1
$\begingroup$

I am looking at the following problem and solution from Casella and Berger's Statistical Inference.

Let $X_1, \dots, X_n$ be a random sample, i.e. iid random variables with finite fourth moment. Show that $$\operatorname{Var} S^2 = \frac{1}{n}(\theta_4 - \frac{n-3}{n-1}\theta_2^2)$$ where $\theta_1 = \operatorname EX_i$, $\theta_j = \operatorname E(X_i - \theta_1)^j, j=2,3,4.$ Here, we use the following identity:

$$S^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i - \bar{X})^2 = \frac{1}{2n(n-1)}\sum_{i=1}^n \sum_{j=1}^n (X_i - X_j)^2$$

enter image description here

In this solution, I don't understand why $E(S^2) = 24\theta_2$. I think this is wrong, as in the final form, where they calculate $\operatorname{Var}(S^2)= \operatorname E(S^4)- \operatorname E(S^2)^2$, we need $\frac{1}{12} \theta_2^2$ for $\operatorname E(S^2)^2$. However, this means $\operatorname E(S^2)= \frac{1}{\sqrt{12}} \theta_2$, but I can't see how a square root of $12$ appears here. Shouldn't we have

$$\operatorname E(S^2) = \frac{1}{24} \sum_i \sum_j 2\theta_2^2 = \frac{32}{24} \theta_2^2?$$

Finally, I don't understand why we get $112=4\times 16 + 4 \times 16 - 4^2$ terms of zero, and $24, 96, 24$ each for the three remaining terms below. I would greatly appreciate it if anyone helps me understand this.

$\endgroup$
  • $\begingroup$ How are the $X_i$´s are distributed? $\endgroup$ – callculus Jul 26 '18 at 17:40
  • 1
    $\begingroup$ @callculus $X_i$'s are iid with a finite fourth moment. The solution assumes that $E(X_i)=0$. $\endgroup$ – nomadicmathematician Jul 26 '18 at 17:41
  • $\begingroup$ So you don´t know how they are distributed? But from where do the $\theta_i$´s come from? That´s really a mystery. uhh... $\endgroup$ – callculus Jul 26 '18 at 17:43
  • $\begingroup$ @callculus Yes $\theta_i = E(X^i)$, and why would we need a specific distribution here? $\endgroup$ – nomadicmathematician Jul 26 '18 at 17:46
  • $\begingroup$ But from where do the $θ$ i´s come from? That´s my question. $\endgroup$ – callculus Jul 26 '18 at 17:46
1
$\begingroup$

Partial answer:

The $E(S^2)=\theta_2=\operatorname{Var}(X_1)$. This follows easily from the identity $$\sum_i(X_i-\theta_1)^2=\sum_i(X_i-\bar X)^2+n(\bar X-\theta_1)^2,$$ since $$E\left(\sum_i(X_i-\theta_1)^2\right)=\sum_i E(X_i-\theta_1)^2=n\theta_2$$ and $$E(\bar X-\theta_1)^2=\operatorname{Var}(\bar X)=\frac{\theta_2}n,$$ because $E(\bar X)=\theta_1$, too.

So, $$n\theta_2=E\left(\sum_i(X_i-\bar X)^2\right)+n\frac{\theta_2}n,$$ that is $$E\left(\sum_i(X_i-\bar X)^2\right)=(n-1)\theta_2;$$ and then $$E\left(S^2\right)=E\left(\frac1{n-1}\sum_i(X_i-\bar X)^2\right)=\theta_2.$$

$\endgroup$

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.