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Say we have iid sample of size $n$ with $X_i \sim Exp(\lambda)$ and the task is to find asymptotic distribution of the statistic $$T_n := \frac{\bar{X}}{s}$$, where $s^2$ is the unbiased sample variance.

Now, it is easy to show that $E(\bar{X}) = E(X) =: \mu$, so (skipping some details) if we let $Y_i := X_i - \mu$, then $\frac{1}{\sqrt{n}} \sum_{k=1}^n Y_k \overset{D}{\to} N(0, \sigma^2_Y)$, where $\sigma^2_Y$ is the variance of $Y$

Now I know how to also show, that $s \overset{p}{\to} \sigma_X$, so that (?) $Var(T_n) \overset{p}{\to} 1$.

So if I'm correct up until here, it feels that well, we have an estimator $T_n$ with variance 1 so its asymptotic distribution should be standard normal. But then I haven't used the fact that the sample has exponential distribution.

Basically, I would be very grateful for tips/scheme on who I should approach/think/solve this problem as it would be the most beneficial in terms of learning.

Note: I know Slutsky theorem, Continuous Mapping Theorem and (classical) CLT.

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    $\begingroup$ I think it does not matter what distribution the sample has as long as the observations are i.i.d. The CLT which kicks in in the asymptotics does not assume a particular distribution of the underlying observations; i.i.d. is enough. $\endgroup$ Commented Mar 24, 2015 at 9:11
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    $\begingroup$ The only aspects of the exponential that matter are those that come into applying the CLT in some way. $\endgroup$
    – Glen_b
    Commented Mar 24, 2015 at 13:12
  • $\begingroup$ Are you allowed to use delta method? $\endgroup$
    – KOE
    Commented Mar 24, 2015 at 14:41
  • $\begingroup$ yes, forgot to mention that $\endgroup$
    – Sarunas
    Commented Mar 24, 2015 at 14:57
  • $\begingroup$ I've added a hint. Note also in your question that $\mathrm{Var}T_n$ is a sequence of numbers, so they converge as numbers, not in probability. $\endgroup$
    – KOE
    Commented Mar 24, 2015 at 15:09

2 Answers 2

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Here is a hint. Let me know if you need more and I'll expand.

Show that (in particular, make sure to check the assumptions; I didn't): $$ \sqrt{n}\left( (\bar{X}_n,S_n^2)' - (\mu, \nu)'\right)\overset{d}{\to}N(0,\Sigma), $$

for some constants $\mu,\nu$ and covariance matrix $\Sigma$. Consider then the function $g(x,y)=x/\sqrt{y}$. What can you say about $g(\bar{X}_n,S_n^2)$, and why?

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  • $\begingroup$ would I be correct in saying that I need some kind of 'multivariate' CLT to show that $ \sqrt{n}\left( (\bar{X}_n,S_n^2)' - (\mu, \nu)'\right)\overset{d}{\to}N(0,\Sigma), $ ? I realize that the rest is just trivial Delta method application. but I'm not sure on how to show this multivariate convergence in distribution $\endgroup$
    – Sarunas
    Commented Mar 24, 2015 at 17:37
  • $\begingroup$ @Sarunas yes you would need multivariate CLT for this. If you don't have that, this may not be the way to go. I'll see if I can update with something based on univariate results when I have time. $\endgroup$
    – KOE
    Commented Mar 25, 2015 at 1:54
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    $\begingroup$ I got it, thank you :) I think we can use multivariate CLT too, so that works. just noticed the professor has it in his notes. $\endgroup$
    – Sarunas
    Commented Mar 25, 2015 at 8:40
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As you note yourself, $\mathrm{Var}[T_n]$ becomes 1 in the limit, so $T_n$ becomes deterministically equal to $\mu/\sigma$ in the limit. For the exponential distribution, we have that $\mu=\sigma=1/\lambda$, so $T_n\to 1$.

But that is "the limit value" of $T_n$, i.e. for $n=\infty$. When you ask for "the asymptotic distribution", you probably want to know how $T_n$ is distributed for extremely large $n$ but still $n<\infty$. In that case, the CLT is of limited use and you will have to work out the distribution of $\bar{X}/s$ for general $n$ from scratch. The main difficulty here will most likely be the fact that $\bar{X}$ and $s$ are not independent random variables. Sample mean and sample variance are independent if and only if $X$ is normally distributed.

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    $\begingroup$ What does it mean to "become deterministically equal to in the limit"? If it means almost surely (or something even stronger), then what you say is generally not true. Moreover, CLT is perhaps the most common tool used to find asymptotic distributions. $\endgroup$
    – KOE
    Commented Mar 24, 2015 at 13:00
  • $\begingroup$ Yes, I mean almost sure convergence to the deterministic value 1. That is because the variance of $T_n$ approaches 0. Yes, CLT is the most common tool for asymptotic distributions but here it gives a strange answer, $T_n \overset{a.s.}{\to} 1$, that may not be very useful. If it is for you, so much the better. $\endgroup$ Commented Mar 24, 2015 at 13:08
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    $\begingroup$ The reason that the limit is degenerate is that the scaling $\sqrt{n}$ is missing. Under appropriate scaling, the limit is a random variable. $\endgroup$
    – MånsT
    Commented Mar 24, 2015 at 13:12
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    $\begingroup$ @StijnDeVuyst, variance approaching 0 does not imply almost sure convergence, that's what I'm getting at. $\endgroup$
    – KOE
    Commented Mar 24, 2015 at 13:42
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    $\begingroup$ I'm sorry if you feel attacked, that's not anyone's intention I dare to say. $\endgroup$
    – KOE
    Commented Mar 24, 2015 at 15:42

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