I came across this 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 $n\bar{X}_n= X_1+X_2+\cdots+X_n$. I need to prove that $$T_n\stackrel{\text{dist}}{\stackrel{n\to\infty}{\longrightarrow}}N(0,{1}/{\mu})$$ 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{\mu\sqrt{n}}{\sqrt{n}-t}) \end{align*}

Now, I should do the integration $\int_0^\frac{\mu\sqrt{n}}{\sqrt{n}-t}$ 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 $T_n\stackrel{\text{d}}{\longrightarrow}Normal$, not finding the distribution of $\bar{X}_n$.

Any help will be appreciated!

  • $\begingroup$ @Xi'an Is it $Ga(\mu,n)$? $\endgroup$
    – iTurki
    Dec 27, 2014 at 20:24
  • $\begingroup$ @Xi'an Can you clarify, please? How to check if it applies or not? $\endgroup$
    – iTurki
    Dec 27, 2014 at 20:27
  • $\begingroup$ @Xi'an I came to the conclusion that to integral is not solvable, or it won't help me do the next step. CLT is close to what I have, but not the same. The problem is the $\bar{X}$ in the dominator. $\endgroup$
    – iTurki
    Dec 27, 2014 at 20:48
  • $\begingroup$ @Xi'an Sorry, but I still can't see how Slutsky's theorem would help. The theorem assumes one variable to converge to a constant. In my case, both nominator and dominator should converge to a variable. $\endgroup$
    – iTurki
    Dec 27, 2014 at 20:57
  • $\begingroup$ @Xi'an That's what I meant in my previous comment. Both of them should converge to a variable. Not a variable and a constant, as stated by Slutsky's theorem. $\endgroup$
    – iTurki
    Dec 27, 2014 at 21:10

1 Answer 1


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, $$\sqrt{n}(\bar{X}_n-\mu)\stackrel{\text{dist}}{\stackrel{n\to\infty}{\longrightarrow}}\mathcal{N}(0,\text{var}(X_i))$$ and here $\text{var}(X_i)=\mu$, hence $$\sqrt{n}(\bar{X}_n-\mu)\stackrel{\text{dist}}{\stackrel{n\to\infty}{\longrightarrow}}\mathcal{N}(0,\mu)\,.$$ 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, $$\dfrac{\sqrt{n}(\bar{X}_n-\mu)}{\bar{X}_n}\stackrel{\text{dist}}{\stackrel{n\to\infty}{\longrightarrow}}\mathcal{N}(0/\mu,\mu/\mu^2) $$i.e. $$\dfrac{\sqrt{n}(\bar{X}_n-\mu)}{\bar{X}_n}\stackrel{\text{dist}}{\stackrel{n\to\infty}{\longrightarrow}}\mathcal{N}(0,1/\mu)$$ Q.E.D.

  • $\begingroup$ Thanks for the answer. I though $\bar{X}$ will converge to a variable. I was Wrong. Thanks again! $\endgroup$
    – iTurki
    Dec 27, 2014 at 21:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.