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

Are the following steps correct?

\begin{align}\operatorname{Var}\left\{(\hat{m}-m)^2\right\} &= \mathrm E\left\{(\hat{m}-m)^4\right\} - \mathrm E^2\left\{(\hat{m}-m)^2\right\}\\&= 3\mathrm E^2\left\{(\hat{m}-m)^2\right\} - \mathrm E^2\left\{(\hat{m}-m)^2\right\}\\&= 2\mathrm E^2\left\{(\hat{m}-m)^2\right\}\end{align}

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

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

However the textbook says (without any proving) that

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

Where am I going wrong?

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

Therefore, the variance is

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

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

  • 2
    $\begingroup$ 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. $\endgroup$
    – whuber
    Dec 8, 2010 at 22:55
  • $\begingroup$ @whuber. yes you are right again :D $\endgroup$
    – Isaac
    Dec 9, 2010 at 10:19

2 Answers 2


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

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

  • $\begingroup$ 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. $\endgroup$
    – shabbychef
    Dec 8, 2010 at 19:43
  • 3
    $\begingroup$ 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. $\endgroup$
    – shabbychef
    Dec 8, 2010 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.


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