I have two problems in which the limiting distribution needs to be found. Both require a different method to solve and I do not know how to determine the right method. The first problem is the "standard" method, in which I can solve most of the problems. The second problem requires a different method than the first one.
First Problem
If $Y \sim \mathrm{N}\left(0, \sigma^2\right)$, then $X=|Y|$ will have a distribution with pdf:
$$
f(x ; \sigma)=\sqrt{\frac{2}{\pi \sigma^2}} \exp \left(-\frac{x^2}{2 \sigma^2}\right), \quad x \geq 0,
$$
where $\sigma>0$ is a parameter. In addition, we have $\mathbb{E}(X)=\sigma \sqrt{\frac{2}{\pi}}$ and $\mathrm{E}\left(X^2\right)=\sigma^2$.
Q: $\text { Prove that } \sqrt{n}(\widehat{\sigma}-\sigma) \stackrel{d}{\rightarrow} N\left(0, \frac{\sigma^2}{2}\right) \text { as } n \rightarrow \infty$
S:
\begin{aligned}
-\mathbb{E}\left[\frac{d^2}{d \sigma^2} \ln f(X ; \sigma)\right] &=-\mathbb{E}\left[\frac{d^2}{d \sigma^2}\left(-\ln (\sigma)-\frac{X^2}{2 \sigma^2}\right)\right]=-\mathbb{E}\left[\frac{1}{\sigma^2}-\frac{3 X^2}{\sigma^4}\right] \\
&=-\frac{1}{\sigma^2}+\frac{3 \mathbb{E}\left(X^2\right)}{\sigma^4}=\frac{2}{\sigma^2} .
\end{aligned}
The variance is the reciprocal of the last term.
Second problem
Consider a random sample $X_1, X_2, \ldots, X_n$ from a distribution with the following pdf:
$$
f(x ; \sigma)=\frac{1}{2 \sigma} \exp \left(-\frac{|x|}{\sigma}\right), \quad-\infty<x<\infty, \quad \sigma>0 .
$$
If we denote by $X$ a random variable with the pdf as stated above, then $\mathbb{E}(X)=0$, $\operatorname{Var}(X)=2 \sigma^2$, and $\mathbb{E}\left(X^4\right)=24 \sigma^4$
Q: $\text { Prove that } \sqrt{n}\left(\hat{\sigma}_{M M}-\sigma\right) \stackrel{d}{\rightarrow} N\left(0, \frac{5}{4} \sigma^2\right) \text { as } n \rightarrow \infty$
S: We will derive the limiting distribution in two steps. We first use the CLT to get the limiting distribution of $\sqrt{n}\left(\frac{1}{n} \sum_{i=1}^n X^2-\mathbb{E}\left(X^2\right)\right)$. In the second step we apply the delta method using the transformation $g(y)=\sqrt{\frac{1}{2} y}$.
We have $\operatorname{Var}\left(X^2\right)=\mathbb{E}\left(X^4\right)-\left[\mathbb{E}\left(X^2\right)\right]^2=24 \sigma^4-\left[2 \sigma^2\right]^2=20 \sigma^4$ (by the previous results and exercise header). Hence, the CLT tells us that
$$
\sqrt{n}\left(\frac{1}{n} \sum_{i=1}^n X^2-2 \sigma^2\right) \stackrel{d}{\rightarrow} N\left(0,20 \sigma^4\right) .
$$
To apply the delta method, note the following results: $g\left(2 \sigma^2\right)=\sigma, g^{\prime}(y)=$ $\sqrt{\frac{1}{2}} \frac{d}{d y} \sqrt{y}=\frac{1}{\sqrt{8 y}}$, and $g^{\prime}\left(2 \sigma^2\right)=\frac{1}{4 \sigma}$. Combining these results gives
$$
\sqrt{n}\left(\hat{\sigma}_{M M}-\sigma\right)=\sqrt{n}\left(g\left(\frac{1}{n} \sum_{i=1}^n X_i^2\right)-g\left(\mathbb{E}\left(X^2\right)\right)\right) \stackrel{d}{\rightarrow} N\left(0, \frac{5}{4} \sigma^2\right) .
$$
Attempt
I will futher state "method I" and "method II" for the soluton of the according problem. In the last problem, the problem is solved by "plugging in" the second order method of moments equation and then applying the delta method. My observation is that the latter is MME and therefore requires a different treatment to find the limiting distribution. I am not sure about this, since I only encountered one case with a MME.
Questions:
- Is it correct to apply method II when finding the limiting distribution of a MME?
- Is it correct to apply method I when finding the limiting distribution of a MLE?
- If else, how to determine when to apply method I or method II?
- When is it necesarry to apply the delta method?
Best Answer
That means not it is not specifically tied to the method of momement estimator