4th moment of the sample mean estimator I got a following setup:
$(X_i)_{i \geq 1}$ are iid random variables with values in $\mathbb{R}$ and finite second moment.
By the weak law of large numbers: $\sqrt{n}(\bar{X} - E(X))$ converges in distribution to $N(0, Var(X))$, so we could deploy the delta method with a transformation $g(x) = x^4$, implying that $\sqrt{n}g(\bar{X} - E(X))$ should be normally distributed.
I was trying to convince myself about it:
$$
\sqrt{n}g(\bar{X} - E(X)) = \sqrt{n}(\bar{X} - E(X))^4 = \sqrt{n}(Var(\bar{X}))^2 = \sqrt{n} \text{Var}(X)^2 \to \infty \text{ for } n \to \infty$$
so $\sqrt{n}g(\bar{X} - E(X))$ can't be normally distributed
Am I right? If I am, why does the delta method doesn't work in this case? Thanks!
 A: Comming from Mastodon here.
It seems the isue is you are not correctly applying the method. The theorem states that, if $g'(\theta)\neq0$:
$$ \sqrt{n} \cdot \left( X_i - \theta  \right) \overset {D}{ \rightarrow} \mathit{N}(0,\sigma^{2}) \Rightarrow  \sqrt{n} \cdot \left( g(X_i) - g(\theta)  \right) \overset {D}{ \rightarrow} \mathit{N}(0,\sigma^{2}\cdot \left[g'(\theta)\right]^2) $$
In case you are not seeing the difference, the expresion covered by the theorem is $ \sqrt{n} \cdot \left(X_i^4-E^4(X) \right)$ and not yours.
The expression you are evaluating is not deduced from this theorem, and therefore won't help you disprove it as you are attempting.
I strongly suggest reading the proof if you want to convince yourself. It is based on elemental probability and calculus and it is quite directly leading to the expression. I know working with probability can be disheartening as there are so many concepts yo juggle with, but you will get used and it will become clearer.
There are errors in your procedure.
The most important one is you are not properly applying the definition of the variance. The definition is: $V(X)=E\left[\left(X-E(X)\right)^2 \right]$, you seem to have applied "$V(X)=\left(X-E(X)\right)^2$", which is wrong.
Another error here is when you work on the the relationship between the variance and the variante of the mean. It actually is $Var(\bar{X})=\frac{Var(X)}{n^2}$
Good luck with your studies! Let me know if you see any error on my part or want to keep discussing the issue.
