asymptotic distribution of a statistic 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.
 A: 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?
A: 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.
