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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.

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    $\begingroup$ 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 $\theta$. $\endgroup$ – Xi'an Nov 19 '18 at 19:10
  • $\begingroup$ 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. $\endgroup$ – Ranir123 Nov 19 '18 at 19:28
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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 $$\frac{\sum_{i=1}^n (2X_i - \theta)}{\sqrt{n}\,X_{n:n}}$$

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    $\begingroup$ 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 $\theta/2$. $\endgroup$ – Ranir123 Nov 19 '18 at 22:54
  • $\begingroup$ Your expression as given diverges almost surely. $\endgroup$ – Xi'an Nov 20 '18 at 5:34
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    $\begingroup$ Great thank you very much, this confirms my belief that there's indeed a typo in the question, or else it's not possible to use the CLT. $\endgroup$ – Ranir123 Nov 20 '18 at 20:57

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