Central Limit theorem with extremely skewed population If $X$~$N(1,0.1)$, and $Y=X^n$ where $n$ is very large (e.g. 200).
Due to the asymetric distribution, $E(Y)$ will be skewed very far towards the extreme upper end of the distribution, dominated by values of $X$ that do not even appear in most samples.
The Central Limit Theorem states that the distribution of sample means, for a large enough sample size and number of samples, follows a normal distribution with mean equal to $E(Y)$. However, given in this case the population mean is dominated by values that will almost never in practice feature in the sample, it will be extremely rare for the sample mean to be as large as the population mean.
Therefore, the distribution of sample means is not normally distributed around the population mean (which I have checked empirically).
How does this fit in with the laws of the Central Limit Theorem?
Thanks
 A: I think the crucial factor here is that the Central Limit Theorem's statement depends on the variance of the distribution from which you sample.
The expectations of powers of a normal random variable are known. Instead of thinking of $X\sim N(1, 0.1)$, we can think of $Z=X-1$ because then $Z\sim N(0, 0.1)$ which makes the moments simpler to compute. We then have $Y=X^n = (Z+1)^n$. Now:
$$
E(Y)=E[(Z+1)^n]
=\sum_{k=0}^n {n \choose k}E(Z^k)
=\sum_{k=0}^n \delta(k\mod2) {n \choose k} \sigma^k(k-1)!!,
$$
which is non-negative and finite but eventually grows very rapidly with $n$ because the double factorial grows faster than $\sigma^k$ decays. We can compute the variance of $Y$,
$$
V(Y) = E(Y^2)-E(Y)^2,
$$
which is also non-negative and finite but will grow very fast with $n$ as well (the positive term overwhelms the negative, again because of the double factorial).
The Central Limit Theorem states that for a random sample $A_1, A_2, \dots, A_p$, where $A_i$ are independent, identically distributed with mean $\mu_A$ and variance $\sigma_A^2$, a normalised error of the sample mean as an estimator of the mean converges in distribution to a normal variable; precisely:
$$
\sqrt{p}\,(S_p-\mu_A) \xrightarrow[]{d} N(0,\sigma_A^2),
$$
where $p$ is the number of samples. Not very formally, this means that the error, the random variable $(S_p-\mu_A)$, is approximately distributed like $N(0, \frac{\sigma_A^2}{p})$. As the number of samples $p$ grows, the distribution of the error is both closer to a normal distribution and has a lower variance. 
However, if the variance $\sigma_A^2$ is much larger than $p$, the variange of the error, approximately $\frac{\sigma_A^2}{p}$, can still have a large variance, i.e. $S_p$ could be very different from $\mu_A$ with large probability. I think this is exactly what is happening here, because the variance of $Y$ is enormous for large $n$.
