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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:

enter image description here

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 $m$?

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  • $\begingroup$ you could consider using numerics $\endgroup$
    – fGDu94
    Commented Jul 8 at 18:52
  • $\begingroup$ @fgdu94 I need to get the complete expression, not approximations. $\endgroup$
    – Cardstdani
    Commented Jul 8 at 19:19
  • $\begingroup$ Stirling numbers of the second kind appear in occupancy problems like the coupon collectors problem and the birthday problem, which often have simple expressions for expectations by using linearity of expectation rather than sums like yours (presumably of a discrete distribution on the integers rather than a continuous density). What does your distribution represent? $\endgroup$
    – Henry
    Commented Jul 9 at 0:27
  • $\begingroup$ Incidentally, $m$ and $n$ only seem to appear as their product $mn$ in your expression. $\endgroup$
    – Henry
    Commented Jul 9 at 0:28
  • $\begingroup$ As a further thought, the values in your chart do not sum to $1$ (some individual values look as if they are over $40$) and I am not sure $\sum\limits_k \frac{(m n)! \mathcal{S}_k^{(j)}}{(m n)^k (m n-j)!}=1$ either (my attempt seemed to get a sum of $30$ for your example). What is this trying to represent? $\endgroup$
    – Henry
    Commented Jul 9 at 14:40

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