Limiting distribution of a ratio using Basu's theorem

Question

Edit: there's seems to be a typo in original question.

This is a past exam question that I'm trying to solve. Suppose that $X_1,\ldots, X_n$ are i.i.d. Uniform (0, $\theta$) random variables. Let $X_{n:n} = \max_{1 \leq i \leq n} X_i.$

Find the limiting distribution of: $$\frac{2\sum_{i=1}^n X_i - \theta}{\sqrt{n}X_{n:n}},$$ as $n \rightarrow \infty$. Here's what I've been able to do:

• 1) Can show that the limiting distribution of $X_{n:n}$ is $\theta$ by working on the CDF of $X_{n:n}$ as $n$ tends to infinity;
• 2) Using Basu's Theorem, I was able to show that $\sum_{i=1}^n X_i/\theta$ is independent of $X_{n:n}$ since the distribution of $\sum_{i=1}^n X_i/\theta$ is independent of $\theta$ and is thus an ancillary statistic. I was also able to show that $X_{n:n}$ was a boundedly complete sufficient statistic.

I can thus work on the initial term to get the following: $$\frac{2\sum_{i=1}^n X_i - \theta}{\sqrt{n}X_{n:n}} = \frac{\theta}{X_{n:n}}\frac{\left(2 \sum_{i=1}^n X_i/\theta - 1\right)}{\sqrt{n}}.$$ From 1), I know that $\theta/X_{n:n}$ converges in distribution to 1. But, for the second term $\frac{\left(2 \sum_{i=1}^n X_i/\theta - 1\right)}{\sqrt{n}}$, I'm not able to find a limiting distribution. I tried to use the CLT, but without any success. Is my work so far wrong? Thanks for your help.

Answer 1

If one applies the CLT to $$\bar{X}_n=\frac{1}{n}\sum_{i=1}^n X_i$$it implies that $$\sqrt{n}\left(\bar{X}_n-\frac{\theta}{2}\right)\longrightarrow \mathcal{N}(0,\theta^2/12)$$ Further,$$\frac{X_{n:n}}{\theta}\stackrel{\text{a.s.}}{\longrightarrow}1$$This should be enough to conclude. I believe the question as expressed $$\frac{2\sum_{i=1}^n X_i - \theta}{\sqrt{n}\,X_{n:n}}$$contains a typo and it should instead be $$\frac{\sum_{i=1}^n (2X_i - \theta)}{\sqrt{n}\,X_{n:n}}$$