4
$\begingroup$

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.

$\endgroup$
  • 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 '10 at 22:55
  • $\begingroup$ @whuber. yes you are right again :D $\endgroup$ – Isaac Dec 9 '10 at 10:19
7
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ 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 '10 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 '10 at 20:14
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.