Find the variance of $(\bar X^2)$ without using Moment Generating Function 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 $Var(\bar X^2)$.
 A: 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.
A: Couldn't you use the standard rules for variance calculations? Given that your X:s are IID

We get this many terms

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

We put it all together

