Skellam distribution: Deep connection between Poisson distributions and Bessel function? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:32:56Z http://mathoverflow.net/feeds/question/27255 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27255/skellam-distribution-deep-connection-between-poisson-distributions-and-bessel-fu Skellam distribution: Deep connection between Poisson distributions and Bessel function? vonjd 2010-06-06T17:44:36Z 2011-01-09T15:13:19Z <p>The probability mass function for the <a href="http://en.wikipedia.org/wiki/Skellam_distribution" rel="nofollow">Skellam distribution</a> for a count difference $k=n_1-n_2$ from two <a href="http://en.wikipedia.org/wiki/Poisson_distribution" rel="nofollow">Poisson-distributed</a> variables with means $\mu_1$ and $\mu_2$ is given by:</p> <p>$$f(k;\mu_1,\mu_2)= e^{-(\mu_1+\mu_2)} \left({\mu_1\over\mu_2}\right)^{k/2}I_{|k|}(2\sqrt{\mu_1\mu_2})$$</p> <p>where $I_k(z)$ is the modified <a href="http://en.wikipedia.org/wiki/Bessel_function" rel="nofollow">Bessel function</a> of the first kind.</p> <p><strong>My question</strong>: Is it just for convenience to get to grips with the resulting infinite summation terms that the Bessel function appears in this formula or is there a deeper mathematical reason connecting Poisson distributions and Bessel functions or even the Poisson distribution with Bessel differential equations? Is there perhaps even some physical interpretation or intution?</p> http://mathoverflow.net/questions/27255/skellam-distribution-deep-connection-between-poisson-distributions-and-bessel-fu/51549#51549 Answer by Didier Piau for Skellam distribution: Deep connection between Poisson distributions and Bessel function? Didier Piau 2011-01-09T15:13:19Z 2011-01-09T15:13:19Z <p>A mathematical reason is as follows.</p> <p>On the one hand, the Laurent series for the modified Bessel functions of the first kind $I_k$ can be deduced from the Laurent series for the Bessel functions of the first kind $J_k$ given <a href="http://en.wikipedia.org/wiki/Bessel_function" rel="nofollow">here</a>. It reads $$\sum_{k\in\mathbb{Z}}I_k(x)t^k=\mathrm{e}^{(x/2)(t+1/t)}.$$ On the other hand, the characteristic function of a Poisson random variable $Y$ with mean $\mu$ is $E(z^{Y})=\mathrm{e}^{-\mu(1-z)}$, at least for every complex number $z$ of modulus $1$. Hence the characterization, which you recalled in your post, of Skellam distribution as the distribution of $X=Y_1-Y_2$ for independent Poisson random variables $Y_1$ and $Y_2$ with means $\mu_1$ and $\mu_2$ shows that $$E(z^X)=\mathrm{e}^{-(\mu_1+\mu_2)}\mathrm{e}^{\mu_1 z+\mu_2/z}.$$ Solving $(x/2)(t+1/t)=\mu_1 z+\mu_2/z$ for $x$ fixed and $t$ depending on $z$ yields $$x=2\sqrt{\mu_1\mu_2},\qquad t=z\sqrt{\mu_1/\mu_2}.$$ The value of $\mathbb{P}(X=k)$ for every integer $k$ follows, which involves $I_k(2\sqrt{\mu_1\mu_2})$. </p> <p>Finally, one should not worry too much about the appearance of $I_k$ in this answer versus $I_{|k|}$ in the OP's post because $J_{-k}(-x)=(-1)^kJ_k(x)$ for every integer $k$, hence $I_{-k}(x)=I_k(x)$ (a relation which is also a consequence of the invariance by $t\to1/t$ of the Laurent series for the functions $I_k$).</p>