expected valuerandom variablevariance
Suppose $X$ is a random variable with mean $0$ and variance $\sigma_x^2$.
How can I calculate mean and variance of $X^2$?
I calculated the mean like this
\begin{equation*}
\operatorname{Var}(X) = \operatorname{E}(X^{2}) – \operatorname{E}^{2}(X) \rightarrow \operatorname{E}(X^{2}) = \sigma_{x}^2
\end{equation*}
but am stuck at the variance.
Can you not use $e^x \approx 1+ x+ \frac{X^2}{2} $ then $Var(e^X) \approx Var(X)+ \frac{Var(X^2)}{2} $ ?
It seems to me you want 2nd order approximation ..
The distribution of the sample variance is slightly tricky, particularly because of the way the sample mean comes into it.
Note that
it has a discrete distribution,
by taking deviations from the sample mean, the sizes of the positive and negative deviations will vary from sample to sample and will generally not be of the same sizes (e.g. imagine with $n=10$ that the mean is $1.9$; then deviations ($x_i-\bar{x}$) for value above the mean, will be $0.1$ or $1.1$ or $2.1$, while those below will be $-0.9$ or $-1.9$; but in the next sample the mean might be $1.7$, so the deviations will be values like $0.3$ or $-1.7$)
the squaring makes both the deviations above and below the mean of different sizes even within one sample (consider deviations about the mean of -1.9, -0.9, 0.1, 1.1, 2.1; their squares are 3.61, 0.81, 0.01, 1.21 and 4.41, so the gaps between adjacent values of those jump in different increments ... and then these are "averaged" - but with n-1 denominator - to produce a sample variance)
As a result you have a discrete distribution over a pretty complicated set of values (this set is countably infinite in size). The set of values taken also varies with sample size (n=3 yields a different set of possible values than n=10). Here's an example via simulation (though the simulation is so large that the distribution displayed is essentially the population cdf -- it's accurate to within about a pixel):
We can clearly see "clumpiness" in the distribution - uneven spacing, as well as an intermingled jumble of large and small probabilities.
The distributions are of course different at different values of $n$ and the Poisson mean, but the general impression (a clumpy discrete distribution, with uneven spacing and irregular progression of probabilities) is - unsurprisingly - similar across a range of values.
The issue is even trickier if you want to back out some form of confidence interval for the population variance, since I am pretty sure you won't have a pivotal quantity to work with.
However, you might be able to get somewhere via approximation. The upper tail in particular is a fair bit smoother than the lower tail and might be amenable to a continuous approximation.
It looks like either large $\lambda$ or large samples will give smooth outcomes that may have scaled chi-square approximations.
Best Answer
If you have only the mean and variance of $X$ as 0 and $\sigma_x^2$, then there is insufficient information to calculate the variance of $X^2$, which is
$$ E[X^4] - E[X^2]^2 = E[X^4] - \sigma_x^4. $$
For a normal R.V. $\sim N(0, \sigma_x^2)$, $E[X] = 0, E[X^2] = \sigma_x^2, E[X^4] = 3 \sigma_x^4$.
For a uniform R.V. $\sim U(-1.5 \sigma_x, 1.5 \sigma_x)$, $E[X] = 0, E[X^2] = \sigma_x^2, E[X^4] = 9 / 5 \sigma_x^4$.
It's easy to build distributions for which this is infinite.