Assume $X_1,...,X_n$, are i.i.d, $E(X_i)=\mu, Var(X_i) = \sigma^2, E(X^k)=\mu_k, k=3,4$ Then sample variance satisfies $\sqrt{n}(\hat\sigma^2-\sigma^2)\to_D N(0,\alpha^2)$, find $\alpha$
the question doesnt say but I assume it means sample variance $\frac{1}{n}\sum_{i=1}^n (X_i-\bar X)^2$, not divided by $n-1$.
So Central Limit Theorem gives me that $\sqrt{n}(\frac{1}{n}\sum_{i=1}^n X_i - \mu)\to_D N(0,\Sigma)$
And so from here I want to use delta method, to find some function $f$, to take $f(\bar X)-f(\mu)=\hat\sigma^2-\sigma^2$ and then use the delta method to compute $\alpha$
I expanded $\hat\sigma^2=\frac{1}{n}\sum_{i=1}^n (X_i^2 - 2X_i \bar X +\bar X^2)=\frac{1}{n}\sum_{i=1}^n X_i^2 - \bar X^2$
I know that $\sigma^2=E(X^2)-\mu^2$
From here I'm not sure what to do, I also don't see where the condition that $E(X^k)=\mu_k, k=3,4$ would be needed.
Here is a sketch for how to approach this:
Multivariate CLT tells us
$$W_i\equiv(X_i,X_i^2)'\text{ iid}\implies \sqrt n (\frac{1}{n}\sum_i W_i-E[W_1])\to_d N(0,V(W_1)).$$
Note how the finite moment assumptions are used in the above!
Now write
$$\sqrt n(\hat \sigma^2-\sigma^2)=\sqrt n \left(g\left(\frac{1}{n}\sum_i W_i\right)-g\left(E[W_1]\right)\right)$$
where $$g:\mathbb{R}^2\to\mathbb{R}, g(a,b)=b-a^2$$ and use delta method.