Calculate estimator's asymptotic distribution Let $W_1,\ldots,W_n \sim f_w(w;\lambda)=\frac{3w^2}{\lambda}e^{\frac{-w^3}{\lambda}}\mathbf{1}_{w>0}$ for some $\lambda>0$.

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*Give the asymptotic behavior of $\sqrt{n}(\lambda_{\text{MoM}}-\lambda)$ as $n\rightarrow \infty$


*Give the asymptotic behavior of $\sqrt{n}(\lambda_{\text{MLE}}-\lambda)$ as $n\rightarrow \infty$


*Give the asymptotic $(1-\alpha)\cdot100 \% $ confidence interval for $\log(\lambda)$
Note: MoM means method of moment, MLE means maximum likelihood estimator
Try: I have use MoM: $\frac{\sum{Xi}}{n}=E[X]$ and $l=\operatorname{Log}(\text{Likelihood})$ to calculate the estimators, but don't know how to solve the asymptoic distribution $\sqrt{n}(\lambda_{\text{estimator}}-\lambda)$.
 A: The idea is to use the central limit theorem: note that
$$
\sqrt{n}(\lambda_{\text{MoM}}-\lambda)=\frac 1{\sqrt n}\sum_{i=1}^n(W_i-\mathbb E[W_i]),
$$
which converges to a centered distribution normal distribution with variance $\operatorname{Var}(W_1)$.
For the maximum likelihood estimator, you will get something like $n^{-1}\sum_{i=1}^nW_i^3$ and you can again use the central limit theorem.
A: As commented in the previous post, $\lambda_{\text{MLE}}=\frac{1}{n}\sum_{i=1}^nW_i^3$. Notice $\mathbb{E}(W_1^3)=\lambda$ and $\mathbb{V}(W_1^3)=\lambda^2$.
From the central limit theorem we know $\lambda_{\text{MLE}}$ is approximately $\mathcal{N}\left(\lambda,\frac{\lambda^2}{n}\right)$ when $n$ is large. Put more formally, $$\sqrt{n}\Big(\lambda_{\text{MLE}}-\lambda\Big)\longrightarrow_d\mathcal{N}\left(0,\lambda^2\right)$$ Using the delta method with $g(x)=\log(x)$ we have $$\sqrt{n}\left(\log\left(\lambda_{\text{MLE}}\right)-\log(\lambda)\right)\longrightarrow _d \mathcal{N}(0,1)$$ So for large $n$ $$\begin{eqnarray*}\mathbb{P}\left(-z_{\alpha/2}<\sqrt{n}\left(\log(\lambda_{\text{MLE}})-\log(\lambda)\right)<z_{\alpha/2}\right)\approx 1-\alpha\end{eqnarray*}$$ Evidently $$\log(\lambda_{\text{MLE}})\pm \frac{z_{\alpha/2}}{\sqrt{n}}$$ is a confidence interval that you seek. You may obtain a different confidence interval for $\log(\lambda)$ by using your MOM estimator for $\lambda$.
