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Let $X_1,\ldots,X_n$ be independent $\mathcal{N}(\mu, \sigma^2).$

I got stumbled in one term of a task. I know

$$\mathbb E\big[\sum_{i=1}^{n} (X_i-\bar X)^2\big]= \sum_{i=1}^{n}\mathbb E\big[(X_i-\bar X)^2\big] =\sum_{i=1}^{n}\mathbb V(X_i)=n \sigma^2.$$

But how can I simplify the following term by an expression that only involved $n$ and $\sigma^2$?

$$\mathbb E\big[\big\{\sum_{i=1}^{n} (X_i-\bar X)^2\big\}^2\big]=?$$

I tried by the following: $$\mathbb E\big[\big\{\sum_{i=1}^{n} (X_i-\bar X)^2\big\}^2\big]$$ $$=\mathbb E\big[\big\{\sum_{i=1}^{n} (X_i^2-2X_i\bar X+\bar X^2)\big\}^2\big]$$ $$=\mathbb E\big[\big\{\sum_{i=1}^{n}X_i^2-n\bar X^2)\big\}^2\big]$$ $$=\mathbb E\big[\big(\sum_{i=1}^{n}X_i^2\big)^2-2n\bar X^2\sum_{i=1}^{n}X_i^2+n^2\bar X^4]$$

Then, I don't know how is to simplify it more.

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  • $\begingroup$ $E[(X_i-\overline X)^2]\ne V(X_i)$. Finding the first expectation does not require normality. For the second expectation, it is easier to use the fact that $(n-1)s^2/\sigma^2$ has a $\chi^2_{n-1}$ distribution where $s^2=\frac{1}{n-1}\sum (X_i-\overline X)^2$. $\endgroup$ Nov 2, 2019 at 11:36
  • $\begingroup$ A chi-squared distribution with $n-1$ degrees of freedom has mean $n-1$ and variance $2n-2$ so a second moment about zero of $n^2-1$. You need to scale this appropriately, i.e. by $\sigma^4$ $\endgroup$
    – Henry
    Nov 2, 2019 at 13:21

1 Answer 1

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Your first expression is not correct, the result is $(n-1)\sigma^2$. To see this, observe that the sample variance can be written as $(X-\mu)^\top M (X-\mu)$ where $M$ is an idempotent matrix with rank $n-1$. This also shows that the distribution of the sample variance is proportional to $\chi^2_{n-1}$. Now all you need is the mean and variance of $\chi^2$ distribution which one can find, for example, in Mood, Graybill, and Boes.

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