Find var$(\overline{X_n}^2)$ given that $EX_1=\mu,E(X_1-\mu)^k=\alpha_k$

Question

$X_i\sim F$ are $n$ iid observations. $\overline{X_n}$ is their mean. I want to find var$(\overline{X_n}^2)$ given that $EX_1=\mu,E(X_1-\mu)^k=\alpha_k$.

What I found out are:

• var$(\overline{X_n})=\frac{\alpha_2}{n}$
• var$(\overline{X_n}^2)=E(\overline{X_n}^4)-[E(\overline{X_n}^2)]^2$
• $E(\overline{X_n}^2)=\frac{\alpha_2}{n}+\mu^2$

I found a hint that

• $E[\overline{X}_n^4]={1\over n^4}\sum E[X_i X_j X_k X_\ell]$ but I can't figure how to get this or how to use this.

I can also find $E(\overline{X_1}^4)$ in terms of $\alpha_k$'s and $\mu$ but how do I use these to get $E(\overline{X_n}^4)$? Any help?

Answer 1

Let $Y_n=X_n-\mu$; then you need to compute $E(\bar Y_n+\mu)^4$. Now open the brackets and observe that you can derive $E \bar Y_n^k=n^{-k} E[(Y_1+\dots+Y_n)^k]$ by opening the brackets again, as $E Y_i^k=\alpha_k$.

Answer 2

Let's use the MGF. Define $m(s):=E\left[ e^{s(X_1-\mu)}\right]$ and note $m^{(k)}(0)=\alpha_k.$ Observe

$$M_{\bar X_n}(t):=E\left[e^{t\bar X _n}\right]=E\left[\prod_{i=1}^n e^{{t \over n}(X_i-\mu)}e^{{t \over n}\mu} \right]=e^{\mu t}\prod_{i=1}^n m(t/n)=e^{\mu t}m(t/n)^n.$$

Your answer is $M_{\bar X_n}^{(4)}(0)$. Computational software can help with derivatives.