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I'm trying to find a neater expression for this sequence:

$$ \sum_{s \geq 0} e^{-\lambda}\frac{\lambda^{s}}{s!} sf(w)F(w)^{s-1} $$

where $f(w) = F'(w)$.

The following is my approach, but I think I did something horribly wrong with the sequence - I yield a factorial of $-1$ and don't know how to deal with that...

$$ \\ e^{-\lambda} f(w)\sum_{s \geq 0} \frac{\lambda^{s}}{(s-1)!} F(w)^{s-1} \\ e^{-\lambda} f(w)\sum_{s \geq -1} \frac{\lambda^{s+1}}{s!} F(w)^{s} \\ \frac{1}{\lambda}e^{-\lambda} f(w)\sum_{s \geq -1} \frac{\lambda^{s}}{s!} F(w)^{s} \\ \frac{1}{\lambda}e^{-\lambda} f(w)\left[ \frac{1}{-1!}F(w)^{-1} + \sum_{s \geq 0} \frac{\lambda^{s}}{s!} F(w)^{s} \right]\\ \frac{1}{\lambda}e^{-\lambda} f(w)\left[ \frac{1}{-1!}F(w)^{-1} + e^{\lambda F(w)}\right] \\ \frac{1}{\lambda}e^{-\lambda}f(w) \frac{1}{-1!}F(w)^{-1} + \int \frac{e^{-\lambda (1 - F(w))}}{\lambda}f(w)\\ $$

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  • $\begingroup$ What are $f$ and $F$? Is $F$ an antiderivative of $f$ or something? And what's $E$, is it expected value? $\endgroup$
    – Ian
    Sep 6, 2016 at 11:44
  • $\begingroup$ @Ian Sorry, $E$ should've been removed from this for simplicity. If it matters, $F$ is the antiderivative of $f$. I thought I'd keep the description short as I didn't think there would be much done using that. $\endgroup$
    – FooBar
    Sep 6, 2016 at 11:47
  • $\begingroup$ Which antiderivative? It matters in this context. Or is it more that you are given $F$ and $f=F'$? In that case you should write it that way. $\endgroup$
    – Ian
    Sep 6, 2016 at 11:48
  • $\begingroup$ Basically, this is the expected maximum of drawing $s$ times from a distribution $F(w)$, where $s$ is Poisson distributed. $\endgroup$
    – FooBar
    Sep 6, 2016 at 11:48
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    $\begingroup$ You should skip over the $s=0$ term which is just zero. Then you won't have any $(-1)!$. This is a very familiar trick (for example, it can be used to compute moments of the binomial distribution). $\endgroup$
    – Ian
    Sep 6, 2016 at 11:49

1 Answer 1

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One may observe that the first term in your initial sum vanishes $$ \begin{align} \sum_{s \geq 0} e^{-\lambda}\frac{\lambda^{s}}{s!} sf(w)F(w)^{s-1}&=\sum_{s \geq 1} e^{-\lambda}\frac{\lambda^{s}}{s!} sf(w)F(w)^{s-1} \\\\&= f(w)\: \lambda e^{-\lambda}\sum_{s \geq 0}\frac{\lambda^{s}}{s!}F(w)^{s} \\\\&=f(w)\: \lambda e^{-\lambda}e^{\lambda F(w)} \end{align} $$ with a change of index in the last step.

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