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Let $z_i|\mu\sim N(\mu,1)$ and $\mu\sim N(B,A)$ for $i=1,\dots,N$, the implication is that $z_i|B\sim N(B,A+1)$.

Define $S=\|\textbf{z}\|^2$, and let $B=0$, then $S\sim(A+1)\chi^2_N$ since $\|\textbf{z}\|^2/(A+1)\sim \chi^2_N$.

Question: How do I show that: $$E\Big\{\frac{N-2}{S}\Big\}=\frac{1}{A+1}?$$

I do not see how this is possible since:

$E(S)=(A+1)N$ and $\frac{1}{A+1}=E\Big\{\frac{N}{S}\Big\}$.

I do not know where the $N-2$ is coming from.

Can anybody help?

Context: $\newcommand{\E}{\mathbb{E}}$This question originates from page 4 of Efron's book (Large Scale Inference, Empirical Bayes methods for estimation, Testing and Prediction).

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For any chisquare random variable with $\nu > 2$ degrees of freedom, $X\sim \chi^2_\nu$, the expected value of the inverse variable is $$E\left(\frac{1}{X}\right)=\frac{1}{\nu-2}$$ See for example the Wikipedia article on the inverse chisquare distribution

The result you ask about follows immediately.

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