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}

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

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

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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][1] 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][2] is also correct and i prefer to chose it as the best answer.


  [1]: https://stats.stackexchange.com/users/919/whuber
  [2]: https://stats.stackexchange.com/users/2116/mpiktas