Discrete Statistical Distributions¶

Discrete random variables take on only a countable number of values. The commonly used distributions are included in SciPy and described in this document. Each discrete distribution can take one extra integer parameter: $L.$ The relationship between the general distribution $p$ and the standard distribution $p_{0}$ is

$p\left(x\right)=p_{0}\left(x-L\right)$

which allows for shifting of the input. When a distribution generator is initialized, the discrete distribution can either specify the beginning and ending (integer) values $a$ and $b$ which must be such that

$p_{0}\left(x\right)=0\quad x<a\textrm{ or }x>b$

in which case, it is assumed that the pdf function is specified on the integers $a+mk\leq b$ where $k$ is a non-negative integer ( $0,1,2,\ldots$ ) and $m$ is a positive integer multiplier. Alternatively, the two lists $x_{k}$ and $p\left(x_{k}\right)$ can be provided directly in which case a dictionary is set up internally to evaulate probabilities and generate random variates.

Probability Mass Function (PMF)¶

The probability mass function of a random variable X is defined as the probability that the random variable takes on a particular value.

$p\left(x_{k}\right)=P\left[X=x_{k}\right]$

This is also sometimes called the probability density function, although technically

$f\left(x\right)=\sum_{k}p\left(x_{k}\right)\delta\left(x-x_{k}\right)$

is the probability density function for a discrete distribution [1] .

[1]	XXX: Unknown layout Plain Layout: Note that we will be using $p$ to represent the probability mass function and a parameter (a XXX: probability). The usage should be obvious from context.

Cumulative Distribution Function (CDF)¶

The cumulative distribution function is

$F\left(x\right)=P\left[X\leq x\right]=\sum_{x_{k}\leq x}p\left(x_{k}\right)$

and is also useful to be able to compute. Note that

$F\left(x_{k}\right)-F\left(x_{k-1}\right)=p\left(x_{k}\right)$

Survival Function¶

The survival function is just

$S\left(x\right)=1-F\left(x\right)=P\left[X>k\right]$

the probability that the random variable is strictly larger than $k$ .

Percent Point Function (Inverse CDF)¶

The percent point function is the inverse of the cumulative distribution function and is

$G\left(q\right)=F^{-1}\left(q\right)$

for discrete distributions, this must be modified for cases where there is no $x_{k}$ such that $F\left(x_{k}\right)=q.$ In these cases we choose $G\left(q\right)$ to be the smallest value $x_{k}=G\left(q\right)$ for which $F\left(x_{k}\right)\geq q$ . If $q=0$ then we define $G\left(0\right)=a-1$ . This definition allows random variates to be defined in the same way as with continuous rv’s using the inverse cdf on a uniform distribution to generate random variates.

Inverse survival function¶

The inverse survival function is the inverse of the survival function

$Z\left(\alpha\right)=S^{-1}\left(\alpha\right)=G\left(1-\alpha\right)$

and is thus the smallest non-negative integer $k$ for which $F\left(k\right)\geq1-\alpha$ or the smallest non-negative integer $k$ for which $S\left(k\right)\leq\alpha.$

Hazard functions¶

If desired, the hazard function and the cumulative hazard function could be defined as

$h\left(x_{k}\right)=\frac{p\left(x_{k}\right)}{1-F\left(x_{k}\right)}$

and

$H\left(x\right)=\sum_{x_{k}\leq x}h\left(x_{k}\right)=\sum_{x_{k}\leq x}\frac{F\left(x_{k}\right)-F\left(x_{k-1}\right)}{1-F\left(x_{k}\right)}.$

Moments¶

Non-central moments are defined using the PDF

$\mu_{m}^{\prime}=E\left[X^{m}\right]=\sum_{k}x_{k}^{m}p\left(x_{k}\right).$

Central moments are computed similarly $\mu=\mu_{1}^{\prime}$

$\begin{eqnarray*} \mu_{m}=E\left[\left(X-\mu\right)^{m}\right] & = & \sum_{k}\left(x_{k}-\mu\right)^{m}p\left(x_{k}\right)\\ & = & \sum_{k=0}^{m}\left(-1\right)^{m-k}\left(\begin{array}{c} m\\ k\end{array}\right)\mu^{m-k}\mu_{k}^{\prime}\end{eqnarray*}$

The mean is the first moment

$\mu=\mu_{1}^{\prime}=E\left[X\right]=\sum_{k}x_{k}p\left(x_{k}\right)$

the variance is the second central moment

$\mu_{2}=E\left[\left(X-\mu\right)^{2}\right]=\sum_{x_{k}}x_{k}^{2}p\left(x_{k}\right)-\mu^{2}.$

Skewness is defined as

$\gamma_{1}=\frac{\mu_{3}}{\mu_{2}^{3/2}}$

while (Fisher) kurtosis is

$\gamma_{2}=\frac{\mu_{4}}{\mu_{2}^{2}}-3,$

so that a normal distribution has a kurtosis of zero.

Moment generating function¶

The moment generating funtion is defined as

$M_{X}\left(t\right)=E\left[e^{Xt}\right]=\sum_{x_{k}}e^{x_{k}t}p\left(x_{k}\right)$

Moments are found as the derivatives of the moment generating function evaluated at $0.$

Fitting data¶

To fit data to a distribution, maximizing the likelihood function is common. Alternatively, some distributions have well-known minimum variance unbiased estimators. These will be chosen by default, but the likelihood function will always be available for minimizing.

If $f_{i}\left(k;\boldsymbol{\theta}\right)$ is the PDF of a random-variable where $\boldsymbol{\theta}$ is a vector of parameters ( e.g. $L$ and $S$ ), then for a collection of $N$ independent samples from this distribution, the joint distribution the random vector $\mathbf{k}$ is

$f\left(\mathbf{k};\boldsymbol{\theta}\right)=\prod_{i=1}^{N}f_{i}\left(k_{i};\boldsymbol{\theta}\right).$

The maximum likelihood estimate of the parameters $\boldsymbol{\theta}$ are the parameters which maximize this function with $\mathbf{x}$ fixed and given by the data:

$\begin{eqnarray*} \hat{\boldsymbol{\theta}} & = & \arg\max_{\boldsymbol{\theta}}f\left(\mathbf{k};\boldsymbol{\theta}\right)\\ & = & \arg\min_{\boldsymbol{\theta}}l_{\mathbf{k}}\left(\boldsymbol{\theta}\right).\end{eqnarray*}$

Where

$\begin{eqnarray*} l_{\mathbf{k}}\left(\boldsymbol{\theta}\right) & = & -\sum_{i=1}^{N}\log f\left(k_{i};\boldsymbol{\theta}\right)\\ & = & -N\overline{\log f\left(k_{i};\boldsymbol{\theta}\right)}\end{eqnarray*}$

Standard notation for mean¶

We will use

$\overline{y\left(\mathbf{x}\right)}=\frac{1}{N}\sum_{i=1}^{N}y\left(x_{i}\right)$

where $N$ should be clear from context.

Combinations¶

Note that

$k!=k\cdot\left(k-1\right)\cdot\left(k-2\right)\cdot\cdots\cdot1=\Gamma\left(k+1\right)$

and has special cases of

$\begin{eqnarray*} 0! & \equiv & 1\\ k! & \equiv & 0\quad k<0\end{eqnarray*}$

and

$\left(\begin{array}{c} n\\ k\end{array}\right)=\frac{n!}{\left(n-k\right)!k!}.$

If $n<0$ or $k<0$ or $k>n$ we define $\left(\begin{array}{c} n\\ k\end{array}\right)=0$

Bernoulli¶

A Bernoulli random variable of parameter $p$ takes one of only two values $X=0$ or $X=1$ . The probability of success ( $X=1$ ) is $p$ , and the probability of failure ( $X=0$ ) is $1-p.$ It can be thought of as a binomial random variable with $n=1$ . The PMF is $p\left(k\right)=0$ for $k\neq0,1$ and

$\begin{eqnarray*} p\left(k;p\right) & = & \begin{cases} 1-p & k=0\\ p & k=1\end{cases}\\ F\left(x;p\right) & = & \begin{cases} 0 & x<0\\ 1-p & 0\le x<1\\ 1 & 1\leq x\end{cases}\\ G\left(q;p\right) & = & \begin{cases} 0 & 0\leq q<1-p\\ 1 & 1-p\leq q\leq1\end{cases}\\ \mu & = & p\\ \mu_{2} & = & p\left(1-p\right)\\ \gamma_{3} & = & \frac{1-2p}{\sqrt{p\left(1-p\right)}}\\ \gamma_{4} & = & \frac{1-6p\left(1-p\right)}{p\left(1-p\right)}\end{eqnarray*}$

$M\left(t\right)=1-p\left(1-e^{t}\right)$

$\mu_{m}^{\prime}=p$

$h\left[X\right]=p\log p+\left(1-p\right)\log\left(1-p\right)$

Binomial¶

A binomial random variable with parameters $\left(n,p\right)$ can be described as the sum of $n$ independent Bernoulli random variables of parameter $p;$

$Y=\sum_{i=1}^{n}X_{i}.$

Therefore, this random variable counts the number of successes in $n$ independent trials of a random experiment where the probability of success is $p.$

$\begin{eqnarray*} p\left(k;n,p\right) & = & \left(\begin{array}{c} n\\ k\end{array}\right)p^{k}\left(1-p\right)^{n-k}\,\, k\in\left\{ 0,1,\ldots n\right\} ,\\ F\left(x;n,p\right) & = & \sum_{k\leq x}\left(\begin{array}{c} n\\ k\end{array}\right)p^{k}\left(1-p\right)^{n-k}=I_{1-p}\left(n-\left\lfloor x\right\rfloor ,\left\lfloor x\right\rfloor +1\right)\quad x\geq0\end{eqnarray*}$

where the incomplete beta integral is

$I_{x}\left(a,b\right)=\frac{\Gamma\left(a+b\right)}{\Gamma\left(a\right)\Gamma\left(b\right)}\int_{0}^{x}t^{a-1}\left(1-t\right)^{b-1}dt.$

Now

$\begin{eqnarray*} \mu & = & np\\ \mu_{2} & = & np\left(1-p\right)\\ \gamma_{1} & = & \frac{1-2p}{\sqrt{np\left(1-p\right)}}\\ \gamma_{2} & = & \frac{1-6p\left(1-p\right)}{np\left(1-p\right)}.\end{eqnarray*}$

$M\left(t\right)=\left[1-p\left(1-e^{t}\right)\right]^{n}$

Boltzmann (truncated Planck)¶

$\begin{eqnarray*} p\left(k;N,\lambda\right) & = & \frac{1-e^{-\lambda}}{1-e^{-\lambda N}}\exp\left(-\lambda k\right)\quad k\in\left\{ 0,1,\ldots,N-1\right\} \\ F\left(x;N,\lambda\right) & = & \left\{ \begin{array}{cc} 0 & x<0\\ \frac{1-\exp\left[-\lambda\left(\left\lfloor x\right\rfloor +1\right)\right]}{1-\exp\left(-\lambda N\right)} & 0\leq x\leq N-1\\ 1 & x\geq N-1\end{array}\right.\\ G\left(q,\lambda\right) & = & \left\lceil -\frac{1}{\lambda}\log\left[1-q\left(1-e^{-\lambda N}\right)\right]-1\right\rceil \end{eqnarray*}$

Define $z=e^{-\lambda}$

$\begin{eqnarray*} \mu & = & \frac{z}{1-z}-\frac{Nz^{N}}{1-z^{N}}\\ \mu_{2} & = & \frac{z}{\left(1-z\right)^{2}}-\frac{N^{2}z^{N}}{\left(1-z^{N}\right)^{2}}\\ \gamma_{1} & = & \frac{z\left(1+z\right)\left(\frac{1-z^{N}}{1-z}\right)^{3}-N^{3}z^{N}\left(1+z^{N}\right)}{\left[z\left(\frac{1-z^{N}}{1-z}\right)^{2}-N^{2}z^{N}\right]^{3/2}}\\ \gamma_{2} & = & \frac{z\left(1+4z+z^{2}\right)\left(\frac{1-z^{N}}{1-z}\right)^{4}-N^{4}z^{N}\left(1+4z^{N}+z^{2N}\right)}{\left[z\left(\frac{1-z^{N}}{1-z}\right)^{2}-N^{2}z^{N}\right]^{2}}\end{eqnarray*}$

$M\left(t\right)=\frac{1-e^{N\left(t-\lambda\right)}}{1-e^{t-\lambda}}\frac{1-e^{-\lambda}}{1-e^{-\lambda N}}$

Planck (discrete exponential)¶

Named Planck because of its relationship to the black-body problem he solved.

$\begin{eqnarray*} p\left(k;\lambda\right) & = & \left(1-e^{-\lambda}\right)e^{-\lambda k}\quad k\lambda\geq0\\ F\left(x;\lambda\right) & = & 1-e^{-\lambda\left(\left\lfloor x\right\rfloor +1\right)}\quad x\lambda\geq0\\ G\left(q;\lambda\right) & = & \left\lceil -\frac{1}{\lambda}\log\left[1-q\right]-1\right\rceil .\end{eqnarray*}$

$\begin{eqnarray*} \mu & = & \frac{1}{e^{\lambda}-1}\\ \mu_{2} & = & \frac{e^{-\lambda}}{\left(1-e^{-\lambda}\right)^{2}}\\ \gamma_{1} & = & 2\cosh\left(\frac{\lambda}{2}\right)\\ \gamma_{2} & = & 4+2\cosh\left(\lambda\right)\end{eqnarray*}$

$M\left(t\right)=\frac{1-e^{-\lambda}}{1-e^{t-\lambda}}$

$h\left[X\right]=\frac{\lambda e^{-\lambda}}{1-e^{-\lambda}}-\log\left(1-e^{-\lambda}\right)$

Poisson¶

The Poisson random variable counts the number of successes in $n$ independent Bernoulli trials in the limit as $n\rightarrow\infty$ and $p\rightarrow0$ where the probability of success in each trial is $p$ and $np=\lambda\geq0$ is a constant. It can be used to approximate the Binomial random variable or in it’s own right to count the number of events that occur in the interval $\left[0,t\right]$ for a process satisfying certain “sparsity “constraints. The functions are

$\begin{eqnarray*} p\left(k;\lambda\right) & = & e^{-\lambda}\frac{\lambda^{k}}{k!}\quad k\geq0,\\ F\left(x;\lambda\right) & = & \sum_{n=0}^{\left\lfloor x\right\rfloor }e^{-\lambda}\frac{\lambda^{n}}{n!}=\frac{1}{\Gamma\left(\left\lfloor x\right\rfloor +1\right)}\int_{\lambda}^{\infty}t^{\left\lfloor x\right\rfloor }e^{-t}dt,\\ \mu & = & \lambda\\ \mu_{2} & = & \lambda\\ \gamma_{1} & = & \frac{1}{\sqrt{\lambda}}\\ \gamma_{2} & = & \frac{1}{\lambda}.\end{eqnarray*}$

$M\left(t\right)=\exp\left[\lambda\left(e^{t}-1\right)\right].$

Geometric¶

The geometric random variable with parameter $p\in\left(0,1\right)$ can be defined as the number of trials required to obtain a success where the probability of success on each trial is $p$ . Thus,

$\begin{eqnarray*} p\left(k;p\right) & = & \left(1-p\right)^{k-1}p\quad k\geq1\\ F\left(x;p\right) & = & 1-\left(1-p\right)^{\left\lfloor x\right\rfloor }\quad x\geq1\\ G\left(q;p\right) & = & \left\lceil \frac{\log\left(1-q\right)}{\log\left(1-p\right)}\right\rceil \\ \mu & = & \frac{1}{p}\\ \mu_{2} & = & \frac{1-p}{p^{2}}\\ \gamma_{1} & = & \frac{2-p}{\sqrt{1-p}}\\ \gamma_{2} & = & \frac{p^{2}-6p+6}{1-p}.\end{eqnarray*}$

$\begin{eqnarray*} M\left(t\right) & = & \frac{p}{e^{-t}-\left(1-p\right)}\end{eqnarray*}$

Negative Binomial¶

The negative binomial random variable with parameters $n$ and $p\in\left(0,1\right)$ can be defined as the number of extra independent trials (beyond $n$ ) required to accumulate a total of $n$ successes where the probability of a success on each trial is $p.$ Equivalently, this random variable is the number of failures encoutered while accumulating $n$ successes during independent trials of an experiment that succeeds with probability $p.$ Thus,

$\begin{eqnarray*} p\left(k;n,p\right) & = & \left(\begin{array}{c} k+n-1\\ n-1\end{array}\right)p^{n}\left(1-p\right)^{k}\quad k\geq0\\ F\left(x;n,p\right) & = & \sum_{i=0}^{\left\lfloor x\right\rfloor }\left(\begin{array}{c} i+n-1\\ i\end{array}\right)p^{n}\left(1-p\right)^{i}\quad x\geq0\\ & = & I_{p}\left(n,\left\lfloor x\right\rfloor +1\right)\quad x\geq0\\ \mu & = & n\frac{1-p}{p}\\ \mu_{2} & = & n\frac{1-p}{p^{2}}\\ \gamma_{1} & = & \frac{2-p}{\sqrt{n\left(1-p\right)}}\\ \gamma_{2} & = & \frac{p^{2}+6\left(1-p\right)}{n\left(1-p\right)}.\end{eqnarray*}$

Recall that $I_{p}\left(a,b\right)$ is the incomplete beta integral.

Hypergeometric¶

The hypergeometric random variable with parameters $\left(M,n,N\right)$ counts the number of “good “objects in a sample of size $N$ chosen without replacement from a population of $M$ objects where $n$ is the number of “good “objects in the total population.

$\begin{eqnarray*} p\left(k;N,n,M\right) & = & \frac{\left(\begin{array}{c} n\\ k\end{array}\right)\left(\begin{array}{c} M-n\\ N-k\end{array}\right)}{\left(\begin{array}{c} M\\ N\end{array}\right)}\quad N-\left(M-n\right)\leq k\leq\min\left(n,N\right)\\ F\left(x;N,n,M\right) & = & \sum_{k=0}^{\left\lfloor x\right\rfloor }\frac{\left(\begin{array}{c} m\\ k\end{array}\right)\left(\begin{array}{c} N-m\\ n-k\end{array}\right)}{\left(\begin{array}{c} N\\ n\end{array}\right)},\\ \mu & = & \frac{nN}{M}\\ \mu_{2} & = & \frac{nN\left(M-n\right)\left(M-N\right)}{M^{2}\left(M-1\right)}\\ \gamma_{1} & = & \frac{\left(M-2n\right)\left(M-2N\right)}{M-2}\sqrt{\frac{M-1}{nN\left(M-m\right)\left(M-n\right)}}\\ \gamma_{2} & = & \frac{g\left(N,n,M\right)}{nN\left(M-n\right)\left(M-3\right)\left(M-2\right)\left(N-M\right)}\end{eqnarray*}$

where (defining $m=M-n$ )

$\begin{eqnarray*} g\left(N,n,M\right) & = & m^{3}-m^{5}+3m^{2}n-6m^{3}n+m^{4}n+3mn^{2}\\ & & -12m^{2}n^{2}+8m^{3}n^{2}+n^{3}-6mn^{3}+8m^{2}n^{3}\\ & & +mn^{4}-n^{5}-6m^{3}N+6m^{4}N+18m^{2}nN\\ & & -6m^{3}nN+18mn^{2}N-24m^{2}n^{2}N-6n^{3}N\\ & & -6mn^{3}N+6n^{4}N+6m^{2}N^{2}-6m^{3}N^{2}-24mnN^{2}\\ & & +12m^{2}nN^{2}+6n^{2}N^{2}+12mn^{2}N^{2}-6n^{3}N^{2}.\end{eqnarray*}$

Zipf (Zeta)¶

A random variable has the zeta distribution (also called the zipf distribution) with parameter $\alpha>1$ if it’s probability mass function is given by

$\begin{eqnarray*} p\left(k;\alpha\right) & = & \frac{1}{\zeta\left(\alpha\right)k^{\alpha}}\quad k\geq1\end{eqnarray*}$

where

$\zeta\left(\alpha\right)=\sum_{n=1}^{\infty}\frac{1}{n^{\alpha}}$

is the Riemann zeta function. Other functions of this distribution are

$\begin{eqnarray*} F\left(x;\alpha\right) & = & \frac{1}{\zeta\left(\alpha\right)}\sum_{k=1}^{\left\lfloor x\right\rfloor }\frac{1}{k^{\alpha}}\\ \mu & = & \frac{\zeta_{1}}{\zeta_{0}}\quad\alpha>2\\ \mu_{2} & = & \frac{\zeta_{2}\zeta_{0}-\zeta_{1}^{2}}{\zeta_{0}^{2}}\quad\alpha>3\\ \gamma_{1} & = & \frac{\zeta_{3}\zeta_{0}^{2}-3\zeta_{0}\zeta_{1}\zeta_{2}+2\zeta_{1}^{3}}{\left[\zeta_{2}\zeta_{0}-\zeta_{1}^{2}\right]^{3/2}}\quad\alpha>4\\ \gamma_{2} & = & \frac{\zeta_{4}\zeta_{0}^{3}-4\zeta_{3}\zeta_{1}\zeta_{0}^{2}+12\zeta_{2}\zeta_{1}^{2}\zeta_{0}-6\zeta_{1}^{4}-3\zeta_{2}^{2}\zeta_{0}^{2}}{\left(\zeta_{2}\zeta_{0}-\zeta_{1}^{2}\right)^{2}}.\end{eqnarray*}$

$\begin{eqnarray*} M\left(t\right) & = & \frac{\textrm{Li}_{\alpha}\left(e^{t}\right)}{\zeta\left(\alpha\right)}\end{eqnarray*}$

where $\zeta_{i}=\zeta\left(\alpha-i\right)$ and $\textrm{Li}_{n}\left(z\right)$ is the $n^{\textrm{th}}$ polylogarithm function of $z$ defined as

$\textrm{Li}_{n}\left(z\right)\equiv\sum_{k=1}^{\infty}\frac{z^{k}}{k^{n}}$

$\mu_{n}^{\prime}=\left.M^{\left(n\right)}\left(t\right)\right|_{t=0}=\left.\frac{\textrm{Li}_{\alpha-n}\left(e^{t}\right)}{\zeta\left(a\right)}\right|_{t=0}=\frac{\zeta\left(\alpha-n\right)}{\zeta\left(\alpha\right)}$

Logarithmic (Log-Series, Series)¶

The logarimthic distribution with parameter $p$ has a probability mass function with terms proportional to the Taylor series expansion of $\log\left(1-p\right)$

$\begin{eqnarray*} p\left(k;p\right) & = & -\frac{p^{k}}{k\log\left(1-p\right)}\quad k\geq1\\ F\left(x;p\right) & = & -\frac{1}{\log\left(1-p\right)}\sum_{k=1}^{\left\lfloor x\right\rfloor }\frac{p^{k}}{k}=1+\frac{p^{1+\left\lfloor x\right\rfloor }\Phi\left(p,1,1+\left\lfloor x\right\rfloor \right)}{\log\left(1-p\right)}\end{eqnarray*}$

where

$\Phi\left(z,s,a\right)=\sum_{k=0}^{\infty}\frac{z^{k}}{\left(a+k\right)^{s}}$

is the Lerch Transcendent. Also define $r=\log\left(1-p\right)$

$\begin{eqnarray*} \mu & = & -\frac{p}{\left(1-p\right)r}\\ \mu_{2} & = & -\frac{p\left[p+r\right]}{\left(1-p\right)^{2}r^{2}}\\ \gamma_{1} & = & -\frac{2p^{2}+3pr+\left(1+p\right)r^{2}}{r\left(p+r\right)\sqrt{-p\left(p+r\right)}}r\\ \gamma_{2} & = & -\frac{6p^{3}+12p^{2}r+p\left(4p+7\right)r^{2}+\left(p^{2}+4p+1\right)r^{3}}{p\left(p+r\right)^{2}}.\end{eqnarray*}$

$\begin{eqnarray*} M\left(t\right) & = & -\frac{1}{\log\left(1-p\right)}\sum_{k=1}^{\infty}\frac{e^{tk}p^{k}}{k}\\ & = & \frac{\log\left(1-pe^{t}\right)}{\log\left(1-p\right)}\end{eqnarray*}$

Thus,

$\mu_{n}^{\prime}=\left.M^{\left(n\right)}\left(t\right)\right|_{t=0}=\left.\frac{\textrm{Li}_{1-n}\left(pe^{t}\right)}{\log\left(1-p\right)}\right|_{t=0}=-\frac{\textrm{Li}_{1-n}\left(p\right)}{\log\left(1-p\right)}.$

Discrete Uniform (randint)¶

The discrete uniform distribution with parameters $\left(a,b\right)$ constructs a random variable that has an equal probability of being any one of the integers in the half-open range $[a,b).$ If $a$ is not given it is assumed to be zero and the only parameter is $b.$ Therefore,

$\begin{eqnarray*} p\left(k;a,b\right) & = & \frac{1}{b-a}\quad a\leq k<b\\ F\left(x;a,b\right) & = & \frac{\left\lfloor x\right\rfloor -a}{b-a}\quad a\leq x\leq b\\ G\left(q;a,b\right) & = & \left\lceil q\left(b-a\right)+a\right\rceil \\ \mu & = & \frac{b+a-1}{2}\\ \mu_{2} & = & \frac{\left(b-a-1\right)\left(b-a+1\right)}{12}\\ \gamma_{1} & = & 0\\ \gamma_{2} & = & -\frac{6}{5}\frac{\left(b-a\right)^{2}+1}{\left(b-a-1\right)\left(b-a+1\right)}.\end{eqnarray*}$

$\begin{eqnarray*} M\left(t\right) & = & \frac{1}{b-a}\sum_{k=a}^{b-1}e^{tk}\\ & = & \frac{e^{bt}-e^{at}}{\left(b-a\right)\left(e^{t}-1\right)}\end{eqnarray*}$

Discrete Laplacian¶

Defined over all integers for $a>0$

$\begin{eqnarray*} p\left(k\right) & = & \tanh\left(\frac{a}{2}\right)e^{-a\left|k\right|},\\ F\left(x\right) & = & \left\{ \begin{array}{cc} \frac{e^{a\left(\left\lfloor x\right\rfloor +1\right)}}{e^{a}+1} & \left\lfloor x\right\rfloor <0,\\ 1-\frac{e^{-a\left\lfloor x\right\rfloor }}{e^{a}+1} & \left\lfloor x\right\rfloor \geq0.\end{array}\right.\\ G\left(q\right) & = & \left\{ \begin{array}{cc} \left\lceil \frac{1}{a}\log\left[q\left(e^{a}+1\right)\right]-1\right\rceil & q<\frac{1}{1+e^{-a}},\\ \left\lceil -\frac{1}{a}\log\left[\left(1-q\right)\left(1+e^{a}\right)\right]\right\rceil & q\geq\frac{1}{1+e^{-a}}.\end{array}\right.\end{eqnarray*}$

$\begin{eqnarray*} M\left(t\right) & = & \tanh\left(\frac{a}{2}\right)\sum_{k=-\infty}^{\infty}e^{tk}e^{-a\left|k\right|}\\ & = & C\left(1+\sum_{k=1}^{\infty}e^{-\left(t+a\right)k}+\sum_{1}^{\infty}e^{\left(t-a\right)k}\right)\\ & = & \tanh\left(\frac{a}{2}\right)\left(1+\frac{e^{-\left(t+a\right)}}{1-e^{-\left(t+a\right)}}+\frac{e^{t-a}}{1-e^{t-a}}\right)\\ & = & \frac{\tanh\left(\frac{a}{2}\right)\sinh a}{\cosh a-\cosh t}.\end{eqnarray*}$

Thus,

$\mu_{n}^{\prime}=M^{\left(n\right)}\left(0\right)=\left[1+\left(-1\right)^{n}\right]\textrm{Li}_{-n}\left(e^{-a}\right)$

where $\textrm{Li}_{-n}\left(z\right)$ is the polylogarithm function of order $-n$ evaluated at $z.$

$h\left[X\right]=-\log\left(\tanh\left(\frac{a}{2}\right)\right)+\frac{a}{\sinh a}$

Discrete Gaussian*¶

Defined for all $\mu$ and $\lambda>0$ and $k$

$p\left(k;\mu,\lambda\right)=\frac{1}{Z\left(\lambda\right)}\exp\left[-\lambda\left(k-\mu\right)^{2}\right]$

where

$Z\left(\lambda\right)=\sum_{k=-\infty}^{\infty}\exp\left[-\lambda k^{2}\right]$

$\begin{eqnarray*} \mu & = & \mu\\ \mu_{2} & = & -\frac{\partial}{\partial\lambda}\log Z\left(\lambda\right)\\ & = & G\left(\lambda\right)e^{-\lambda}\end{eqnarray*}$

where $G\left(0\right)\rightarrow\infty$ and $G\left(\infty\right)\rightarrow2$ with a minimum less than 2 near $\lambda=1$

$G\left(\lambda\right)=\frac{1}{Z\left(\lambda\right)}\sum_{k=-\infty}^{\infty}k^{2}\exp\left[-\lambda\left(k+1\right)\left(k-1\right)\right]$

Discrete Statistical Distributions¶

Probability Mass Function (PMF)¶

Cumulative Distribution Function (CDF)¶

Survival Function¶

Percent Point Function (Inverse CDF)¶

Inverse survival function¶

Hazard functions¶

Moments¶

Moment generating function¶

Fitting data¶

Standard notation for mean¶