Skellam distribution

The probability mass function of a Poisson-distributed random variable with mean μ is given by p(k;\mu) = {\mu^k\over k!} e^{-\mu}.\, for k \ge 0 (and zero otherwise). The Skellam probability mass function for the difference of two independent counts K = N_1 - N_2 is the convolution of two Poisson distributions: (Skellam, 1946) \begin{align} p(k;\mu_1,\mu_2) & =\sum_{n=-\infty}^\infty p(k+n;\mu_1)p(n;\mu_2) \\ & =e^{-(\mu_1+\mu_2)}\sum_{n=\max(0,-k)}^\infty {{\mu_1^{k+n}\mu_2^n}\over{n!(k+n)!}} \end{align} Since the Poisson distribution is zero for negative values of the count (p(N, the second sum is only taken for those terms where n \ge 0 and n+k \ge 0. It can be shown that the above sum implies that \frac{p(k;\mu_1,\mu_2)}{p(-k;\mu_1,\mu_2)} = \left(\frac{\mu_1}{\mu_2}\right)^k so that: p(k;\mu_1,\mu_2)= e^{-(\mu_1+\mu_2)} \left({\mu_1\over\mu_2}\right)^{k/2} I_(2\sqrt{\mu_1\mu_2}) where I k(z) is the modified Bessel function of the first kind. The special case for \mu_1 = \mu_2 (= \mu) is given by Irwin (1937): p{\left(k;\mu,\mu\right)} = e^{-2\mu}I_(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 \mu_2 = 0. == Properties ==

As it is a discrete probability function, the Skellam probability mass function is normalized: \sum_{k=-\infty}^\infty p(k;\mu_1,\mu_2) = 1. We know that the probability generating function (pgf) for a Poisson distribution is: G\left(t;\mu\right)= e^{\mu(t-1)}. It follows that the pgf, G(t;\mu_1,\mu_2), for a Skellam probability mass function will be: \begin{align} G(t;\mu_1,\mu_2) & = \sum_{k=-\infty}^\infty p(k;\mu_1,\mu_2)t^k \\[4pt] & = G\left(t;\mu_1\right)G\left(1/t;\mu_2\right) \\[4pt] & = e^{-(\mu_1+\mu_2)+\mu_1 t+\mu_2/t}. \end{align} Notice that the form of the probability-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 \pm 1 would change the support of the distribution and alter the pattern of moments in a way that no Skellam distribution can satisfy. The moment-generating function is given by: M\left(t;\mu_1,\mu_2\right) = G(e^t;\mu_1,\mu_2) = \sum_{k=0}^\infty { t^k \over k!}\,m_k which yields the raw moments mk . Define: \Delta\ \stackrel{\mathrm{def}}{=}\ \mu_1-\mu_2 \mu\ \stackrel{\mathrm{def}}{=}\ \tfrac{1}{2}(\mu_1+\mu_2). Then the raw moments mk are \begin{align} m_1 &= \Delta \\ m_2 &= 2\mu + \Delta^2 \\ m_3 &= \Delta \left(1 + 6\mu + \Delta^2\right) \end{align} The central moments M k are \begin{align} M_2 &= 2\mu, \\ M_3 &= \Delta, \\ M_4 &= 2\mu + 12\mu^2.\, \end{align} The mean, variance, skewness, and kurtosis excess are respectively: \begin{align} \operatorname E(n) & = \Delta, \\[4pt] \sigma^2 & = 2\mu, \\[4pt] \gamma_1 & = \Delta/(2\mu)^{3/2}, \\[4pt] \gamma_2 & = 1/2. \end{align} The cumulant-generating function is given by: K(t;\mu_1,\mu_2)\ \stackrel{\mathrm{def}}{=}\ \ln(M(t;\mu_1,\mu_2)) = \sum_{k=0}^\infty \frac{ t^k }{ k!}\,\kappa_k which yields the cumulants: \begin{align} \kappa_{2k} &= 2\mu, \\ \kappa_{2k+1} &= \Delta . \end{align} For the special case when μ1 = μ2, an asymptotic expansion of the modified Bessel function of the first kind yields for large μ: p(k;\mu,\mu)\sim \frac{1}{\sqrt{4\pi\mu}} \left[1 + \sum_{n=1}^\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, when k is also large, and of order of the square root of 2μ, the distribution tends to a normal distribution: 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. Bounds on weight above zero If {{nowrap|X \sim \operatorname{Skellam} (\mu_1, \mu_2) ,}} with then \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} \leq \Pr\{X \geq 0\} \leq \exp \left[- \left(\sqrt{\mu_1} -\sqrt{\mu_2}\right)^2\right] Details can be found in == See also ==