Skellam
Probability mass function Examples of the probability mass function for the Skellam distribution. The horizontal axis is the indexk. (The function is only defined at integer values ofk. The connecting lines do not indicate continuity.)
Parameters	${\displaystyle {\mu }_1\ge 0,{\mu }_2\ge 0}$
Support	${\displaystyle k\in \{\dots ,-2,-1,0,1,2,\dots \}}$
PMF	${\displaystyle e^{-({\mu }_1+{\mu }_2)}{\left(\frac{{\mu }_1}{{\mu }_2}\right)}^{k/2}{I}_k(2\sqrt{{\mu }_1{\mu }_2})}$
Mean	${\displaystyle {\mu }_1-{\mu }_2}$
Median	N/A
Variance	${\displaystyle {\mu }_1+{\mu }_2}$
Skewness	${\displaystyle \frac{{\mu }_1-{\mu }_2}{({\mu }_1+{\mu }_2{)}^{3/2}}}$
Excess kurtosis	${\displaystyle \frac{1}{{\mu }_1+{\mu }_2}}$
MGF	${\displaystyle e^{-({\mu }_1+{\mu }_2)+{\mu }_1{e}^t+{\mu }_2{e}^{-t}}}$
CF	${\displaystyle e^{-({\mu }_1+{\mu }_2)+{\mu }_1{e}^{it}+{\mu }_2{e}^{-it}}}$

TheSkellam distribution is thediscrete probability distribution of the difference${\displaystyle N_1-N_2}$ of twostatistically independentrandom variables${\displaystyle N_1}$ and${\displaystyle N_2,}$ eachPoisson-distributed with respectiveexpected values${\displaystyle {\mu }_1}$ and${\displaystyle {\mu }_2}$. It is useful in describing the statistics of the difference of two images with simplephoton noise, as well as describing thepoint spread distribution in sports where all scored points are equal, such asbaseball,hockey andsoccer.

The distribution is also applicable to a special case of the difference of dependent Poisson random variables, but just the obvious case where the two variables have a common additive random contribution which is cancelled by the differencing: see Karlis & Ntzoufras (2003) for details and an application.

Theprobability mass function for the Skellam distribution for a difference${\displaystyle K=N_1-N_2}$ between two independent Poisson-distributed random variables with means${\displaystyle {\mu }_1}$ and${\displaystyle {\mu }_2}$ is given by:

$${\displaystyle p(k;{\mu }_1,{\mu }_2)=Pr\{K=k\}=e^{-({\mu }_1+{\mu }_2)}{\left(\frac{{\mu }_1}{{\mu }_2}\right)}^{k/2}{I}_k(2\sqrt{{\mu }_1{\mu }_2})}$$

whereI_k(z) is themodified Bessel function of the first kind. Sincek is an integer we have thatI_k(z) = I_|k|(z).

Theprobability mass function of aPoisson-distributed random variable with mean μ is given by

$${\displaystyle p(k;\mu )=\frac{{\mu }^k}{k!}{e}^{-\mu }.}$$

for${\displaystyle k\ge 0}$ (and zero otherwise). The Skellam probability mass function for the difference of two independent counts${\displaystyle K=N_1-N_2}$ is theconvolution of two Poisson distributions: (Skellam, 1946)

Since the Poisson distribution is zero for negative values of the count, the second sum is only taken for those terms where${\displaystyle n\ge 0}$ and${\displaystyle n+k\ge 0}$. It can be shown that the above sum implies that

$${\displaystyle \frac{p(k;{\mu }_1,{\mu }_2)}{p(-k;{\mu }_1,{\mu }_2)}={\left(\frac{{\mu }_1}{{\mu }_2}\right)}^k}$$

so that:

$${\displaystyle p(k;{\mu }_1,{\mu }_2)=e^{-({\mu }_1+{\mu }_2)}{\left(\frac{{\mu }_1}{{\mu }_2}\right)}^{k/2}{I}_{|k|}(2\sqrt{{\mu }_1{\mu }_2})}$$

whereI _k(z) is themodified Bessel function of the first kind. The special case for${\displaystyle {\mu }_1={\mu }_2(=\mu )}$ is given by Irwin (1937):

$${\displaystyle p\left(k;\mu ,\mu \right)=e^{-2\mu }{I}_{|k|}(2\mu ).}$$

Using the limiting values of the modified Bessel function for small arguments, we can recover the Poisson distribution as a special case of the Skellam distribution for${\displaystyle {\mu }_2=0}$.

As it is a discrete probability function, the Skellam probability mass function is normalized:

$${\displaystyle \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}p(k;{\mu }_1,{\mu }_2)=1.}$$

We know that theprobability generating function (pgf) for aPoisson distribution is:

$${\displaystyle G\left(t;\mu \right)=e^{\mu (t-1)}.}$$

It follows that the pgf,${\displaystyle G(t;{\mu }_1,{\mu }_2)}$, for a Skellam probability mass function will be:

Notice that the form of theprobability-generating function implies that the distribution of the sums or the differences of any number of independent Skellam-distributed variables are again Skellam-distributed. It is sometimes claimed that any linear combination of two Skellam distributed variables are again Skellam-distributed, but this is clearly not true since any multiplier other than${\displaystyle \pm 1}$ would change thesupport of the distribution and alter the pattern ofmoments in a way that no Skellam distribution can satisfy.

Themoment-generating function is given by:

$${\displaystyle M\left(t;{\mu }_1,{\mu }_2\right)=G(e^t;{\mu }_1,{\mu }_2)=\sum _{k=0}^{\mathrm{\infty }}\frac{t^k}{k!}{m}_k}$$

which yields the raw momentsm_k . Define:

$${\displaystyle \mathrm{\Delta }\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}{\mu }_1-{\mu }_2}$$

$${\displaystyle \mu \stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}\frac{1}{2}({\mu }_1+{\mu }_2).}$$

Then the raw momentsm_k are

Thecentral momentsM_k are

Themean,variance,skewness, andkurtosis excess are respectively:

Thecumulant-generating function is given by:

$${\displaystyle K(t;{\mu }_1,{\mu }_2)\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}\ln (M(t;{\mu }_1,{\mu }_2))=\sum _{k=0}^{\mathrm{\infty }}\frac{t^k}{k!}{\kappa }_k}$$

which yields thecumulants:

For the special case whenμ₁ =μ₂, anasymptotic expansion of themodified Bessel function of the first kind yields for largeμ:

$${\displaystyle p(k;\mu ,\mu )\sim \frac{1}{\sqrt{4\pi \mu }}\left[1+\sum _{n=1}^{\mathrm{\infty }}{\left(-1\right)}^n\frac{\left\{4k^2-1^2\right\}\left\{4k^2-3^2\right\}\cdots \left\{4k^2-(2n-1{)}^2\right\}}{n!2^{3n}(2\mu {)}^n}\right].}$$

(Abramowitz & Stegun 1972, p. 377). Also, for this special case, whenk is also large, and oforder of the square root of 2μ, the distribution tends to anormal distribution:

$${\displaystyle p(k;\mu ,\mu )\sim \frac{e^{-k^2/4\mu }}{\sqrt{4\pi \mu }}.}$$

These special results can easily be extended to the more general case of different means.

If${\displaystyle X\sim \mathrm{Skellam}({\mu }_1,{\mu }_2)}$, with, then

$${\displaystyle \frac{\exp \left[-{\left(\sqrt{{\mu }_1}-\sqrt{{\mu }_2}\right)}^2\right]}{{\left({\mu }_1+{\mu }_2\right)}^2}-\frac{e^{-({\mu }_1+{\mu }_2)}}{2\sqrt{{\mu }_1{\mu }_2}}-\frac{e^{-({\mu }_1+{\mu }_2)}}{4{\mu }_1{\mu }_2}\le Pr\{X\ge 0\}\le \exp \left[-{\left(\sqrt{{\mu }_1}-\sqrt{{\mu }_2}\right)}^2\right]}$$

Details can be found inPoisson distribution § Poisson races

• Ratio distribution for (truncated) Poisson distributions

• Discrete distributions
• Poisson distribution
• Infinitely divisible probability distributions

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