Showing $E(S^2\mid \bar X)=\bar X$ for i.i.d Poisson random variables $X_i$

Solution 1:

$\newcommand{\e}{\operatorname{E}}$ \begin{align} & \Pr(X_1=x_1 \mid \overline X = x/n) = \Pr(X_1=x_1\mid X_1+\cdots+X_n = x) \\[10pt] = {} & \frac{\Pr(X_1=x_1\ \&\ X_1+\cdots+X_n = x)}{\Pr(X_1+\cdots+X_n = x)} \\[10pt] = {} & \frac{\Pr(X_1=x_1\ \&\ X_2+\cdots+X_n = x-x_1)}{\Pr(X_1+\cdots+X_n = x)} \\[10pt] = {} & \frac{\dfrac{\lambda^{x_1} e^{-\lambda}}{(x-x_1)!} \cdot \dfrac{((n-1)\lambda)^{x-x_1} e^{-((n-1)\lambda)}} {x_1!}}{\left( \dfrac{(n\lambda)^x e^{-n\lambda}} {x!} \right)} = \binom x {x_1} \left( \frac 1 n \right)^x \left( 1 - \frac 1 n \right)^{x-x_1} \end{align} In other words, $$ X_1\mid \overline X \sim \operatorname{Binomial} \left(n\overline X, \frac 1 n \right). $$ Therefore $$ \operatorname E\left((X_1-\overline X)^2 \mid \overline X\right) = \overline X\left( 1 - \frac 1 n \right). $$ Consequently $$ \operatorname E\left( (X_1-\overline X)^2 + \cdots + (X_n-\overline X)^2 \mid \overline X\right) = (n-1)\overline X. $$