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

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 $$E[N^2 | N>2]$$?

• $\sum\limits_{k=2}^\infty 1/(e^1k!) = 1-2e^{-1} \approx 0.26424$ so that would not be a probability mass function Jul 21 '20 at 0:05
• What I did was $\sum_{k=1}^{\infty} \frac{c}{k!} = 1$ which using exp series gave $ce^1=1$ Did I do anything wrong here? Jul 21 '20 at 0:11
• You need to decide whether your sum starts with $0$ or $1$ or $2$ or $3$. Though it does not really matter if you are conditioning on $N \gt 2$ Jul 21 '20 at 0:23
• Did I get the pmf wrong? The c value I got was 1/e Jul 21 '20 at 0:28

\begin{align} \mathbb{E}[N^2 | N > 2] &= \sum_{n=0}^\infty n^2 Pr(N=n|N>2) \\ &= \frac1{1-P(N\le 2)}\sum_{n=3}^\infty n^2\cdot \frac{1}{(e-2)n!} \\ &=\frac1{1-\frac1{2(e-2)}}\sum_{n=3}^\infty n\cdot \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}