I am looking at trying to derive an expression for the asymptotic distribution. We have $X_1,\ldots, X_n$ i.i.d $N(\mu, σ^2)$.

So we have defined $\hat \sigma^2 = \frac 1n \sum_{i=1}^n(X_i-\mu)^2$. (MLE of $σ^2$) If further we are told that $\hat\sigma^2$ is a sample mean, how would I go about deriving an expression for the asymptotic distribution for:

$$\sqrt n( \hat\sigma^2 -σ^2).$$

I initially had the thought of using a chi squared argument and ultimately proving that $\sqrt n( \hat\sigma^2 -σ^2)$ -> $N(0, 2σ^4)$. However, I now think that my argument is flawed.

How would you guys go about it?

  • $\begingroup$ This generalization, stats.stackexchange.com/q/105337/28746 may be of interest to you. $\endgroup$ – Alecos Papadopoulos Nov 27 '18 at 21:29
  • $\begingroup$ Not going to lie that example made me very confused! $\endgroup$ – Olivia Roberts Nov 27 '18 at 22:33
  • $\begingroup$ Why? It is the general case where the variables are not necessarily normal. Then, one just notes that the 4th central moment of the Normal is $3\sigma^4$ and arrives at the $2\sigma^4$ variance that holds for the Normal case. $\endgroup$ – Alecos Papadopoulos Nov 27 '18 at 22:56
  • $\begingroup$ So in the case for i.i.d normal Xis, is the var((Xi−μ)^2) =2σ^4 $\endgroup$ – Olivia Roberts Nov 27 '18 at 22:59
  • $\begingroup$ Yes. In general it is $E(x^4) - [E(x^2)]^2$. $\endgroup$ – Alecos Papadopoulos Nov 28 '18 at 0:48

$\dfrac 1 n \sum_{i=1}^n (X_i-\mu)^2$ is the MLE of $\sigma^2$ only if $\mu$ is known.

You have $$ \operatorname E((X_i-\mu)^2) = \sigma^2 $$ and $$ \operatorname{var}((X_i-\mu)^2) = 2\sigma^4 $$ and $(X_i-\mu)^2,\,\, i = 1,\ldots,n$ are independent.

Therefore the central limit theorem is applicable and you get $$ \frac{\left( \sum_{i=1}^n (X_i - \mu)^2 \right) - n\sigma^2}{\sqrt2\,\,\sigma^2\cdot\sqrt n} \overset{L}{\longrightarrow} N(0,1) $$ or in other words $$ \frac{\left( \frac 1 n \sum_{i=1}^n (X_i-\mu)^2\right) - \sigma^2}{\sqrt2\,\,\sigma^2/\sqrt n} \overset{L}{\longrightarrow} N(0,1). $$

If $\mu$ is not known, you have a more complicated problem since $(X_i-\overline X)^2,\,\,i=1,\ldots,n$ are not independent. But it can be shown that $\frac 1 {\sigma^2} \sum_{i=1}^n (X_i - \overline X)^2 \sim \chi^2_{n-1},$ and thus it has the same distribution as $Z_1^2+\cdots + Z_{n-1}^2$ where $Z_i,\,\,i=1,\ldots,n-1$ are i.i.d. $N(0,1),$ and then the central limit theorem can again be applied.

  • $\begingroup$ Thank you Michael! Mu is known so it would be the first case. I follow completely, apart from how did you get the var((Xi−μ)^2)=2σ^2. Ive read online that it is this, however when it comes to deriving it myself I get stuck! How did you come to this conclusion? $\endgroup$ – Olivia Roberts Nov 27 '18 at 18:44
  • $\begingroup$ @AlecosPapadopoulos : Sorry -- I needed a fourth power there: $$\operatorname{var}((X_i-\mu)^2) = 2\sigma^4.$$ $\endgroup$ – Michael Hardy Nov 28 '18 at 1:34
  • $\begingroup$ @OliviaRoberts : I should have had $\operatorname{var}((X_i - \mu)^2) = 2\sigma^4$ rather than $2\sigma^2.$ Deriving it is elementary but takes some work. Could you post a separate question on that? $\endgroup$ – Michael Hardy Nov 28 '18 at 1:42
  • $\begingroup$ Yes I will do that! $\endgroup$ – Olivia Roberts Nov 28 '18 at 8:19

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