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:

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

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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.

  • $\begingroup$ 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

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 $$E\left(S^2\right)=E\left(\frac1{n-1}\sum_i(X_i-\bar X)^2\right)=\theta_2.$$


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