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- #1
That's exactly it for the formula !Something like this?
$g(B)=e^{-B}\frac{B^k}{k!}$
$\Rightarrow E(g(B))=\displaystyle\int_{-\infty}^{\infty}g(b)f_B(b)db$
$=\displaystyle\int_{0}^{\infty}e^{-b}\frac{b^k}{k!}\mu e^{-\mu b}db$
$=\frac{\mu}{k!}\displaystyle\int_{0}^{\infty}b^ke^{-b(1+\mu)} db$
$=\frac{\mu}{k!}(0-(-1))=\frac{\mu}{k!}$
erm... ?
Well if you want to compute it without mentioning gamma function, it's possible, but you'd have to do successive integrations by parts.Gamma functions?! Haven't done this before...
$=\frac{\mu}{k!}\displaystyle\int_{0}^{\infty}b^ke ^{-(1+\mu)b} db$
$=\frac{u}{k!} \frac {\Gamma (k+1)}{(1+u)^{k+1}}$
$=\frac{u}{(1+u)^{k+1}}$
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