I came across this problem:

**Problem**

If I have $X_1, X_2, …, X_n$ $n$ iid random variables which pdf is

$$

f_X(x) = begin{cases} dfrac{x^{mu-1} e^{-x}}{Gamma{(mu)}} &0<x<infty, \0&text{elsewhere}end{cases}

$$

namely a $text{Gamma}(alpha=mu, beta=1)$ distribution. Let $nbar{X}_n= X_1+X_2+cdots+X_n$. I need to prove that $$T_nstackrel{text{dist}}{stackrel{ntoinfty}{longrightarrow}}N(0,{1}/{mu})$$ where

$$T_n= frac{sqrt{n}(bar{X}_n-mu)}{bar{X}_n}$$

**My Solution:**

I thought of a possible solution in two steps:

- First, we need to find the pdf of $bar{X}_n$ and then of $T_n$.
- Then we take the limit of it and if we get a Normal distribution then, we solved the question.

begin{align*}

F_T(t) &= P(T le t)

\&

= P(frac{sqrt{n}(bar{X}_n-mu)}{bar{X}_n} le t)

\&

= P(frac{(bar{X}_n-mu)}{bar{X}} le frac{t}{sqrt{n}})

\&

= P(1- frac{mu}{bar{X}_n} le frac{t}{sqrt{n}})

\&

= P(frac{mu}{bar{X}_n} ge 1- frac{t}{sqrt{n}})

\&

= P(frac{bar{X}_n}{mu} le frac{sqrt{n}}{sqrt{n}-t})

\&

= P(bar{X}_n le frac{musqrt{n}}{sqrt{n}-t})

end{align*}

Now, I should do the integration $int_0^frac{musqrt{n}}{sqrt{n}-t}$ of the pdf of $bar{X}_n$. But it is not the same distribution as $X_i$. It is something else. This is where I stuck in my solution.

To clarify: *My goal is to prove $T_nstackrel{text{d}}{longrightarrow}Normal$, not finding the distribution of $bar{X}_n$.*

Any help will be appreciated!

**Contents**hide

#### Best Answer

This problem is a direct application of Slutsky's theorem:

If $X_n$ converges in distribution to a random element $X$ and if $Y_n$ converges in probability to a constant $c$, then $X_n/Y_n$ converges in distribution to $X/c$.

In your setting, the theorem numerator "$X_n$" is replaced with $sqrt{n}(bar{X}_n-mu)$, since, by virtue of the Central Limit Theorem, $$sqrt{n}(bar{X}_n-mu)stackrel{text{dist}}{stackrel{ntoinfty}{longrightarrow}}mathcal{N}(0,text{var}(X_i))$$ and here $text{var}(X_i)=mu$, hence $$sqrt{n}(bar{X}_n-mu)stackrel{text{dist}}{stackrel{ntoinfty}{longrightarrow}}mathcal{N}(0,mu),.$$ And the theorem denominator "$Y_n$" is replaced with $bar{X}_n$ which converges in probability to its expectation $mu$ by the weak law of large numbers. Hence, $$dfrac{sqrt{n}(bar{X}_n-mu)}{bar{X}_n}stackrel{text{dist}}{stackrel{ntoinfty}{longrightarrow}}mathcal{N}(0/mu,mu/mu^2) $$i.e. $$dfrac{sqrt{n}(bar{X}_n-mu)}{bar{X}_n}stackrel{text{dist}}{stackrel{ntoinfty}{longrightarrow}}mathcal{N}(0,1/mu)$$ *Q.E.D.*

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