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Is there an easier way to find the variance of sample average squared $(\bar X^2)$ without using the moment generating function? $X\sim N(\mu, \sigma^2)$

I know that $Var(\bar X^2)= E(\bar X^4) -(E(\bar X^2))^2$. Thus I can find the $E(\bar X^4)$ using the moment generating function, and hence the $Var(\bar X^2)$.

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For the hard part, one can expand $(X_1+\cdots+X_n)^4$ and find the expectation. A bit tedious. –  André Nicolas Feb 7 '13 at 3:39
    
You don't need to expand the fourth power of a sum of $n$ terms, but only of two terms. I'll post an answer. –  Michael Hardy Feb 7 '13 at 5:05

2 Answers 2

up vote 1 down vote accepted

You have $$ \bar X \sim N\left( \mu, \frac{\sigma^2}{n} \right). $$ The density of that distribution is $$ x\mapsto\frac{\sqrt{n}}{\sqrt{2\pi\,{}}\,\sigma} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2/n} \right). $$ Therefore $$ \mathbb E(\bar X^4) = \frac{\sqrt{n}}{\sqrt{2\pi\,{}}\,\sigma} \int_{-\infty}^\infty x^4 \exp\left( -\frac{(x-\mu)^2}{2\sigma^2/n} \right) \, dx. $$ So write \begin{align} y & = \frac{x-\mu}{\sigma/\sqrt{n}}, \\[10pt] dy & = \sqrt{n}\frac{dx}{\sigma}. \end{align} Then the integral becomes $$ \frac{\sqrt{n}}{\sqrt{2\pi\,{}}\,\sigma} \int_{-\infty}^\infty \left(\frac{\sigma}{\sqrt{n}} y + \mu\right)^4 \exp\left( -\frac{y^2}{2} \right) \left( \frac{\sigma}{\sqrt{n}}\,dy\right). $$

When the $4$th power is expanded as the sum of five terms, one gets $$ = \frac{\sigma^4}{n^2\sqrt{2\pi\,{}}} \int_{-\infty}^\infty y^4 e^{-y^2/2} \, dy. $$ Next: $$ \int_{-\infty}^\infty y^3 e^{-y^2/2} \Big( y \, dy \Big) = 2\int_0^\infty \text{ditto} = 2\int_0^\infty u^{3/2} e^{-u} \, du $$ $$ 2\Gamma\left(\frac52\right) = s\cdot\frac32\Gamma\left(\frac12\right) = 2\sqrt{\pi}. $$

One then does the same thing with the other terms in the expansion of the fourth power.

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Couldn't you use the standard rules for variance calculations? Given that your X:s are IID

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We get this many terms

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and for Var[xi*xj] we can use Goodman's expression

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We put it all together

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Please use proper syntax and not photos of various sizes. –  user88595 May 14 at 15:04

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