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I was reading some lecture notes when I saw a simplification I didn't understand. Specifically, we have $V_n\sim\chi^2(n)$. It was then written then $$E\left(\frac{1}{n+2}V_n-1\right)^2=\frac{2}{n+2}.$$ Is there an easy way to see this that I am just missing? I know that the mean and variance of $V_n$ are $n$ and $2n$ respectively, but I don't see how the equality follows from this.

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Let's expand the square and take the expectation of the expansion:

$$\begin{eqnarray} \mathrm{E}\left[{V_n^2 \over (n+2)^2}-{2\over n+2}V_n+1 \right] &=& {\mathrm{E}V_n^2 \over (n+2)^2}-{2\mathrm{E}V_n \over n+2}+1 \\ &=& {n^2 + 2n \over (n+2)^2}-{2n\over n+2}+1 \\ &=&{n^2 + 2n - 2n(n+2) + (n+2)^2\over (n+2)^2} \end{eqnarray}$$

In line 2 we have used the relationship $\mathrm{E}V_n^2 = (\mathrm{E}V_n)^2 + \mathrm{Var}V_n. $Rearranging terms in the numerator gives us:

$${n^2-2n^2+n^2+2n-4n+4n+4 \over (n+2)^2} ={2n+4 \over (n+2)^2} = {2 \over n+2}$$

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  • $\begingroup$ Thank you! That makes a lot of sense! $\endgroup$
    – Anon
    Commented May 6 at 3:38

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