My question is relatively simple: what is the limiting distribution of the sample mean? But there are some technicalities I want to discuss.
context: I was asked this problem in an exam, and I feel my answer is correct. Professor doesn't think so and gives me a roundabout explanation that doesn't really address the issue.
By the Weak Law of Large Numbers, we know that $\bar{X}$ converges to $\mu$ in distribution, where $\bar{X}$ denotes the sample mean and $\mu$ denotes the true mean.
I was asked to find the limiting distribution of $\bar{X}$. I used the idea that convergence in probability to a constant $\mu$ implies convergence in distribution to that constant. So, I specified the limiting distribution of $\bar{X}$ as:
$F_{\bar{X}}(\bar{X} \leq x) = 1$ if $x \geq \mu$ and 0 otherwise.
The answer I was expected to give was:
$\sqrt{n}(\bar{X}-\mu) \to N(0, \sigma^2)$ in distribution.
My problem is that this isn't the limiting distribution of $\bar{X}$ itself — it's the limiting distribution of a function of $\bar{X}$. Am I correct in stating that $\bar{X}$ converges to $\mu$ in distribution, or did I miss the point of what a limiting distribution is?
Thanks in advance.
Best Answer
You are correct that convergence in probability implies convergence in distribution as a weaker property. If the sample mean $\bar{X} \rightarrow_p \mu$ by the WLLN we know that $\bar{X} \rightarrow_d $ a constant. A different way to frame a similar question is to say, what is an approximating distribution of $\bar{X}_n$ ($n$ being the sample size in question). Then it would be right to say $\bar{X}_n \dot{\sim} \mathcal{N} \left( \mu, \sigma^2/n \right)$
I think it's sloppy notation and the professor should have been clearer. In fact, in my theory classes, our professor had the deepest ire for what he considered a serious deficiency of understanding if students found limiting distributions that were functions of the $n$.