# Limiting distribution of a ratio using Basu's theorem

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.

• the scaling of the sum is incorrect... and the call to Basu's theorem does not seem right, since $\sum_{i=1}^{n}X_{i}/\theta$ is not a statistic, being also a function of $\theta$. – Xi'an Nov 19 '18 at 19:10
• What do you mean the scaling of the sum is incorrect? I don't see my mistake, I'm sorry. You have a good point as for the call of the Basu's theorem, $\sum_{i=1}^{n} X_i/\theta$ can't be a statistic since it depends on $\theta$. I'm quite lost on what I should do. How would go about computing the following expectation: $E(\sum_{i=1}^{n}X_i/X_{n:n}).$ I used a similar method than in my question in order to be able to use Basu's theorem. – Ranir123 Nov 19 '18 at 19:28

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 $$\frac{\sum_{i=1}^n (2X_i - \theta)}{\sqrt{n}\,X_{n:n}}$$
• Thank you for your response. I'm sorry but I don't understand how this solves the problem. I understand that the CLT implies the convergence in distribution of $\sqrt{n}\left(\bar{X}_{n} - \theta/2\right)$ to a gaussian of zero mean and variance of $\theta^{2}/12$. What I don't understand is how to use the CLT in my problem. For example: $\frac{\left(2 \sum_{i=1}^{n}X_i - \theta \right)}{\sqrt{n}X_{n:n}} = ... = \frac{\sqrt{n}2\left( \bar{X}_{n} - \theta/2n\right)}{X_{n:n}}$. My issue is with the term $\theta/2n$, I'm not able to get only $\theta/2$. – Ranir123 Nov 19 '18 at 22:54