Mean of probability distribution

Question

I have a probability distribution defined by the following density function:

$f(k,j,n,m)=\frac{(m n)! \mathcal{S}_k^{(j)}}{(m n)^k (m n-j)!}$ (With $\mathcal{S}_k^{(j)}$ being the Stirling number of the second kind.)

Here you can see a sample plot for $j=29,n=30,m=1$, with $k$ being the horizontal axis:

My goal is to calculate its mean to get an expected value for $k$, but when I apply the definition I get the following expression:

$\sum _{k=1}^{\infty } \frac{k (m n)! \mathcal{S}_k^{(j)} (m n)^{-k}}{(m n-j)!}$

How can I solve this summation so that it can provide a resulting expression as a function of $j,n$ and $m$?

Answer 1

Denote $p=mn$, then the mean $\bar{k}$ you seek equals $\bar{k}=\frac{p!}{(p-j)!}f_j(p)$, with $$f_j(p)=\sum _{k=1}^{\infty } \frac{k \mathcal{S}_k^{(j)} }{p^{k}}=\frac{F_j(p)}{\prod_{n=1}^j(p-n)^2}.$$ The numerator $F_j(p)$ is a polynomial in $p$ of degree $j$, the coefficients are given by the OEIS sequence A196837, for example: _{$$\{F_1(p),F_2(p),\ldots F_5(p)\}=\left\{p,2 p^2-3 p,3 p^3-12 p^2+11 p,4 p^4-30 p^3+70 p^2-50 p,5 p^5-60 p^4+255 p^3-450 p^2+274 p\right\}.$$}

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In response to a comment by Michael Engelhardt, I note that $$\frac{p!}{(p-j)!}\sum _{k=1}^{\infty } \frac{ \mathcal{S}_k^{(j)} }{p^k}=\frac{p}{p-j},$$ so a normalization factor $1-j/p$ is missing in the OP. The normalized probability function is therefore $$P_{j,p}(k)=\frac{(p-1)!}{(p-j-1)!}\frac{ \mathcal{S}_k^{(j)} }{p^k},$$ plotted here for $j=29$, $p=mn=30$.