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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 $n\bar{X}_n= X_1+X_2+\cdots+X_n$. 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!

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  • $\begingroup$ @Xi'an Is it $Ga(\mu,n)$? $\endgroup$ – iTurki Dec 27 '14 at 20:24
  • $\begingroup$ @Xi'an Can you clarify, please? How to check if it applies or not? $\endgroup$ – iTurki Dec 27 '14 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 '14 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 '14 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 '14 at 21:10
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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.

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  • $\begingroup$ Thanks for the answer. I though $\bar{X}$ will converge to a variable. I was Wrong. Thanks again! $\endgroup$ – iTurki Dec 27 '14 at 21:28

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