# Calculating $Var\left\{(\hat{m}-m)^2\right\}$ for a univariate normal distribution

Suppose $\hat{m} = \frac{1}{N}\sum_{i=1}^{N}(X_i)$ where $X_i \sim N(m,\sigma)$.

Are the following steps correct?

$Var\left\{(\hat{m}-m)^2\right\} = E\left\{(\hat{m}-m)^4\right\} - E^2\left\{(\hat{m}-m)^2\right\}$

$= 3E^2\left\{(\hat{m}-m)^2\right\} - E^2\left\{(\hat{m}-m)^2\right\}$

$= 2E^2\left\{(\hat{m}-m)^2\right\}$

and I know that $E\left\{(\hat{m}-m)^2\right\} = \frac{1}{N^2}\sigma$. (I was wrong here. Read the Update)

Then, $Var\left\{(\hat{m}-m)^2\right\} = 2\frac{1}{N^4}\sigma^2$

However the textbook says (without any proving) that

$Var\left\{(\hat{m}-m)^2\right\} \tilde{} \frac{1}{N^2}$

Where am I going wrong?

Update: as whuber told in the comments, i was wrong about $E\left\{(\hat{m}-m)^2\right\}$. This expectation equla to $\frac{1}{N}\sigma$ and not $\frac{1}{N^2}\sigma$.

Therefore, the variance is

$Var\left\{(\hat{m}-m)^2\right\} = 2E^2\left\{(\hat{m}-m)^2\right\} = 2\frac{1}{N^2}\sigma^2 \tilde{} \frac{1}{N^2}$

Anyway, the answer provided by mpiktas is also correct and i prefer to chose it as the best answer.

• The variance of the sample mean ("and I know that...") scales as 1/N, not 1/N^2. That's the only change you need. – whuber Dec 8 '10 at 22:55
• @whuber. yes you are right again :D – Isaac Dec 9 '10 at 10:19

## 2 Answers

If $\hat{m}=\frac{1}{n}\sum_{i=1}^nX_i$, where $X_i$ is iid normal sample, then $\hat{m}\sim N(m,\frac{\sigma^2}{n})$. Then $(\hat{m}-m)\sim N(0,\frac{\sigma^2}{n})$ and we can apply the results about normal distribution. We have

\begin{align*} Var((\hat{m}-m)^2)&=Var((N(0,\frac{\sigma^2}{n}))^2)\\ &=E(N(0,\frac{\sigma^2}{n}))^4-(E(N(0,\frac{\sigma^2}{n}))^2)^2\\ &=3\frac{\sigma^4}{n^2}-\frac{\sigma^4}{n^2}=2\frac{\sigma^4}{n^2} \end{align*}

• exactly. The error is in the first line of the question, where the OP defines $\hat{m}$ as the sum of the $X_i$ not the mean, as you do. – shabbychef Dec 8 '10 at 19:43
• The quantity $(N(0,\frac{\sigma^2}{n}))^2$ is a scaled Chi-square. As a check, the variance of a Chi-square with $k$ dof is $2k$, which gives the same result you get. – shabbychef Dec 8 '10 at 20:14

I think you intended to take the mean of the $x_i$, instead you took the sum in the definition of $\hat{m}$. This makes the quantity $\hat{m} - m$ look weird.