Find $E[N^2 | N 2]$ for a frequency distribution

Question

N has probability mass function: $p_o = p_1 =0$ and $p_k = frac{1}{(e^1-2)k!}$ for $k=2,3,4,...$ I Solved for the pgf of N and got $G(t) = frac{e^t}{e^1-2}$
How do I calculate $E[N^2 | N>2]$?

Siong Thye Goh · Answer

begin{align}
mathbb{E}[N^2 | N > 2] &= sum_{n=0}^infty n^2 Pr(N=n|N>2) \
&= frac1{1-P(Nle 2)}sum_{n=3}^infty n^2cdot frac{1}{(e-2)n!} \
&=frac1{1-frac1{2(e-2)}}sum_{n=3}^infty ncdot frac{1}{(e-2)(n-1)!} \
&= frac1{1-frac1{2(e-2)}}sum_{n=3}^infty (n-1+1)cdot frac{1}{(e-2)(n-1)!} \
&=frac{2}{2(e-2)-1} left(sum_{n=3}^infty frac{1}{(n-2)!}  + sum_{n=3}^infty frac{1}{(n-1)!} right)
end{align}
I am leaving the last step for you to simplify.

Find $E[N^2 | N > 2]$ for a frequency distribution

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