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Consider independent $X_1,\ldots, X_n\sim\mathcal{N}(\mu_0,\sigma^2)$ with a known $\mu_0\in\mathbb{R}$ and unknown $\sigma^2\in(0,\infty)$. I already know that

$$\DeclareMathOperator{\Var}{Var}\frac{1}{n}\sum_{i=1}^n (X_i-\mu_0)^2$$

is variance minimising and that

$$E\Big[\frac{1}{n}\sum_{i=1}^n (X_i-\mu_0)^2\Big]=E\Big[\frac{1}{n-1} \sum_{i=1}^n(X_i-\overline{X})^2\Big]=\sigma^2$$

What I want to show is that

$$\frac{1}{n^2}\Var[\sum_{i=1}^n (X_i-\mu_0)^2]=\Var[\frac{1}{n}\sum_{i=1}^n (X_i-\mu_0)^2]<\Var[\frac{1}{n-1}\sum_{i=1}^n(X_i-\overline{X})^2]=\frac{1}{(n-1)^2}\Var[\sum_{i=1}^n(X_i-\overline{X})^2]$$

Since the first moments are the same it would be enough, to know, that

$$E\Big[\Big(\sum_{i=1}^n (X_i-\mu_0)^2\Big)^2\Big]=E\Big[\Big(\sum_{i=1}^n(X_i-\overline{X})^2\Big)^2\Big]$$

Is this true? There might be an easy argument, but I do not find it.

Thanks in advance!

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Recall that if $X_1,X_2,...,X_n$ iid $N(\mu, \sigma ^ 2)$, then $$ \DeclareMathOperator{\Var}{Var} \frac{\sum ( X_i - \mu ) ^ 2}{ \sigma ^ 2} \sim \chi^2_n, \quad \frac{\sum ( X_i - \bar{X}_n ) ^ 2}{ \sigma ^ 2} \sim \chi^2_{n-1}, $$ and $$ \Var(\chi^2_n)=2n, \quad \Var(\chi^2_{n-1})=2(n-1). $$

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  • $\begingroup$ Where $\overline{X}_n=\frac{1}{n}\sum_{i=1}^nX_i$? $\endgroup$ – user408858 Dec 1 '18 at 14:10
  • $\begingroup$ @user408858 Right $\endgroup$ – V. Vancak Dec 1 '18 at 14:10
  • $\begingroup$ So I get $Var[\sum_{i=1}^n (X_i-\mu_0)^2]= 2n\sigma^4$ and $Var[ \sum_{i=1}^n(X_i-\overline{X})^2]=2(n-1)\sigma^4$, right? $\endgroup$ – user408858 Dec 1 '18 at 14:13
  • $\begingroup$ Yup, that is right $\endgroup$ – V. Vancak Dec 1 '18 at 14:18

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