Let $X_1, X_2, ..., X_n$ be a random sample from a distribution with mean $\mu \in \mathbb{R}$, variance $\sigma^{2} \gt 0$ and $E\{X^4\} \lt \infty$. For the sample variance $S^{2}_{n}$, we have shown the following fact: $\sqrt{n}(S^{2}_{n} - \sigma^{2}) \rightarrow^{D} N(0,\psi^2)$, as $n \rightarrow \infty$, where $\psi^2 = Var\{(X - \mu)^2\}$. Construct an estimator $\hat{\psi^2}$ for $\psi^2$, and show that $\hat{\psi^2}$ is a consistent estimator.
$\textbf{My approach:}$
$Var\{(X - \mu)^2\} = E\{(X-\mu)^4\} - [E\{(X-\mu)^2\}]^2 = E\{X^4 - 4X^3\mu + 6X^2\mu^2 - 4X\mu^3 + \mu^4\} - \sigma^4$ $= E\{X^4\}-4\mu E\{X^3\} + 6\mu^2 E\{X^2\} - 4\mu^3 E\{X\} + \mu^4 - \sigma^4 =$ $E\{X^4\} - 4\mu E\{X^3\} + 6\mu^2(\sigma^2 + \mu^2) - 4\mu^4 + \mu^4 - \sigma^4 =$ $E\{X^4\} - 4\mu E\{X^3\} + 6\mu^2\sigma^2 + 6\mu^4 - 4\mu^4 + \mu^4 - \sigma^4 =$ $E\{X^4\} - 4\mu E\{X^3\} + 6\mu^2\sigma^2 + 3\mu^4 - \sigma^4$.
Using the method of moments and sample variance $S^2_n$, I get the following estimator:
$\hat{\psi^2} = \frac{\sum X^4_i}{n} - 4(\frac{\sum X_i}{n})(\frac{\sum X^3_i}{n})+6\Bigl(\frac{\sum X_i}{n}\Bigl)^2 S^2_n + 3\Bigl(\frac{\sum X_i}{n}\Bigl)^{4} - (S^2_n)^2.$
I am stuck in proving whether this estimator is consistent or not. I have tried using the Strong Law of Large Numbers but I do not know how to find $E\{X^4\}$ and $E\{X^3\}.$
Does anyone have any ways to prove this or a better approach?
Many thanks.