Determine the Asymptotic Distribution of the Method of Moments Estimator of $\theta$, $\tilde{\theta}$

Question

I am having difficulty understanding what it means to find the asymptotic distribution of a statistic. I have the correct answer (as far as I know), but I am unconvinced that I understand the process of finding the asymptotic dist.

Determine the Asymptotic Distribution of the MME of $\theta$, $\tilde{\theta}$

$X_1...X_n$ is a randomly selected sample from $f(x,\theta)= 4\dfrac{\theta^4}{x^5}$, for $x>\theta$.

# Finding the MME is relatively simple. $$ E(x)= \int_{\theta}^{\infty}{\dfrac{4\theta^4}{x^5} x \ dx }= \dfrac{4\theta}{3} = \dfrac{\sum{x_i}}{n}= \bar{x}$$ $$\dfrac{4\theta}{3}=\bar{x}, \Longrightarrow \tilde{\theta}=\dfrac{3\bar{x}}{4}$$ Unfortunately I am struggling to understand the process of finding the asymptotic distribution. Here is my work, the solution was given. $$E\bigg(\dfrac{3x}{4}\bigg)= \dfrac{3}{4} E(x)=\theta$$ $$Var\bigg(\dfrac{3x}{4}\bigg)=\dfrac{9}{16}Var(x)$$ $$Var(x)=E(x^2)-[E(x)]^2 $$ $$E(x^2)= \int_{\theta}^{\infty}{\dfrac{4\theta^4}{x^5} x^2 \ dx}= 2\theta^2 $$ $$Var(x)= \dfrac{16 \theta^2}{9}-2\theta^2= \dfrac{-2\theta^2}{9} $$ $$Var\bigg(\dfrac{3x}{4}\bigg)= \dfrac{\theta^2}{8}$$ The answer given is... $$N\sim \bigg(\theta, \ \dfrac{\theta^2}{8n}\bigg)$$

So I had a few questions...

1. If I were doing this on my own I would have calculated $E\bigg(\dfrac{3\bar{x}}{4}\bigg)$ instead of $E\bigg(\dfrac{3x}{4}\bigg)$ (same goes with the variance), why just $x$?

2. Is it safe to assume the process of finding the asymptotic distribution of an estimator involves finding the expectation and variance of said estimator? Why is it in this case that I can assume normality? What would I do if I couldn't assume normality?

3. In the denominator of the variance of the normalized distribution, where did the multiple of $n$ come from?

Thanks so much for taking the time to look at my question, I appreciate any feedback and help.

Answer 1

1. It is easier to calculate the mean and variance of $\tilde \theta$ for a single observation $X_1$ ($n = 1$): ${\rm E}[\tilde \theta] = {\rm E}\left[\frac{3X_1}{4}\right] = \theta$. Notice your variance is obviously incorrect, because it cannot be negative! The correct value is ${\rm Var}[\tilde \theta] = \frac{\theta^2}{8}.$ If we had calculated the expectation and variance for $\tilde \theta$ for some general sample size $n$, then we would find ${\rm E}[\tilde \theta] = \theta$ as before, but now ${\rm Var}[\tilde\theta] = \frac{\theta^2}{8n}$. We can also see this because we recall that if $X_1, \ldots, X_n$ are IID with variance $\sigma^2$, then ${\rm Var}[\bar X] = \frac{1}{n^2}{\rm Var}[\sum_i X_i] = \frac{1}{n^2}\sum_i {\rm Var}[X_i] = \frac{\sigma^2}{n}.$ This should make intuitive sense: the variance of the MME of $\theta$ should decrease with larger sample size, because you have more information from which to calculate an estimator, thus the estimator will be more precise.

2. The reason why we only calculated $\tilde \theta$ for a single observation, and why we can assume asymptotic normality, is because we can use the central limit theorem, which loosely states that if $X_1, X_2, \ldots, X_n$ are IID from a 'well behaved' distribution with mean $\mu$ and variance $\sigma^2$, then $\bar X = \frac{1}{n} \sum_i X_i \sim {\rm Normal}(\mu, \frac{\sigma^2}{n})$.

3. The multiple of $n$ comes from the CLT: from parts 1 and 2 above, the mean of $\tilde \theta$ is $\mu = \theta$ (i.e., $\tilde \theta$ is an unbiased estimator of $\theta$), and the variance of $\tilde \theta$ is $\sigma^2 = \frac{\theta^2}{8}$. The rest immediately follows. Alternatively, we could simply use the mean and variance of the MME for general $n$ that we calculated above: notice it gives the same result. But even with this latter approach, we still need to appeal to the CLT because without it, all we can say is the mean and variance of the MME are as given above--we would not be able to say that the asymptotic distribution of $\tilde \theta$ is normal.