\mathbb R$ . Therefore, the general solution to Equation 4 is $f\!\left(l,t\right)=\Phi\!\left(l-vt\right)\mathrm e^{-\lambda t}.$

By utilizing Equation 1, we can find $\Phi(\theta)$ for $\theta>0$ . Substitute $t=0$ into Equation 5 and compare with Equation 1, and we have $\Phi(\theta>0)=f_0(\theta).$ This only gives $\Phi(\theta)$ for $\theta>0$ because $f_0$ is not defined on negative numbers. The rest of $\Phi$ , however, may be deduced from Equation 2.

Substitute Equation 5 into Equation 2, and we have $\begin{align*} 1&=\int_0^\infty\Phi\!\left(l-vt\right)\mathrm e^{-\lambda t}\,\mathrm dl\\ &=\mathrm e^{-\lambda t}\left( \int_0^{vt}\Phi\!\left(l-vt\right)\mathrm dl +\int_{vt}^\infty\Phi\!\left(l-vt\right)\mathrm dl \right)\\ &=\mathrm e^{-\lambda t}\left( \int_{-vt}^0\Phi\!\left(\theta\right)\mathrm d\theta +\int_0^\infty f_0\!\left(\theta\right)\mathrm d\theta \right)\\ &=\mathrm e^{-\lambda t}\left(1-\int_0^{-vt}\Phi\!\left(\theta\right)\mathrm d\theta\right). \end{align*}$ Therefore, we have the integral of $\Phi$ on negative intervals: $\int_0^{-vt}\Phi\!\left(\theta\right)\mathrm d\theta =1-\mathrm e^{\lambda t}.$ Find the derivative of both sides of the equation w.r.t. $t$ , and we have $-v\Phi\!\left(-vt\right)=-\lambda\mathrm e^{\lambda t}.$ In other words, $\Phi\!\left(\theta<0\right)=\frac\lambda v\mathrm e^{-\frac\lambda v\theta}.$

Combining Equation 6 and 7 and substituting back to Equation 5, we can find the special solution to Equation 4 subject to restrictions Equation 2 and 1: $f(l,t)=\begin{cases} \frac\lambda v\mathrm e^{-\frac\lambda vl},&0vt. \end{cases}$

This is our final answer.

This result is interesting in that any distribution will finally evolve into an exponential distribution with the rate parameter being $\frac\lambda v$ as $t\to\infty$ no matter what the initial distribution is: $f(l,\infty)=\frac\lambda v\mathrm e^{-\frac\lambda vl}.$ This distribution is the stationary solution to Equation 4. This is actually a normal behavior for first-order PDEs. For example, the thermal equilibrium state is the stationary solution to the heat equation, and any other solution approaches to the stationary solution over time.

This behavior can explain why human body hair tends to grow to only a certain length instead of being indefinitely long. You may try shaving your leg hair and wait for some weeks. You can observe that they grow to approximately the original length but not any longer. It is similar for your hair (on top of your head), but $\lambda$ of hair is so small that it can hardly reach its terminal length if you get haircuts regularly.

Another thing to note is that this may explain a phenomenon that we may observe: the longer your hair is, the more slowly it grows, and your hair no longer seems to grow when it reaches a certain length. If the length of hair that we observe is actually the mean length of the hair, then it is $\mu(t)=\int_0^\infty lf\!\left(l,t\right)\mathrm dl =\left(\mu_0-\frac v\lambda\right)\mathrm e^{-\lambda t}+\frac v\lambda,$ where $\mu_0$ is the mean of the distribution $f_0$ . It can be seen that the growth rate of the mean length of hair varies exponentially.

The longest all- $1$ substring of a random bit string

2022-12-25T12:00:00-08:00

Introduction

As a rhythm game player, I often wonder what my max combo will be in my next play. This is a rather unpredictable outcome, and what I can do is to try to conclude a probability distribution of my max combo.

For those who are not familiar with rhythm games and also to make the question clearer, I state the problem in a more mathematical setting.

Consider a random bit string of length $n\in\mathbb N$ , where each bit is independent and has probability $Y\in[0,1]$ of being $1$ . Let $P_{n,k}(Y)$ be the probability that the length of the longest all- $1$ substring of the bit string is $k\in\mathbb N$ (where obviously $P_{n,k}(Y)$ is nonzero only when $k\le n$ ). What is the expression of $P_{n,k}(Y)$ ?

A more interesting problem to consider is what the probability distribution tends to be when $n\to\infty$ . Define the random variable $\kappa\coloneqq k/n$ where $k$ is the length of the longest all- $1$ substring. Define a parameter $y\coloneqq Y^n$ (this parameter is held constant while $n\to\infty$ ). Define the probability distribution function of $\kappa$ as $f(y,\kappa)\coloneqq\lim_{n\to\infty}\left(n+1\right)P_{n,\kappa n}\!\left(y^{\frac1n}\right).$ What is the expression of $f(y,\kappa)$ ?

Notation

Notation for integer range: $a\ldots b$ denotes the integer range defined by the ends $a$ (inclusive) and $b$ (exclusive), or in other words $\left\{a,a+1,\ldots,b-1\right\}$ . It is defined to be empty if $a\ge b$ . The operator $\ldots$ has a lower precedence than $+$ and $-$ but a higher precedence than $\in$ .

The notation $a\,..b$ denotes the inclusive integer range $\left\{a,a+1,\ldots,b\right\}$ . It is defined to be empty if $a>b$ .

The case for finite $n$

A natural approach to find $P_{n,k}$ is to try to find a recurrence relation of $P_{n,k}$ for different $n$ and $k$ , and then use a dynamic programming (DP) algorithm to compute $P_{n,k}$ for any given $n$ and $k$ .

The first DP approach

For a rhythm game player, the most straightforward way of finding $k$ for a given bit string is to track the current combo, and update the max combo when the current combo is greater than the previous max combo.

To give the current combo a formal definition, denote each bit in the bit string as $b_i$ , where $i\in0\ldots n$ . Define the current combo $r_i$ as the length of the longest all- $1$ substring of the bit string ending before (exclusive) $i$ (so $r_i=0$ if $b_{i-1}=0$ , which is callled a combo break): $r_i\coloneqq\max\left\{r\in0\,..i\,\middle|\,\forall j\in i-r\ldots i:b_j=1\right\},$ where $i\in0\,..n$ .

Now, use three numbers $(n,k,r)$ to define a DP state. Denote $P_{n,k,r}$ to be the probability that the max combo is $k$ and the final combo ( $r_n$ ) is $r$ . Then, consider a transition from state $(n,k,r)$ to state $(n+1,k',r')$ by adding a new bit $b_n$ to the bit string. There are two cases:

If $b_n=0$ (has $1-Y$ probability), then this means a combo break, so we have $r'=0$ and $k'=k$ .
If $b_n=1$ (has $Y$ probability), then the combo continues, so we have $r'=r+1$ . The max combo needs to be updated if needed, so we have $k'=\max(k,r')$ .

However, in actual implementation of the DP algorithm, we need to reverse this transition by considering what state can lead to the current state $(n,k,r)$ (to use the bottom-up approach).

First, obviously in any possible case $r\in0\,..k$ (currently we only consider the cases where $n>k>0$ ). Divide all those cases into three groups:

If $r=0$ , this is means a combo break, so the last bit is $0$ , and the previous state can have any possible final combo $r'$ . Therefore, it can be transitioned from any $(n-1,k,r')$ where $r'\in0\,..k$ . For each possible previous state, the probability of the transition to this new state is $1-Y$ .
If $r\in1\,..k-1$ , this means the last bit is $1$ , the previous final combo is $r-1$ , and the previous max combo is already $k$ . Therefore, the previous state is $(n-1,k,r-1)$ , and the probability of the transition is $Y$ .
If $r=k$ , this means the max combo may (or may not) have been updated. In either case, the previous final combo is $r-1=k-1$ .

If the max combo is updated, the previous max combo must be $k-1$ because it must not be less than the previous final combo $k-1$ and must be less than the new max combo $k$ . Therefore, the previous state is $(n-1,k-1,k-1)$ , and the probability of the transition is $Y$ .
If the max combo is not updated, the previous max combo is the same as the new one, which is $k$ . Therefore, the previous state is $(n-1,k,k-1)$ , and the probability of the transition is $Y$ .

Therefore, we can write a recurrence relation that is valid when $n>k>0$ : $P_{n,k,r}=\begin{cases} \left(1-Y\right)\sum_{r'=0}^kP_{n-1,k,r'},&r=0\\ YP_{n-1,k,r-1},&r\in1\,..k-1\\ Y\left(P_{n-1,k-1,k-1}+P_{n-1,k,k-1}\right),&r=k. \end{cases}$

However, there are also other cases (mostly edge cases) because we assumed $n>k>0$ . Actually, in the meaningfulness condition $n\ge k\ge r\ge0$ (necessary condition for $P_{n,k,r}$ to be nonzero), there are three inequality that can be altered between a less-than sign or an equal sign, so there are totally $2^3=8$ cases. Considering all those cases (omitted in this article because of the triviality), we can write a recurrence relation that is valid for all $n,k,r$ , covering all the edge cases: $P_{n,k,r}=\begin{cases} 1,& n=k=r=0,\\ YP_{n-1,n-1,n-1},& n=k=r>0,\\ 0,& n=k>r>0,\\ 0,& n=k>r=0,\\ Y\left(P_{n-1,k-1,k-1}+P_{n-1,k,k-1}\right),& n>k=r>0,\\ YP_{n-1,k,r-1},& n>k>r>0,\\ \left(1-Y\right)\sum_{r'=0}^kP_{n-1,k,r'},& n>k>r=0,\\ \left(1-Y\right)P_{n-1,0,0},& n>k=r=0. \end{cases}$

Note that the probabilities related to note count $n$ only depend on those related to note count $n-1$ and that the probabilities related to max combo $k$ and final combo $r$ only depend on those related to either less max combo than $k$ or less final combo than $r$ (except for the case $n>k>r=0$ , which can be specially treated before the current iteration of $k$ actually starts), so for the bottom-up DP we can reduce the spatial complexity from $O\!\left(n^3\right)$ to $O\!\left(n^2\right)$ by reducing the 3-dimensional DP to a 2-dimensional one. What needs to be taken care of is that the DP table needs to be updated from larger $k$ and $r$ to smaller $k$ and $r$ instead of the other way so that the numbers in the last iteration in $n$ are left untouched while we need to use them in the current iteration.

After the final iteration in $n$ finishes, we need to sum over the index $r$ to get the final answer: $P_{n,k}=\sum_{r=0}^kP_{n,k,r}.$

Writing the code for the DP algorithm is then straightforward. Here is an implementation in Ruby. In the code, dp[k][r] means $P_{n,k,r}$ in the $n$ th iteration.

          
            ## Returns an array of size m+1,

            ## with the k-th element being the probability P_{m,k}.

            def combo m

            	(1..m).each_with_object [[1]] do |n, dp|

            		dp[n] = [0]*n + [Y * dp[n-1][n-1]] # n = k > 0

            		(n-1).downto 1 do |k| # n > k > 0

            			dpk0 = (1-Y) * dp[k].sum

            			dp[k][k] = Y * (dp[k-1][k-1] + dp[k][k-1])        # n > k = r > 0

            			(k-1).downto(1) { |r| dp[k][r] = Y * dp[k][r-1] } # n > k > r > 0

            			dp[k][0] = dpk0                                   # n > k > r = 0

            		end

            		dp[0][0] *= 1-Y # n > k = r = 0

            	end.map &:sum

            end

Because of the three nested loops, the time complexity of the DP algorithm is $O\!\left(n^3\right)$ .

The second DP approach

Here is an alternative way to use DP to solve the problem. Instead of building a DP table with the $k,r$ indices, we can build a DP table with the $n,k$ indices.

First, we need to rewrite the recurrence relation of $P_{n,k}$ instead of that of $P_{n,k,r}$ . We then need to try to express $P_{n,k,r}$ in terms of $P_{n,k}$ terms. The easiest part is the case where $n\ge k=r=0$ . By recursively applying Equation 2 to $P_{n,0,0}$ , we have $\begin{align*} P_{n,0,0}&=\left(1-Y\right)P_{n-1,0,0}\\ &=\left(1-Y\right)^2P_{n-2,0,0}\\ &=\cdots\\ &=\left(1-Y\right)^nP_{0,0,0}. \end{align*}$ Because $P_{0,0,0}=1$ , we have $P_{n,0,0}=\left(1-Y\right)^n.$

For $n>k>r>0$ , we can recursively apply Equation 2 to get $\begin{align*} P_{n,k,r}&=YP_{n-1,k,r-1}\\ &=Y^2P_{n-2,k,r-2}\\ &=\cdots \end{align*}$ This will finally either decend the note count to $k$ or decend the final combo to $0$ , determined by which comes first.

If $n-r\le k$ , we will decend to the term $P_{k,k,r-(n-k)}$ , which must be zero according to the case $n=k>r=0$ and the case $n=k>r>0$ in Equation 2, so $P_{n,k,r}=0$ .
If $n-r>k$ , then we will decend to the term $P_{n-r,k,0}$ , which is equal to $\left(1-Y\right)P_{n-r-1,k}$ according to the case $n>k>r=0$ in Equation 2.

Therefore, for $n>k>r>0$ , we have $P_{n,k,r}=\begin{cases} 0,&n-r\le k,\\ Y^r\left(1-Y\right)P_{n-r-1,k},&n-r>k. \end{cases}$

For the case $n>k=r>0$ , we can also recursively apply Equation 2 to get $\begin{align*} P_{n,k,k}&=Y\left(P_{n-1,k-1,k-1}+P_{n-1,k,k-1}\right)\\ &=Y\left(Y\left(P_{n-2,k-2,k-2}+P_{n-2,k-1,k-2}\right)+P_{n-1,k,k-1}\right)\\ &=\cdots\\ &=Y\left(Y\left(\cdots Y\left(P_{n-k,0,0}+P_{n-k,1,0}\right)+\cdots\right)+P_{n-1,k,k-1}\right)\\ &=Y^kP_{n-k,0,0}+\sum_{j=1}^kY^jP_{n-j,k-j+1,k-j}. \end{align*}$ We can then substitute Equation 3 and 4 into the above equation. The substitution of Equation 3 can be done without a problem, but the substitution of Equation 4 requires some care because of the different cases.

If $n-k>k$ , then only the case $n-r>k$ in Equation 4 will be involved in the summation.
If $n-k\le k$ , then both cases in Equation 4 will be involved in the summation. To be specific, for $j\in1\,..\,2k-n+1$ , we need the case $n-r\le k$ in Equation 4 (where the summed terms are just zero and can be omitted); for other terms in the summation, we need the other case.

Considering both cases, we may realize that we can just modify the range of the summation to $j\in\max(1,2k-n+1)\,..k$ and adopt the case $n-r>k$ in Equation 4 for all terms in the summation. Therefore, we have $\begin{align*} P_{n,k,k}&=Y^k\left(1-Y\right)^{n-k}+\sum_{j=\max(1,2k-n+1)}^{k}Y^jY^{k-j}\left(1-Y\right)P_{n-j-(k-j)-1,k-j+1}\\ &=Y^k\left(1-Y\right)^{n-k}+Y^k\left(1-Y\right)\sum_{k'=1}^{\min(k,n-k-1)}P_{n-k-1,k'}, \end{align*}$ where in the last line we changed the summation index to $k'\coloneqq k-j+1$ to simplify it. Because $P_{n-k-1,0}=P_{n-k-1,0,0}=\left(1-Y\right)^{n-k-1}$ according to Equation 3, we can combine the two terms into one summation to get the final result for $n>k=r>0$ : $P_{n,k,k}=Y^k\left(1-Y\right)\sum_{k'=0}^{\min(k,n-k-1)}P_{n-k-1,k'}.$ Noticing the obvious fact that $\sum_{k=0}^nP_{n,k}=1$ , the above equation can be simplified, when $k\ge n-k-1$ , to $P_{n,k,k}=Y^k\left(1-Y\right).$ This simplification is not specially useful, but it can be used to simplify the calculation in the program.

Then, for $n>k>0$ , express $P_{n,k}$ in terms of $P_{n,k,r}$ by summing over $r$ , and substitute previous results: $\begin{align*} P_{n,k}&=\sum_{r=0}^kP_{n,k,r}\\ &=P_{n,k,0}+P_{n,k,k}+\sum_{r=1}^{k-1}P_{n,k,r}\\ &=\left(1-Y\right)P_{n-1,k}+Y^k\left(1-Y\right)\sum_{k'=0}^{\min(k,n-k-1)}P_{n-k-1,k'}\\ &\phantom{=~}{}+\sum_{r=1}^{\min(k-1,n-k-1)}Y^r\left(1-Y\right)P_{n-r-1,k}\\ &=\left(1-Y\right)\left( Y^k\sum_{k'=0}^{\min(k,n-k-1)}P_{n-k-1,k'} +\sum_{r=0}^{\min(k-1,n-k-1)}Y^rP_{n-r-1,k} \right) \end{align*}$ where in the last term $r$ is summed to $\min(k-1,n-k-1)$ instead of $k-1$ because of the different cases in Equation 4.

Finally, consider the edge cases where $n=k\ge0$ and $n\ge k=0$ (trivial), we have the complete resursive relation for $P_{n,k}$ : $P_{n,k}=\begin{cases} Y^n,&n=k\ge0,\\ \left(1-Y\right)^n,&n\ge k=0,\\ \displaystyle{\begin{split} \left(1-Y\right)&\,\left( Y^k\sum_{k'=0}^{\min(k,n-k-1)}P_{n-k-1,k'} \right.\\&\left. +\sum_{r=0}^{\min(k-1,n-k-1)}Y^rP_{n-r-1,k} \right), \end{split}}&n>k>0. \end{cases}$

Then, we can write the program to calculate $P_{n,k}$ :

          
            ## Returns an array of size m+1,

            ## with the k-th element being the probability P_{m,k}.

            def combo m

            	(1..m).each_with_object [[1]] do |n, dp|

            		dp[n] = (1..n-1).each_with_object [(1-Y)**n] do |k, dpn|

            			dpn[k] = (1-Y) * (Y**k * (0..[k, n-k-1].min).sum { dp[n-k-1][_1] } + (0..[k-1, n-k-1].min).sum { Y**_1 * dp[n-_1-1][k] })

            		end

            		dp[n][n] = Y**n

            	end.last

            end

This algorithm has the same (asymptotic) space and time complexity as the previous one.

Polynomial coefficients

We have wrote programmes to calculate probabilities $P_{n,k}(Y)$ based on given $Y$ , which we assumed to be a float number. However, float numbers have limited precision, and the calculation may be inaccurate. Actually, the calculation can be done symbolically.

The probability $P_{n,k}$ is a polynomial of degree (at most) $n$ in $Y$ , and the coefficients of the polynomial are integers. This can be easily proven by using mathematical induction and utilizing Equation 7. Therefore, we can calculate the coefficients of the polynomial $P_{n,k}(Y)$ instead of calculate the value directly so that we get a symbolic but accurate result.

Both the two DP algorithms above can be modified to calculate the coefficients of the polynomial. Actually, we can define Y to be a polynomial object that can do arithmetic operations with other polynomials or numbers, and then the programmes can run without any modification. Here, I will modify the second DP algorithm to calculate the coefficients of the polynomial.

We can also utilize Equation 6 to simplify the calculation. Considering the edge cases involved in $\min(k,n-k-1)$ and $\min(k-1,n-k-1)$ , there are three cases we need to consider:

Case $k>n-k-1$ : Equation 6 can be applied, and $r$ is summed to $n-k-1$ .
Case $k=n-k-1$ (can only happen when $n$ is odd): Equation 6 can be applied, and $r$ is summed to $k-1$ .
Case $k$ : Equation 6 cannot be applied, and $r$ is summed to $k-1$ .

Then, use arrays to store the coefficients of the polynomial $P_{n,k}(Y)$ , and we can write the program to calculate the coefficients:

          
            ## Returns a nested array of size m+1 times m+1,

            ## with the j-th element of the k-th element being the coefficient of Y^j in P_{m,k}(Y).

            def combo_pc m

            	(1..m).each_with_object [[[1]]] do |n, dp|

            		dp[n] = Array.new(n+1) { Array.new n+1, 0 }

            

            		# dp[n][0] = (1-Y)**n

            		0.upto(n-1) { dp[n][0][_1] = dp[n-1][0][_1] } # will be multiplied by 1-Y later

            

            		1.upto n/2-1 do |k|

            			# dp[n][k] = (1-Y) * (Y**k * (0..k).sum { |j| dp[n-k-1][j] } + (0..k-1).sum { |r| Y**r * dp[n-r-1][k] })

            			0.upto(k) { |j| 0.upto(n-k-1) { dp[n][k][_1+k] += dp[n-k-1][j][_1] } }

            			0.upto(k-1) { |r| 0.upto(n-r-1) { dp[n][k][_1+r] += dp[n-r-1][k][_1] } }

            		end

            

            		if n % 2 == 1

            			k = n/2

            			# dp[n][k] = (1-Y) * (Y**k + (0..k-1).sum { |r| Y**r * dp[n-r-1][k] })

            			dp[n][k][k] = 1

            			0.upto(k-1) { |r| 0.upto(n-r-1) { dp[n][k][_1+r] += dp[n-r-1][k][_1] } }

            		end

            

            		((n+1)/2).upto n-1 do |k|

            			# dp[n][k] = (1-Y) * (Y**k + (0..n-k-1).sum { |r| Y**r * dp[n-r-1][k] })

            			dp[n][k][k] = 1

            			0.upto(n-k-1) { |r| 0.upto(n-r-1) { dp[n][k][_1+r] += dp[n-r-1][k][_1] } }

            		end

            

            		0.upto(n-1) { |k| n.downto(1) { dp[n][k][_1] -= dp[n][k][_1-1] } } # multiply by 1-Y

            

            		# dp[n][n] = Y**n

            		dp[n][n][n] = 1

            	end.last

            end

Here I list first few polynomials $P_{n,k}(Y)$ calculated by the above program: $\begin{array}{r|llllc} & k=0 & 1 & 2 & 3 & \cdots\\ \hline n=0 & 1\\ 1 & 1-Y & Y\\ 2 & 1-2Y+Y^2 & 2Y-2Y^2 & Y^2\\ 3 & 1-3Y+3Y^2-Y^3 & 3Y-5Y^2+2Y^3 & 2Y^2-2Y^3 & Y^3\\ \vdots \end{array}$

When evaluating the polynomials for large $n$ , the result is inaccurate for $Y$ that is not close to $0$ because of the limited precision of floating numbers. If $Y$ is closer to $1$ , we can first find the coefficients of $P_{n,k}(1-X)$ and then substitute $X\coloneqq1-Y$ .

Plots of the probability distributions

Here are some plots of the probability distribution of max combo $k$ when $n=50$ :

The plots are intuitive as they show that one has higher probability to get a higher max combo when they have a higher success rate.

There is a suspicious jump in $P_{n,k}(Y)$ near $k=n/2$ when $Y$ is close to $1$ . We can look at it closer:

In the zoomed-in plot, we can also see a jump in first derivative (w.r.t. $k$ ) of $P_{n,k}(Y)$ near $k=n/3$ . Actually, the jumps can be modeled in later sections when we talk about the case when $n\to\infty$ .

The case when $n\to\infty$

A natural approach is to try substituting Equation 7 into Equation 1 to get a function w.r.t. the unknown function $f(y,\kappa)$ . First, we can easily write the case when $y=0$ because it means zero success rate, and the only possible max combo is zero: $f(y=0,\kappa)=\delta(\kappa).$ Similarly, we can easily write the case when $y=1$ : $f(y=1,\kappa)=\delta(\kappa-1).$

From now on, we only consider the case when $0$ . First, for the case $\kappa=0$ , according to Equation 7, $\begin{align*} f(y,\kappa=0)&=\lim_{n\to\infty}\left(n+1\right)\left(1-y^{\frac1n}\right)^n\\ &=\begin{cases}0,&0$ The $\infty$ means that there is a Dirac $\delta$ function (shown in Equation 8).

Then, for the case $\kappa=1$ , according to Equation 7, $f(y,\kappa=1)=\lim_{n\to\infty}\left(n+1\right)y=\infty.$ The $\infty$ means that there is a Dirac $\delta$ function. Actually, it is easy to see that there must be a $y\delta(\kappa-1)$ term in the expression of $f(y,\kappa)$ because the probability of getting a max combo ( $\kappa=1$ ) is $y$ .

Define $h(y,\kappa)\coloneqq f(y,\kappa)-y\delta(\kappa-1),$ and then we can get rid of the infinity here.

From now on, we only consider the case when $0$ and $0<\kappa<1$ . According to Equation 7, $\begin{align*} &\phantom{=~}f\!\left(y\in\left(0,1\right),\kappa\in\left(0,1\right)\right)\\ &=\lim_{n\to\infty}\left(n+1\right)\left(1-y^{\frac1n}\right)\left( y^\kappa\sum_{k'=0}^{\min(\kappa n,n-\kappa n-1)}P_{n-\kappa n-1,k'}\!\left(y^{\frac1n}\right) \right.\\&\phantom{=\lim_{n\to\infty}\left(n+1\right)\left(1-y^{\frac1n}\right)}\left. +\sum_{r=0}^{\min(\kappa n-1,n-\kappa n-1)}y^{\frac rn}P_{n-r-1,\kappa n}\!\left(y^{\frac1n}\right) \right)\\ &=\lim_{n\to\infty}n\left(1-y^{\frac1n}\right)\cdot\lim_{n\to\infty}\left( y^\kappa\sum_{t=0,\Delta t=\frac1n}^{\min(\kappa,1-\kappa)}P_{(1-\kappa)n,tn}\!\left(y^{\frac1n}\right) +\sum_{t=0,\Delta t=\frac1n}^{\min(\kappa,1-\kappa)}y^tP_{(1-t)n,\kappa n}\!\left(y^{\frac1n}\right) \right)\\ &=-\ln y\lim_{n\to\infty}\left( y^\kappa\sum_{t=0,\Delta t=\frac1n}^{\min(\kappa,1-\kappa)} \frac1{\left(1-\kappa\right)n}f\!\left(y^{1-\kappa},\frac t{1-\kappa}\right) \right.\\&\phantom{=-\ln y\lim_{n\to\infty}}\left. +\sum_{t=0,\Delta t=\frac1n}^{\min(\kappa,1-\kappa)}y^t \frac1{\left(1-t\right)n}f\!\left(y^{1-t},\frac{\kappa}{1-t}\right) \right)\\ &=-\ln y\lim_{n\to\infty}\sum_{t=0,\Delta t=\frac 1n}^{\min(\kappa,1-\kappa)}\left( \frac{y^\kappa}{1-\kappa}f\!\left(y^{1-\kappa},\frac t{1-\kappa}\right) +\frac{y^t}{1-t}f\!\left(y^{1-t},\frac\kappa{1-t}\right) \right)\Delta t\\ &=-\ln y\int_{t=0}^{\min(\kappa,1-\kappa)}\left( \frac{y^\kappa}{1-\kappa}f\!\left(y^{1-\kappa},\frac t{1-\kappa}\right) +\frac{y^t}{1-t}f\!\left(y^{1-t},\frac\kappa{1-t}\right) \right)\mathrm dt. \end{align*}$ Add back the delta function at $\kappa=1$ , and we have the integral equation $\begin{split} f\!\left(y\in\left(0,1\right),\kappa\right)=-\ln y\int_{t=0}^{\min(\kappa,1-\kappa)}&\,\left( \frac{y^\kappa}{1-\kappa}f\!\left(y^{1-\kappa},\frac t{1-\kappa}\right) \right.\\&\left. +\frac{y^t}{1-t}f\!\left(y^{1-t},\frac\kappa{1-t}\right) \right)\mathrm dt+y\delta(\kappa-1). \end{split}$

There are two terms in the integral. Substitute $u\coloneqq\frac t{1-\kappa}$ in the first term, and we have $\begin{align*} \int_{t=0}^{\min\left(\kappa,1-\kappa\right)}\frac{y^\kappa}{1-\kappa}f\!\left(y^{1-\kappa},\frac t{1-\kappa}\right)\mathrm dt &=\int_0^{\min(\frac\kappa{1-\kappa},1)}y^\kappa f\!\left(y^{1-\kappa},u\right)\mathrm du\\ &=\int_0^{\min(\frac\kappa{1-\kappa},1)}\left(y^\kappa h\!\left(y^{1-\kappa},u\right)+y\delta(u-1)\right)\mathrm du. \end{align*}$ Substitute $v\coloneqq\frac\kappa{1-t}$ in the second term, and we have $\begin{align*} \int_{t=0}^{\min(\kappa,1-\kappa)}\frac{y^t}{1-t}f\!\left(y^{1-t},\frac\kappa{1-t}\right)\mathrm dt &=\int_\kappa^{\min\left(\frac\kappa{1-\kappa},1\right)}y^{1-\frac\kappa v}f\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v\\ &=\int_\kappa^{\min\left(\frac\kappa{1-\kappa},1\right)}\left(y^{1-\frac\kappa v}h\!\left(y^{\frac\kappa v},v\right)+y\delta(v-1)\right)\frac{\mathrm dv}v. \end{align*}$

Further, let (we only consider $y\in\left(0,1\right)$ from now on) $g(y,\kappa)\coloneqq\frac{h(y,\kappa)}{y},$ then the integral equation becomes $\begin{split} g(y,\kappa)=-\ln y&\,\left( \int_0^{\min\left(\frac\kappa{1-\kappa},1\right)}\left(g\!\left(y^{1-\kappa},u\right)+\delta(u-1)\right)\mathrm du \right.\\&\left. +\int_\kappa^{\min\left(\frac\kappa{1-\kappa},1\right)}\left(g\!\left(y^{\frac\kappa v},v\right)+\delta(v-1)\right)\frac{\mathrm dv}v \right). \end{split}$

There is another integral equation for $g$ . Because $\int_0^1f\!\left(y,\kappa\right)\mathrm d\kappa=1$ , we have $\int_0^1g\!\left(y,\kappa\right)\mathrm d\kappa=\frac1y-1.$

Equation 12 and 13 are the equations that we are going to utilize to get the expression for $g(y,\kappa)$ .

The case $\kappa\in\left(\frac12,1\right)$

In this case, we have $\min\!\left(\frac\kappa{1-\kappa},1\right)=1,$ so the Dirac delta functions in Equation 12 should be considered. In this case, it simplifies to $g_1(y,\kappa)\coloneqq g\!\left(y,\kappa\in\left(\frac12,1\right)\right)= -\ln y\left(y^{\kappa-1}+\int_\kappa^1g\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v+1\right),$ where Equation 13 is utilized when finding the first term.

We can try to solve Equation 14 by using Adomian decomposition method (ADM). Suppose $g_1$ can be written in a series $g_1=g_1^{(0)}+g_1^{(1)}+\cdots,$ and substitute it into Equation 14, and we have $g_1^{(0)}(y,\kappa)+\cdots =-\ln y\left(y^{\kappa-1}+1+\int_\kappa^1\left( g^{(0)}\!\left(y^{\frac\kappa v},v\right)+\cdots \right)\frac{\mathrm dv}v\right).$ Assume we may interchange integration and summation (which is OK here because we can verify the solution after we find it using ADM). Then, $\begin{align*} &\phantom{=~}g_1^{(0)}(y,\kappa)+g_1^{(1)}(y,\kappa)+\cdots\\ &=-\ln y\left(y^{\kappa-1}+1\right) -\ln y\int_\kappa^1g^{(0)}\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v-\cdots. \end{align*}$ If we let $\begin{split} g_1^{(0)}(y,\kappa)&\coloneqq-\ln y\left(y^{\kappa-1}+1\right),\\ g_1^{(i+1)}(y,\kappa)&\coloneqq-\ln y\int_\kappa^1g^{(i)}\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v,\quad i\in\mathbb N, \end{split}$ then we can equate each term in the two series. If the sum $g_1=\sum_{i=0}^\infty g_1^{(i)}$ converges, then this is a guess of the solution to Equation 14, which we can verify whether it is correct or not.

Using Equation 15, we can find first few terms in the series by directly integrating. The first few terms are $\begin{split} g_1^{(0)}(y,\kappa)&=-\ln y\left(y^{\kappa-1}+1\right),\\ g_1^{(1)}(y,\kappa)&=-\ln y\left(y^{\kappa-1}-1+\ln y^{\kappa-1}\right),\\ g_1^{(2)}(y,\kappa)&=-\ln y\left(y^{\kappa-1}-1-\ln y^{\kappa-1}+\frac12\left(\ln y^{\kappa-1}\right)^2\right),\\ g_1^{(3)}(y,\kappa)&=-\ln y\left(y^{\kappa-1}-1-\ln y^{\kappa-1}-\frac12\left(\ln y^{\kappa-1}\right)^2+\frac16\left(\ln y^{\kappa-1}\right)^3\right),\\ \vdots& \end{split}$ We may then guess that the terms have general formula $g_1^{(i)}(y,\kappa)=-\ln y\left(y^{\kappa-1}+\frac1{i!}\left(\ln y^{\kappa-1}\right)^i -\sum_{j=0}^{i-1}\frac1{j!}\left(\ln y^{\kappa-1}\right)^j\right).$ Sum up the terms, and we have $\begin{align*} g_1(y,\kappa)&=\sum_{i=0}^\infty g_1^{(i)}(y,\kappa)\\ &=\lim_{q\to\infty}\sum_{i=0}^q-\ln y\left( y^{\kappa-1}+\frac1{i!}\left(\ln y^{\kappa-1}\right)^i -\sum_{j=0}^{i-1}\frac1{j!}\left(\ln y^{\kappa-1}\right)^j \right)\\ &=-\ln y\left(\exp\ln y^{\kappa-1}+\lim_{q\to\infty}\left( \left(q+1\right)y^{\kappa-1} -\sum_{j=0}^q\sum_{i=j+1}^q\frac1{j!}\left(\ln y^{\kappa-1}\right)^j \right)\right)\\ &=-\ln y\left(y^{\kappa-1}+\lim_{q\to\infty}\left( \left(q+1\right)y^{\kappa-1} -\sum_{j=0}^q\frac{q-j}{j!}\left(\ln y^{\kappa-1}\right)^j \right)\right)\\ &=-\ln y\left( y^{\kappa-1}+\lim_{q\to\infty}\left( qy^{\kappa-1}-q\sum_{j=0}^q\frac1{j!}\left(\ln y^{\kappa-1}\right)^j \right) \right.\\&\phantom{=-\ln y}\left.\vphantom{\sum_j^q} +y^{\kappa-1} +\ln y^{\kappa-1}\exp\ln y^{\kappa-1} \right)\\ &=-\ln y\left(2+\ln y^{\kappa-1}\right)y^{\kappa-1}. \end{align*}$

Therefore, we have the final guess of solution $g_1(y,\kappa)=-\ln y\left(2+\ln y^{\kappa-1}\right)y^{\kappa-1}.$ We can substitute it into Equation 14 to verify that it is indeed the solution.

The case $\kappa\in\left(\frac13,\frac12\right)$

In this case, we have $\min\!\left(\frac\kappa{1-\kappa},1\right)=\frac\kappa{1-\kappa}\in\left(\frac12,1\right).$ We can then use the same method as in the previous case to find the solution.

First, by Equation 14, $\begin{align*} g_2(y,\kappa)&\coloneqq g\!\left(y,\kappa\in\left(\frac13,\frac12\right)\right)\\ &=-\ln y\left( \int_0^{\frac\kappa{1-\kappa}}g\!\left(y^{1-\kappa},u\right)\mathrm du +\int_\kappa^{\frac\kappa{1-\kappa}}g\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v \right)\\ &=-\ln y\left( \int_0^1g\!\left(y^{1-\kappa},u\right)\mathrm du -\int_{\frac\kappa{1-\kappa}}^1g_1\!\left(y^{1-\kappa},u\right)\mathrm du \right.\\&\phantom{=-\ln y}\left. +\int_\kappa^{\frac12}g_2\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v +\int_{\frac12}^{\frac\kappa{1-\kappa}}g_1\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v \right). \end{align*}$ Substitute Equation 13 and 16 into the above equation, and we have $\begin{split} g_2(y,\kappa)&=-\ln y\left( y^{\kappa-1} +y^{-\kappa}\left(1+\ln y^{-\kappa}\right) -2y^{2\kappa-1}\left(1+\ln y^{2\kappa-1}\right) \vphantom{\int_\kappa^{\frac12}}\right.\\&\phantom{=-\ln y}\left. +\int_\kappa^{\frac12}g_2\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v \right). \end{split}$

Equation 17 can again be solved by ADM though the calculation is much more complicated than the previous case. We may guess $g_2=\sum_{i=0}^\infty g_2^{(i)}$ is the solution if the series converges, where $\begin{split} g_2^{(0)}(y,\kappa)&\coloneqq-\ln y\left( y^{\kappa-1} +y^{-\kappa}\left(1+\ln y^{-\kappa}\right) -2y^{2\kappa-1}\left(1+\ln y^{2\kappa-1}\right) \right),\\ g_1^{(i+1)}(y,\kappa)&\coloneqq-\ln y\int_\kappa^{\frac12}g^{(i)}\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v, \quad i\in\mathbb N. \end{split}$ The first few terms go too long to be written here before one may find the pattern, so they are omitted here. If you want to see them, use a mathematical software to help you, and you should be able to find the pattern after calculating first six (or so) terms. After looking at first few terms, the guessed general term is $\begin{align*} g_2^{(i)}=-\ln y&\,\left( y^{\kappa-1}+2y^{2\kappa-1}\left(i-1-\ln y^{2\kappa-1}\right) -2\sum_{j=0}^{i-1}\frac{i-j-1}{j!}\left(\ln y^{2\kappa-1}\right)^j \right.\\&\left. {}-y^{-\kappa}\sum_{j=0}^{i-1}\frac1{j!}\left(\ln y^{2\kappa-1}\right)^j +\frac1{i!}y^{-\kappa}\left(1+\ln y^{-\kappa}\right)\left(\ln y^{2\kappa-1}\right)^i \right). \end{align*}$ Then we can sum it to get a guess of $g_2$ .

After some tedious calculation, we have $g_2(y,\kappa)=-\ln y\left( \left(2+\ln y^{\kappa-1}\right)y^{\kappa-1} -\left(2+4\ln y^{2\kappa-1}+\left(\ln y^{2\kappa-1}\right)^2\right)y^{2\kappa-1} \right).$ On may verify that this is indeed the solution by substituting it into Equation 17.

The case $\kappa\in\left(\frac14,\frac13\right)$

By using very similar methods but after very tedious calculation, the solution is $\begin{split} g_3(y,\kappa)&\coloneqq g\!\left(y,\kappa\in\left(\frac14,\frac13\right)\right)\\ &=-\ln y\left( \left(2+\ln y^{\kappa-1}\right)y^{\kappa-1} -\left(2+4\ln y^{2\kappa-1}+\left(\ln y^{2\kappa-1}\right)^2\right)y^{2\kappa-1}\vphantom{\frac12} \right.\\&\phantom{=-\ln y}\left. {}+\left(3\ln y^{3\kappa-1}+3\left(\ln y^{3\kappa-1}\right)^2+\frac12\left(\ln y^{3\kappa-1}\right)^3\right)y^{3\kappa-1} \right). \end{split}$

Other cases

After seeing Equation 16, 18, and 19, one may guess the form of solution for other cases.

Guess the form of solution for $\kappa\in\left(\frac1{q+1},\frac1q\right)$ , where $q\in1\ldots\infty$ , is $g_q(y,\kappa)\coloneqq g\!\left(y,\kappa\in\left(\frac1q,\frac1{q+1}\right)\right)=\sum_{s=1}^q\Delta g_s(y,\kappa),$ where $\Delta g_s(y,\kappa)\coloneqq\left(-1\right)^sy^{s\kappa-1}\ln y\sum_{j=0}^s\frac{A_{s,j}}{j!}\left(\ln y^{s\kappa-1}\right)^j,$ where $A_{s,j}$ are coefficients to be determined.

Now, consider the cases $q\in2\ldots\infty$ . Because $\kappa\in\left(\frac1{q+1},\frac1q\right)$ , $\min\!\left(\frac\kappa{1-\kappa},1\right)=\frac\kappa{1-\kappa}\in\left(\frac1q,\frac1{q-1}\right).$ Therefore, $\begin{align*} &\phantom{=~}\int_0^{\min\left(\frac\kappa{1-\kappa},1\right)}\left(g\!\left(y^{1-\kappa},u\right)+\delta(u-1)\right)\mathrm du\\ &=\int_0^1g\!\left(y^{1-\kappa},u\right)\mathrm du -\sum_{p=1}^{q-2}\int_{\frac1{p+1}}^{\frac1p}g_p\!\left(y^{1-\kappa},u\right)\mathrm du -\int_{\frac\kappa{1-\kappa}}^\frac1{q-1}g_{q-1}\!\left(y^{1-\kappa},u\right)\mathrm du\\ &=y^{\kappa-1}-1 -\sum_{p=1}^{q-2}\int_{\frac1{p+1}}^{\frac1p}\sum_{s=1}^p\Delta g_s\!\left(y^{1-\kappa},u\right)\mathrm du -\int_{\frac\kappa{1-\kappa}}^\frac1{q-1}\sum_{s=1}^{q-1}\Delta g_s\!\left(y^{1-\kappa},u\right)\mathrm du\\ &=y^{\kappa-1}-1 -\sum_{s=1}^{q-1}\left( \sum_{p=s}^{q-2}\int_{\frac1{p+1}}^{\frac1p}+\int_{\frac\kappa{1-\kappa}}^\frac1{q-1} \right)\Delta g_s\!\left(y^{1-\kappa},u\right)\mathrm du\\ &=y^{\kappa-1}-1 -\sum_{s=1}^{q-1}\int_{\frac\kappa{1-\kappa}}^{\frac1s}\Delta g_s\!\left(y^{1-\kappa},u\right)\mathrm du, \end{align*}$ and $\begin{align*} &\phantom{=~}\int_\kappa^{\min\left(\frac\kappa{1-\kappa},1\right)}\left(g\!\left(y^{\frac\kappa v},v\right)+\delta(v-1)\right)\frac{\mathrm dv}v\\ &=\int_\kappa^{\frac1q}g_q\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v +\int_{\frac1q}^{\frac\kappa{1-\kappa}}g_{q-1}\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v\\ &=\int_\kappa^{\frac1q}\sum_{s=1}^q\Delta g_s\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v +\int_{\frac1q}^{\frac\kappa{1-\kappa}}\sum_{s=1}^{q-1}\Delta g_s\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v\\ &=\sum_{s=1}^{q-1}\int_\kappa^{\frac\kappa{1-\kappa}}\Delta g_s\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v +\int_\kappa^{\frac1q}\Delta g_q\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v. \end{align*}$ Substitute into Equation 12, and we have $\begin{split} \sum_{s=1}^q\Delta g_s(y,\kappa)=-\ln y&\,\left( y^{\kappa-1}-1 -\sum_{s=1}^{q-1}\int_{\frac\kappa{1-\kappa}}^{\frac1s}\Delta g_s\!\left(y^{1-\kappa},u\right)\mathrm du \right.\\&\left. {}+\sum_{s=1}^{q-1}\int_\kappa^{\frac\kappa{1-\kappa}}\Delta g_s\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v +\int_\kappa^{\frac1q}\Delta g_q\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v \right). \end{split}$ To simplify later expressions, define $B_{s,l}\coloneqq\left(-1\right)^l\sum_{j=l}^s\left(-1\right)^jA_{s,j}.$

Now, calculate the integrals in Equation 20. Before that, first we introduce a handy integral formula: $\int\left(\ln w\right)^j\,\mathrm dw =\left(-1\right)^jj!\,w\sum_{l=0}^j\left(-1\right)^l\frac{\left(\ln w\right)^l}{l!}+C.$ This formula can be proved by mathematical induction and integration by parts.

Then, we have $\begin{align*} &\phantom{=~}\int_{\frac\kappa{1-\kappa}}^{\frac1s}\Delta g_s\!\left(y^{1-\kappa},u\right)\mathrm du\\ &=\int_{\frac\kappa{1-\kappa}}^{\frac1s}\left(-1\right)^s y^{\left(su-1\right)\left(1-\kappa\right)}\ln y^{1-\kappa} \sum_{j=0}^s\frac{A_{s,j}}{j!}\left(\ln y^{\left(su-1\right)\left(1-\kappa\right)}\right)^j\,\mathrm du\\ &=\frac{\left(-1\right)^s}s\sum_{j=0}^s\frac{A_{s,j}}{j!} \int_{\frac\kappa{1-\kappa}}^{\frac1s}\left(\ln y^{\left(su-1\right)\left(1-\kappa\right)}\right)^j\, \mathrm d\left(y^{\left(su-1\right)\left(1-\kappa\right)}\right)\\ &=\frac{\left(-1\right)^s}s\sum_{j=0}^s\frac{A_{s,j}}{j!} \int_{y^{\left(s+1\right)\kappa-1}}^1\left(\ln w\right)^j\,\mathrm dw\\ &=\frac{\left(-1\right)^s}s\sum_{j=0}^s\frac{A_{s,j}}{j!} \left(-1\right)^jj!\left( 1 -y^{\left(s+1\right)\kappa-1} \sum_{l=0}^j\left(-1\right)^l\frac{\left(\ln y^{\left(s+1\right)\kappa-1}\right)^l}{l!} \right)\\ &=\frac{\left(-1\right)^s}s\sum_{j=0}^s\left(-1\right)^jA_{s,j} -\frac{\left(-1\right)^s}sy^{\left(s+1\right)\kappa-1} \sum_{l=0}^s\left(-1\right)^l\frac{\left(\ln y^{\left(s+1\right)\kappa-1}\right)^l}{l!} \sum_{j=l}^s\left(-1\right)^jA_{s,j}\\ &=\left(-1\right)^s\left( \frac{B_{s,0}}s -\frac1sy^{\left(s+1\right)\kappa-1} \sum_{l=0}^s\frac{B_{s,l}}{l!} \left(\ln y^{\left(s+1\right)\kappa-1}\right)^l \right). \end{align*}$ $\begin{align*} &\phantom{=~}\int_\kappa^{\frac\kappa{1-\kappa}}\Delta g_s\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v\\ &=\int_\kappa^{\frac\kappa{1-\kappa}}\left(-1\right)^s y^{\frac\kappa v\left(sv-1\right)}\ln y^{\frac\kappa v} \sum_{j=0}^s\frac{A_{s,j}}{j!}\left(\ln y^{\frac\kappa v\left(sv-1\right)}\right)^j\frac{\mathrm dv}v\\ &=\left(-1\right)^s\sum_{j=0}^s\frac{A_{s,j}}{j!} \int_\kappa^{\frac\kappa{1-\kappa}}\left(\ln y^{\frac\kappa v\left(sv-1\right)}\right)^j\, \mathrm d\left(y^{\frac\kappa v\left(sv-1\right)}\right)\\ &=\left(-1\right)^s\sum_{j=0}^s\frac{A_{s,j}}{j!} \int_{y^{s\kappa-1}}^{y^{\left(s+1\right)\kappa-1}}\left(\ln w\right)^j\,\mathrm dw\\ &=\left(-1\right)^s\sum_{j=0}^s\frac{A_{s,j}}{j!}\left(-1\right)^jj!\left( y^{\left(s+1\right)\kappa-1}\sum_{l=0}^j\left(-1\right)^l\frac{\left(\ln y^{\left(s+1\right)\kappa-1}\right)^l}{l!} \right.\\&\phantom{=\left(-1\right)^s\sum_{j=0}^s\frac{A_{s,j}}{j!}\left(-1\right)^jj!}\left. {}-y^{s\kappa-1}\sum_{l=0}^j\left(-1\right)^l\frac{\left(\ln y^{s\kappa-1}\right)^l}{l!} \right)\\ &=\left(-1\right)^s\left( y^{\left(s+1\right)\kappa-1}\sum_{l=0}^s\frac{B_{s,l}}{l!} \left(\ln y^{\left(s+1\right)\kappa-1}\right)^l -y^{s\kappa-1}\sum_{l=0}^s\frac{B_{s,l}}{l!} \left(\ln y^{s\kappa-1}\right)^l \right). \end{align*}$ $\begin{align*} &\phantom{=~}\int_\kappa^{\frac1q}\Delta g_q\!\left(y^{\frac\kappa v},v\right)\frac{\mathrm dv}v\\ &=\left(-1\right)^q\left( B_{q,0} -y^{q\kappa-1}\sum_{l=0}^q\frac{B_{q,l}}{l!}\left(\ln y^{q\kappa-1}\right)^l \right). \end{align*}$ Substitute these results into Equation 20, and we have $\begin{split} &\phantom{=~}\sum_{s=1}^q \left(-1\right)^sy^{s\kappa-1}\ln y\sum_{j=0}^s\frac{A_{s,j}}{j!}\left(\ln y^{s\kappa-1}\right)^j\\ &=-\ln y\left(\vphantom{\sum_l^s} y^{\kappa-1}-1 \right.\\&\phantom{=-\ln y}\left. {}-\sum_{s=1}^{q-1}\left(-1\right)^s\left( \frac{B_{s,0}}s -\frac1sy^{\left(s+1\right)\kappa-1} \sum_{l=0}^s\frac{B_{s,l}}{l!} \left(\ln y^{\left(s+1\right)\kappa-1}\right)^l \right) \right.\\&\phantom{=-\ln y}\left. {}+\sum_{s=1}^{q-1}\left(-1\right)^s\left( y^{\left(s+1\right)\kappa-1}\sum_{l=0}^s\frac{B_{s,l}}{l!} \left(\ln y^{\left(s+1\right)\kappa-1}\right)^l -y^{s\kappa-1}\sum_{l=0}^s\frac{B_{s,l}}{l!} \left(\ln y^{s\kappa-1}\right)^l \right) \right.\\&\phantom{=-\ln y}\left. {}+\left(-1\right)^q\left( B_{q,0} -y^{q\kappa-1}\sum_{l=0}^q\frac{B_{q,l}}{l!}\left(\ln y^{q\kappa-1}\right)^l \right) \right). \end{split}$ Cancel factor $\ln y$ on both sides, and we have $\begin{align*} &\phantom{=~}\sum_{s=1}^q \left(-1\right)^sy^{s\kappa-1}\sum_{j=0}^s\frac{A_{s,j}}{j!}\left(\ln y^{s\kappa-1}\right)^j\\ &=-y^{\kappa-1}+1\\ &\phantom{=}{}+\sum_{s=1}^{q-1}\left(-1\right)^s\frac{B_{s,0}}s -\sum_{s=2}^q\frac{\left(-1\right)^{s-1}}{s-1}y^{s\kappa-1}\sum_{l=0}^{s-1}B_{s-1,l}\frac{\left(\ln y^{s\kappa-1}\right)^l}{l!}\\ &\phantom{=}{}-\sum_{s=2}^q\left(-1\right)^{s-1}y^{s\kappa-1}\sum_{l=0}^{s-1}B_{s-1,l}\frac{\left(\ln y^{s\kappa-1}\right)^l}{l!} +\sum_{s=1}^{q-1}\left(-1\right)^sy^{s\kappa-1}\sum_{l=0}^sB_{s,l}\frac{\left(\ln y^{s\kappa-1}\right)^l}{l!}\\ &\phantom{=}{}-\left(-1\right)^qB_{q,0}+\left(-1\right)^qy^{q\kappa-1}\sum_{l=0}^qB_{q,l}\frac{\left(\ln y^{q\kappa-1}\right)^l}{l!}\\ &=1+\sum_{s=1}^{q-1}\frac{\left(-1\right)^s}sB_{s,0}-\left(-1\right)^qB_{q,0}&(*)\\ &\phantom{=}-y^{\kappa-1}\left(1+B_{1,0}-B_{1,1}\ln y^{\kappa-1}\right)&(**)\\ &\phantom{=}{}+\sum_{s=2}^q\left(-1\right)^sy^{s\kappa-1}\left( \sum_{l=0}^{s-1}\left(\frac s{s-1}B_{s-1,l}+B_{s,l}\right)\frac{\left(\ln y^{s\kappa-1}\right)^l}{l!} +B_{s,s}\frac{\left(\ln y^{s\kappa-1}\right)^s}{s!} \right)&(***) \end{align*}$

Equate the coefficients in Line (*) with the corresponding ones on the LHS, and we have $0=1+\sum_{s=1}^{q-1}\frac{\left(-1\right)^s}sB_{s,0}-\left(-1\right)^qB_{q,0}.$ This equation holds for any $q\in 2\ldots\infty$ , so $\begin{cases} \left(-1\right)^qB_{q,0}=1+\sum_{s=1}^{q-1}\frac{\left(-1\right)^s}sB_{s,0},\\ \left(-1\right)^{q+1}B_{q+1,0}=1+\sum_{s=1}^{q}\frac{\left(-1\right)^s}sB_{s,0}, \end{cases}\quad q\in 2\ldots\infty.$ Subtract the two equations, and we have $B_{q+1,0}=-\frac{q+1}qB_{q,0},\quad q\in 2\ldots\infty.$ Equation 22 can determine $B_{q,0}$ for all $q\in 2\ldots\infty$ once $B_{2,0}$ is determined. The relationship between $B_{1,0}$ and $B_{2,0}$ cannot be described by Equation 22, but is given by $B_{2,0}=1-B_{1,0}.$

Equate the coefficients in Line (**) with the corresponding ones on the LHS, and we have $A_{1,0}=1+B_{1,0},\quad A_{1,1}=B_{1,1}.$ By Equation 21, this is equivalent to $A_{1,0}=1+A_{1,0}-A_{1,1},\quad A_{1,1}=A_{1,1}.$ Therefore, $A_{1,1}=1$ , and thus $B_{1,1}=1.$

Equate the coefficients in Line (***) with the corresponding ones on the LHS, and we have $A_{s,l}=\frac s{s-1}B_{s-1,l}+B_{s,l},\quad A_{s,s}=B_{s,s}.$ By Equation 21, $B_{s,l}=A_{s,l}-B_{s,l+1}$ for $l\in 0\ldots s$ , and $A_{s,s}=B_{s,s}$ is always true. Therefore, $0=\frac s{s-1}B_{s-1,l}-B_{s,l+1}.$ This equation is true for any $s\in 2\,..q$ and $l\in0\ldots s$ . Because $q$ is arbitrary, we can change the variable $s$ to $q$ and the equation tells us exactly the same information. Therefore, $B_{q,l}=\frac q{q-1}B_{q-1,l-1},\quad q\in 2\ldots\infty,\quad l\in 1\,..q.$

Equation 22, 23, 24, and 25 are sufficient to determine $B_{q,l}$ for all $q\in 1\ldots\infty$ and $l\in 0\,..q$ up to one arbitrary parameter. Define the arbitrary parameter $b\coloneqq1-B_{1,0},$ then the first few $B_{q,l}$ are $\begin{array}{r|ccccc} &l=0&1&2&3&4\\ \hline q=1&1-b&1\\ 2&2b&2-2b&2\\ 3&-3b&3b&3-3b&3\\ 4&4b&-4b&4b&4-4b&4\\ \vdots&\ddots \end{array}$ The general formula for $B_{q,l}$ is $B_{q,l}=\begin{cases} q,&l=q,\\ q\left(1-b\right),&l=q-1,\\ \left(-1\right)^{q+l}qb,&l\in 0\,..q-2, \end{cases}$ which may be proved by mathematical induction.

Actually, one may find $b=0$ by simply comparing with the results in Equation 16, 18, or 19. Another way to find $b$ is comparing with Eqution 13. Here I wil show the latter approach. $\begin{align*} &\phantom{=~}\int_0^1g\!\left(y,\kappa\right)\mathrm d\kappa\\ &=\sum_{q=1}^\infty\int_{\frac1{q+1}}^{\frac1q}g_q\!\left(y,\kappa\right)\mathrm d\kappa\\ &=\sum_{q=1}^\infty\int_{\frac1{q+1}}^{\frac1q}\sum_{s=1}^q\Delta g_s\!\left(y,\kappa\right)\mathrm d\kappa\\ &=\sum_{s=1}^\infty\sum_{q=s}^\infty\int_{\frac1{q+1}}^{\frac1q}\Delta g_s\!\left(y,\kappa\right)\mathrm d\kappa\\ &=\sum_{s=1}^\infty\int_0^{\frac1s}\Delta g_s\!\left(y,\kappa\right)\mathrm d\kappa\\ &=\sum_{s=1}^\infty\int_0^{\frac1s}\left(-1\right)^sy^{s\kappa-1}\ln y \sum_{j=0}^s\frac{A_{s,j}}{j!}\left(\ln y^{s\kappa-1}\right)^j\,\mathrm d\kappa\\ &=\sum_{s=1}^\infty\frac{\left(-1\right)^s}s\sum_{j=0}^s\frac{A_{s,j}}{j!} \int_0^{\frac1s}\left(\ln y^{s\kappa-1}\right)^j\,\mathrm d\left(y^{s\kappa-1}\right)\\ &=\sum_{s=1}^\infty\frac{\left(-1\right)^s}s\left( B_{s,0}-y^{-1}\sum_{l=0}^sB_{s,l}\frac{\left(\ln y\right)^l}{l!} \right)\\ &=-\left(1-b\right)+y^{-1}\left(\left(1-b\right)-\ln y\right)\\ &\phantom{=}{}+\sum_{s=2}^\infty\frac{\left(-1\right)^s}s\left( \left(-1\right)^ssb-y^{-1}\left( \sum_{l=0}^{s-2}\left(-1\right)^{s+l}sb\frac{\left(\ln y\right)^l}{l!} \right.\right.\\&\phantom{ ={}+\sum_{s=2}^\infty\frac{\left(-1\right)^s}s~\left(\vphantom{\sum_s^\infty}\right.\left(-1\right)^ssb-y^{-1} }\left.\left. {}+s\left(1-b\right)\frac{\left(-\ln y\right)^{s-1}}{\left(s-1\right)!} +s\frac{\left(-\ln y\right)^s}{s!} \right) \right)\\ &=b-1+y^{-1}\left(1-b-\ln y\right) +y^{-1}\sum_{s=2}^\infty\left(\frac{\left(\ln y\right)^{s-1}}{\left(s-1\right)!}-\frac{\left(\ln y\right)^s}{s!}\right)\\ &\phantom{=}{}+b\sum_{s=2}^\infty\left(1-y^{-1}\sum_{l=0}^{s-1}\frac{\left(\ln y\right)^l}{l!}\right)\\ &=b-1+y^{-1}\left(1-b-\ln y\right)+y^{-1}\left(\left(\exp\ln y-1\right)-\left(\exp\ln y-\ln y-1\right)\right)\\ &\phantom{=}{}+b\lim_{q\to\infty}\sum_{s=2}^q\left(1-y^{-1}\sum_{l=0}^{s-1}\left(\ln y\right)^l\right)\\ &=b-1+y^{-1}\left(1-b\right) +b\lim_{q\to\infty}\left(q-1-y^{-1}\left(q-1+\sum_{l=1}^{q-1}\frac{q-l}{l!}\left(\ln y\right)^l\right)\right)\\ &=b-1+y^{-1}\left(1-b\right)+b\left(-1+y^{-1}+\ln y\right)\\ &=y^{-1}-1+b\ln y. \end{align*}$ Compare the result with Equation 13, we have $b=0.$

The table of $B_{q,l}$ is now $\begin{array}{r|ccccc} &k=0&1&2&3&4\\ \hline q=1&1&1\\ 2&0&2&2\\ 3&0&0&3&3\\ 4&0&0&0&4&4\\ \vdots&\ddots \end{array}$ The table of $A_{s,j}$ is then $\begin{array}{r|ccccc} &j=0&1&2&3&4\\ \hline s=1&2&1\\ 2&2&4&2\\ 3&0&3&6&3\\ 4&0&0&4&8&4\\ \vdots&\ddots \end{array}$ The general formula for $A_{s,j}$ is $A_{s,j}=\begin{cases} s,&j\in\left\{s,s-2\right\},\\ 2s,&j=s-1,\\ 0,&j\in0\,..s-3. \end{cases}$

Therefore, the functions $\Delta g_s$ are $\Delta g_s(y,\kappa)=\begin{cases} -y^{\kappa-1}\ln y\left(2+\ln y^{\kappa-1}\right),&s=1\\ \begin{split} &\textstyle{\frac{s\left(-1\right)^s}{\left(s-2\right)!}y^{s\kappa-1}\ln y\left(\ln y^{s\kappa-1}\right)^{s-2} \left(1\vphantom{\left(\ln y^{s\kappa-1}\right)^2}\right.}\\ &\textstyle{\left.\quad{}+\frac2{s-1}\ln y^{s\kappa-1}+\frac1{s\left(s-1\right)}\left(\ln y^{s\kappa-1}\right)^2\right),} \end{split}&s\in2\ldots\infty. \end{cases}$ Therefore, the functions $g_q$ are $\begin{split} &g_q(y,\kappa)=-y^{\kappa-1}\ln y\left(2+\ln y^{\kappa-1}\right) +\ln y\sum_{s=2}^q\frac{s\left(-1\right)^s}{\left(s-2\right)!}y^{s\kappa-1}\left(\ln y^{s\kappa-1}\right)^{s-2} \left(1\vphantom{\frac{\left(-1\right)^s}{s!}}\right.\\ &\phantom{g_q(y,\kappa)=\quad}\left.{}+\frac2{s-1}\ln y^{s\kappa-1}+\frac1{s\left(s-1\right)}\left(\ln y^{s\kappa-1}\right)^2\right) \end{split}$ (the formula is also applicable to $q=1$ ).

Edge cases

Now we have covered almost all cases. The only cases that we have not covered are the cases when $\kappa=\frac1q$ , where $q\in2\ldots\infty$ . The discontinuity in $g$ at $\kappa=\frac1q$ is $\begin{split} &\phantom{=~}g\!\left(y,\frac1q^+\right)-g\!\left(y,\frac1q^-\right)\\ &=-\Delta g_q\!\left(y,\frac1q\right)\\ &=\begin{cases} -2\ln y,&q=2,\\ 0,&q\in3\ldots\infty. \end{cases} \end{split}$ Therefore, for $q\in3\ldots\infty$ , the function $g$ has defined limit at $\kappa=\frac1q$ , and the value of $g$ here should just be the limit value. Now, the only problem is at $\kappa=\frac12$ . We should determine whether the value of $g$ at $\kappa=\frac12$ is its left limit or right limit.

Looking at Equation 12, one may see that the discontinuity at $\kappa=\frac12$ is due to the Dirac $\delta$ function in the integrand. Therefore, whether $g$ at $\kappa=\frac12$ is $g_1$ or $g_2$ depends on whether the Dirac $\delta$ function is within the integrated interval. If it is, then $g$ at $\kappa=\frac12$ is $g_1$ ; otherwise, it is $g_2$ .

The inclusion of the Dirac $\delta$ function in the integrated interval corresponds to the inclusion of the highest term in the summation in Equation 7. Because both $\min(k,n-k-1)$ and $\min(k-1,n-k-1)$ equal $n-k-1$ when $n=2k$ , the highest term in the summation can be reached, so the Dirac $\delta$ function is within the integrated interval. Therefore, $g$ at $\kappa=\frac12$ is $g_1$ .

Therefore, we may conclude that for any $\kappa\in\left(0,1\right)$ , $g(y,\kappa)=g_{\lceil\frac1\kappa\rceil-1}(y,\kappa).$

Another edge case that is interesting to consider is when $\kappa\to0^+$ . However, because the domain of $g$ does not include $\kappa=0$ by definition, so we do not need to consider this case. By some mathematical analysis techniques, one may prove that the limit of $g$ as $\kappa\to0^+$ is $0$ .

The solution

Substitute Equation 27 into Equation 29, and we have $\begin{split} &g(y,\kappa)=-y^{\kappa-1}\ln y\left(2+\ln y^{\kappa-1}\right) +\ln y\sum_{s=2}^{\lceil\frac1\kappa\rceil-1}\frac{s\left(-1\right)^s}{\left(s-2\right)!}y^{s\kappa-1}\left(\ln y^{s\kappa-1}\right)^{s-2} \left(1\vphantom{\frac{\left(-1\right)^s}{s!}}\right.\\ &\phantom{g(y,\kappa)=\quad}\left.{}+\frac2{s-1}\ln y^{s\kappa-1}+\frac1{s\left(s-1\right)}\left(\ln y^{s\kappa-1}\right)^2\right) \end{split}$ Substitute the result into Equation 11 and then Equation 10, and also consider Equation 8 and 9, and we have $f(y,\kappa)=\begin{cases} \delta(\kappa),&y=0,\kappa\in[0,1],\\ 0,&y\in(0,1],\kappa=0,\\ \begin{split} &\textstyle{y\delta(\kappa-1)-y^\kappa\ln y\left(2+\ln y^{\kappa-1}\right)}\\ &\quad\textstyle{ {}+\ln y\sum_{s=2}^{\lceil\frac1\kappa\rceil-1}\frac{s\left(-1\right)^s}{\left(s-2\right)!} y^{s\kappa}\left(\ln y^{s\kappa-1}\right)^{s-2} \left(1\vphantom{\left(\ln y^{s\kappa-1}\right)^2}\right.}\\ &\qquad\textstyle{\left.{}+\frac2{s-1}\ln y^{s\kappa-1}+\frac1{s\left(s-1\right)}\left(\ln y^{s\kappa-1}\right)^2\right)}, \end{split}&y\in(0,1],\kappa\in(0,1]. \end{cases}$

Plots of the probability density functions

Here are plots of the function $f(y,\kappa)$ whose expression is given by Equation 30:

We can compare it with a plot of the distributions when $n$ is finite (say, $100$ ), and we may see that they are very close:

We have not investigated the asymptotic behavior of the error if we approximate the distribution with finite $n$ by the distribution with infinite $n$ , but we may expect that the error is small enough for applicational uses when $n$ is a usual note count in a rhythm game chart (usually at least $500$ ).

Moments

It may be interesting to calculate the moments of the distribution.

We need to evaluate $\mu_\nu\!\left(y\right)\coloneqq\int_0^1\kappa^\nu f\!\left(y,\kappa\right)\mathrm d\kappa.$ First, calculate $\begin{align*} &\phantom{=~}\int_0^{\frac1s}\kappa^\nu\left(\ln y^{s\kappa-1}\right)^jy^{s\kappa-1}\ln y\,\mathrm d\kappa\\ &=\int_{y^{-1}}^1\left(\frac{\log_yw+1}{s}\right)^\nu\left(\ln w\right)^j\,\mathrm dw\\ &=\frac1{s^{\nu+1}}\sum_{p=0}^\nu\binom\nu p\frac1{\left(\ln y\right)^p} \int_{y^{-1}}^1\left(\ln w\right)^{j+p}\,\mathrm dw\\ &=\frac{1}{s^{\nu+1}}\sum_{p=0}^\nu\binom\nu p\frac{\left(-1\right)^{j+p}\left(j+p\right)!}{\left(\ln y\right)^p} \left(1-y^{-1}\sum_{l=0}^{j+p}\frac{\left(\ln y\right)^l}{l!}\right). \end{align*}$ $\begin{align*} &\phantom{=~}\int_0^{\frac1s}\kappa^\nu\Delta g_s\!\left(y,\kappa\right)\mathrm d\kappa\\ &=\sum_{j=0}^s\left(-1\right)^s\frac{A_{s,j}}{j!} \frac{1}{s^{\nu+1}}\sum_{p=0}^\nu\binom\nu p\frac{\left(-1\right)^{j+p}\left(j+p\right)!}{\left(\ln y\right)^p} \left(1-y^{-1}\sum_{l=0}^{j+p}\frac{\left(\ln y\right)^l}{l!}\right)\\ &=\frac{\left(-1\right)^s}{s^{\nu+1}}\sum_{p=0}^\nu\binom\nu p\frac{\left(-1\right)^p}{\left(\ln y\right)^p} \left( \sum_{j=0}^s\frac{\left(j+p\right)!\left(-1\right)^j}{j!}A_{s,j} \right.\\&\qquad\qquad\qquad\qquad\qquad\qquad\left.{} -y^{-1}\sum_{j=0}^s\frac{\left(j+p\right)!\left(-1\right)^j}{j!}A_{s,j}\sum_{l=0}^{j+p}\frac{\left(\ln y\right)^l}{l!} \right). \end{align*}$

Define $B_{s,l,p}\coloneqq\sum_{j=\max(0,l-p)}^s\frac{\left(j+p\right)!\left(-1\right)^j}{j!}A_{s,j},$ $D_{\nu,p,l}\coloneqq\sum_{s=\max(1,l-p)}^\infty\frac{\left(-1\right)^s}{s^{\nu+1}}B_{s,l,p}.$ Then, $\begin{align*} &\phantom{=~}\int_0^{\frac1s}\kappa^\nu\Delta g_s\!\left(y,\kappa\right)\mathrm d\kappa\\ &=\frac{\left(-1\right)^s}{s^{\nu+1}}\sum_{p=0}^\nu\binom\nu p\frac{\left(-1\right)^p}{\left(\ln y\right)^p} \left(B_{s,0,p}-y^{-1}\sum_{l=0}^{s+p}B_{s,l,p}\frac{\left(\ln y\right)^l}{l!}\right). \end{align*}$ $\begin{align*} &\phantom{=~}\int_0^1\kappa^\nu g\!\left(y,\kappa\right)\mathrm d\kappa\\ &=\sum_{s=1}^\infty \frac{\left(-1\right)^s}{s^{\nu+1}}\sum_{p=0}^\nu\binom\nu p\frac{\left(-1\right)^p}{\left(\ln y\right)^p} \left(B_{s,0,p}-y^{-1}\sum_{l=0}^{s+p}B_{s,l,p}\frac{\left(\ln y\right)^l}{l!}\right)\\ &=\sum_{p=0}^\nu\binom\nu p\frac{\left(-1\right)^p}{\left(\ln y\right)^p} \left(D_{\nu,p,0}-y^{-1}\sum_{l=0}^\infty D_{\nu,p,l}\frac{\left(\ln y\right)^l}{l!}\right). \end{align*}$

Now, the only problem is how to get $D_{\nu,p,l}$ . Substitute Equation 26 into Equation 31, and after some calculations, we can get the general formula of $B_{s,l,p}$ : $B_{s,l,p}=\frac{\left(-1\right)^s\max(l,s+p-2)!}{\left(s-1\right)!}\cdot\begin{cases} p\left(p-1\right),&l\in0\,..s+p-2,\\ p-s,&l=s+p-1,\\ 1,&l=s+p. \end{cases}$ Substitute it into Equation 32, and notice the edge cases, we can get $\begin{align*} D_{\nu,p,l}&=\begin{cases} p\left(p-1\right)\sum_{s=1}^\infty\frac{\left(s+p-2\right)!}{s^{\nu+1}\left(s-1\right)!}, &l\in0\,..p-1,\\ p!\left(p-1\right)+p\left(p-1\right) \sum_{s=2}^\infty\frac{\left(s+p-2\right)!}{s^{\nu+1}\left(s-1\right)!}, &l=p,\\ \begin{split} &\textstyle{\frac{l!}{\left(l-p\right)^{\nu+1}\left(l-p-1\right)!} -\frac{l!\left(2p-l-1\right)}{\left(l-p+1\right)^{\nu+1}\left(l-p\right)!}}\\ &\qquad\textstyle{ {}+p\left(p-1\right)\sum_{s=l-p+2}^\infty\frac{\left(s+p-2\right)!}{s^{\nu+1}\left(s-1\right)!},} \end{split} &l\in p+1\ldots\infty \end{cases}\\ &=\begin{cases} p\left(p-1\right)\left(\left(p-1\right)!+S_{\nu,p}\right),& l\in0\,..p,\\ \begin{split} &\textstyle{\frac{l!}{\left(l-p\right)^{\nu+1}\left(l-p-1\right)!} -\frac{l!\left(2p-l-1\right)}{\left(l-p+1\right)^{\nu+1}\left(l-p\right)!}}\\ &\qquad\textstyle{ {}+p\left(p-1\right)\left(S_{\nu,p}-\sum_{s=2}^{l-p+2}\frac{\left(s+p-2\right)!}{s^{\nu+1}\left(s-1\right)!}\right),} \end{split} &l\in p+1\ldots\infty, \end{cases} \end{align*}$ where the infinite sum $S_{\nu,p}\coloneqq\sum_{s=2}^\infty\frac{\left(s+p-2\right)!}{s^{\nu+1}\left(s-1\right)!}.$ There is no closed form for $S_{\nu,p}$ , but we may express it in terms of Stirling numbers of the first kind and the Riemann $\zeta$ function. For $p\in1\,..\nu$ , we have $\begin{align*} S_{\nu,p}&=-\left(p-1\right)!+\sum_{s=1}^\infty\frac{s\left(s+1\right)\cdots\left(s+p-2\right)}{s^{\nu+1}}\\ &=-\left(p-1\right)!+\sum_{s=1}^\infty\frac1{s^{\nu+1}}\sum_{\lambda=0}^{p-1}\begin{bmatrix}p-1\\\lambda\end{bmatrix}s^\lambda\\ &=-\left(p-1\right)!+\sum_{\lambda=0}^{p-1}\begin{bmatrix}p-1\\\lambda\end{bmatrix}\zeta\!\left(\nu-\lambda+1\right), \end{align*}$ where $\begin{bmatrix}\cdot\\\cdot\end{bmatrix}$ denotes (unsigned) Stirling numbers of the first kind. For $p=0$ , we have $\begin{align*} S_{\nu,0} &=\sum_{s=2}^\infty\frac1{s^{\nu+1}\left(s-1\right)}\\ &=\sum_{s=2}^\infty\frac1{s^{\nu+1}\left(s-1\right)}-\sum_{s=2}^\infty\frac1{s\left(s-1\right)}+1\\ &=1-\sum_{s=2}^\infty\frac1{s^{\nu+1}}\frac{s^\nu-1}{s-1}\\ &=\nu+1-\sum_{s=1}^\infty\frac1{s^{\nu+1}}\sum_{\lambda=0}^{\nu-1}s^\lambda\\ &=\nu+1-\sum_{\lambda=0}^{\nu-1}\zeta\!\left(\nu-\lambda+1\right). \end{align*}$

Then, the following steps will be extremely tedious, and I doubt there will be a closed form for our final result, so I will not continue to find the general formula for the moments.

However, we may obtain the first moment (mean) analytically. We have $D_{1,0,l}=\begin{cases}-1,&l=0,\\\frac1l-\frac1{l+1},&l\in1\ldots\infty,\end{cases} \quad D_{1,1,l}=\begin{cases}0,&l=0,1,\\\frac1l+\frac1{l-1},&l\in2\ldots\infty.\end{cases}$ $\begin{align*} \mu_1\!\left(y\right) &\coloneqq\int_0^1\kappa f\!\left(y,\kappa\right)\mathrm d\kappa\\ &=y+y\int_0^1\kappa g\!\left(y,\kappa\right)\mathrm d\kappa\\ &=\frac{\operatorname{li}y-\ln(-\ln y)-\gamma}{\ln y}, \end{align*}$ where $\operatorname{li}$ is the logarithmic integral function, and $\gamma$ is the Euler–Mascheroni constant. The function seems undefined when $y=0$ or $y=1$ , but it has limits at these points: $\mu_1\!\left(y\to0^+\right)=0,\quad\mu_1\!\left(y\to1^-\right)=1,$ which is intuitive. (This function tends to $0$ very slowly when $y\to0^+$ , so slowly that I almost did not believe that when I did the numerical calculation first.)

The plot:

We should also be able to find other statistical quantities like the median, the mode, the variance, etc., but they seem do not have closed forms.

Some interesting observations

The probability distribution of $\kappa$ seems to tend to be a uniform distribution plus a Dirac $\delta$ distribution when $y$ is very close to $1$ . This phenomenon is very visible if we look at the plot of $f(y=0.9,\kappa)$ .

In other words, the distribution seems like $f(y\approx1,\kappa)\approx \left(1-y\right)U\!\left(\frac12,1\right)+y\delta(\kappa-1),$ where $U(a,b)$ denotes the uniform distribution on the interval $[a,b]$ .

This can be justified by expanding $f(y,\kappa)$ in Taylor series of $1-y$ and retaining the first-order terms only. Note that $y^a\left(\ln y\right)^b= \left(y-1\right)^b\left(1+\left(\frac b2-a\right)\left(1-y\right)+\cdots\right),$ so the only case where the Taylor series has a non-zero first-order term is when $b=1$ or $b=0$ . In Equation 30, we can see that the power on $\ln y$ is at least one for each term (because of the general $\ln y$ factor in front), so only the terms with no $\ln y$ factors but the general one will have a first-order term. In this case, the first order term is proportional to $y-1$ , and the proportional coefficient is just the coefficient in the front of the term in $f$ , which is independent of $\kappa$ because $\kappa$ only appears in the power index of $y$ .

Therefore, we may see that only $q=1$ and $q=2$ terms have a non-zero first-order term, and they are respectivey $-2\left(y-1\right)$ and $2\left(y-1\right)$ . This means that when $y$ is very close to $1$ , $f(y\approx 1,\kappa)\approx\begin{cases} 2\left(1-y\right),&\kappa\in\left(\frac12,1\right),\\ 0,&\kappa\in\left(0,\frac12\right). \end{cases}$ This is exactly the uniform distribution.

There is an intuitive way to explain the appearance of the uniform distribution. When $y$ is very close to $1$ , the probability of getting one combo break ( $1-Y$ ) is already very small, so it is very unlikely that there are two or more combo breaks. Assuming there is only one combo break and it may appear anywhere with equal probability. The combo break will cut the string of notes into two pieces, and the length of the larger piece is the max combo, which is uniformly distributed between half note count and full note count.

Every rhythm game player knows: never celebrate too early. You never know whether you will miss near the end. It is then interesting to know what is the probability of getting almost a full combo, i.e. what is the probability of getting $\kappa$ very close to $1$ .

If we find the limit of $f(y,\kappa)$ as $\kappa\to1^-$ , it is $f\!\left(y,\kappa\to1^-\right)=-2y\ln y.$ There is a peak of this probability density at $y=\mathrm e^{-1}$ . Therefore, when $y=\mathrm e^{-1}$ , the probability of getting $\kappa$ very close to $1$ is the largest.

When does $y=\mathrm e^{-1}$ , then? Because $y=Y^n=\left(1-\frac{n_\mathrm b}n\right)^n,$ where $n_\mathrm b$ is the average number of combo breaks, then it tends to $\mathrm e^{-n_\mathrm b}$ when $n\to\infty$ . Therefore, the probability of getting almost a full combo is the highest when your average number of combo breaks is exactly one.

From the plot, it seems that the probability of getting $\kappa$ a little bit higher than $\frac12$ is always higher than the probability of getting $\kappa$ a little bit lower than $\frac12$ . According to Equation 28, the jump in $f(y,\kappa)$ at $\kappa=\frac12$ is $f\!\left(y,\kappa\to\frac12^+\right)-f\!\left(y,\kappa\to\frac12^-\right)=-2y\ln y.$ Interestingly, this coincides with $f\!\left(y,\kappa\to1^-\right)$ .

Define $y_0(\kappa)\coloneqq\mathop{\mathrm{arg\,max}}_{y\in[0,1]}\,f(y,\kappa),$ and then it seems that $y_0:[0,1]\to[0,1]$ is injective but not surjective. It is strictly increasing, and there is a jump at $\kappa=\frac12$ and at $\kappa=1$ .

It has an elementary expression on $\left[\frac12,1\right)$ : $y_0\!\left(\kappa\in\left[\frac12,1\right)\right) =\exp\frac{-2\kappa+1+\sqrt{2\kappa^2-2\kappa+1}}{\kappa\left(\kappa-1\right)}.$

Some applications

In Phigros, one should combo at least $60\%$ of the notes to get a white V () rank. If on average I have one combo break in a chart, which has $1300$ notes, what is the probability of comboing at least $60\%$ of the notes in the chart?

Solution. The success rate is $Y=\frac{1300-1}{1300},\quad y=Y^{1300}\approx\mathrm e^{-1}.$ The probability of comboing more than $60\\%$ of the notes is $\begin{align*} &\phantom{=~}\int_{60\%}^1f\!\left(y,\kappa\right)\mathrm d\kappa\\ &=y+\int_{0.6}^1-y^\kappa\ln y\left(2+\ln y^{\kappa-1}\right)\mathrm d\kappa\\ &\approx\mathrm e^{-1}+\int_{0.6}^1\mathrm e^{-\kappa}\left(3-\kappa\right)\mathrm d\kappa\\ &=\frac75\mathrm e^{-\frac35}\\ &\approx0.768. \end{align*}$

Oh, my god! It is hard to come up with application problems. I hope readers find out the applications themselves.

Ulysses’ trip

Multi-pass Gaussian blur filter

The role of particle indistinguishability in statistical mechanics

The distribution when indistinguishable balls are put into boxes

A measure-theoretic formulation of statistical ensembles (part 2)

Introduction

Introducing non-thermal contact with an example

Non-thermal contact with a bath

Non-thermal ensembles (bath version)

Non-thermal ensembles (non-bath version)

Summary

A measure-theoretic formulation of statistical ensembles (part 1)

Mathematical tools and notations

Relationship between the Gini coefficient and the variance

This is what will happen after you get a haircut

The longest all-11 substring of a random bit string

Introduction

Notation

The case for finite nn

The first DP approach

The second DP approach

Polynomial coefficients

Plots of the probability distributions

The case when n→∞n\to\infty

The case κ∈(12,1)\kappa\in\left(\frac12,1\right)

The case κ∈(13,12)\kappa\in\left(\frac13,\frac12\right)

The case κ∈(14,13)\kappa\in\left(\frac14,\frac13\right)

Other cases

Edge cases

The solution

Plots of the probability density functions

Moments

Some interesting observations

Some applications

The longest all- $1$ substring of a random bit string

The case for finite $n$

The case when $n\to\infty$

The case $\kappa\in\left(\frac12,1\right)$

The case $\kappa\in\left(\frac13,\frac12\right)$

The case $\kappa\in\left(\frac14,\frac13\right)$