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

Are the following steps correct?

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

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

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

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.

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}

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.

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

Are the following steps correct?

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

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

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

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.

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

Are the following steps correct?

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

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

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

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.

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$. 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.

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

Are the following steps correct?

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

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

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

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.