Given $\bar{X_n}$ is mean of random sample with size $n$ from Gamma distribution with parameter $\alpha=\mu$ and $\beta=1$. I wanna find the limiting distribution of $\frac{\sqrt{n}\left(\bar{X_n}-\mu\right)}{\sqrt{\bar{X_n}}}$.

Can I use the Central Limit Theorem for doing this? Because it is $\sqrt{\bar{X_n}}$ in the denominator, not $\sqrt{S^2}$.

I am doing this problem with this method :

Consider that the Moment Generating Function (MGF) of $\bar{X_n}$ is $M_\bar{x}\left(t\right)=\left(1-\frac{t}{n}\right)^{-n\mu}$ for $t<n$. So that, the pdf of $\bar{X_n}$ is

$f_n\left(\bar{x}\right)=\frac{1}{\Gamma\left(n\mu\right)\left(\frac{1}{n}\right)^{n\mu}}\bar{x}^{n\mu-1}e^{-n\bar{x}}$ for $0<\bar{x}<\infty$

From the MGF, we know that the mean of $\bar{X_n}$ is $\mu$ and the variance of $\bar{X_n}$ is $\frac{\mu}{n^2}$.

Then, consider the transformation $Y_n=\frac{\sqrt{n}\left(\bar{X_n}-\mu\right)}{\sqrt{\bar{X_n}}}$ so that,


Then I get stuck to define $\bar{x}$ in function of $y$ to find pdf and CDF of $Y$

Any simpler method? Or I do mistake when I define the transformation? Any ideas?


1 Answer 1


Just write

\begin{align} \frac{\sqrt{n} (\bar{X} - \mu)}{\sqrt{\bar{X}}} &= \frac{\sqrt{n} (\bar{X} - \mu)}{\sqrt{\mu}} \cdot \sqrt{\frac{\mu}{\bar{X}}} . \end{align}

By the central limit theorem the first term has a limiting normal$(0, 1)$ distribution and we also know $\bar{X} \to \mu$ with probability one by the strong law of large numbers, so the second factor converges to one almost surely. Then appealing to Slutsky's theorem the entire thing goes to a normal$(0, 1)$.

  • $\begingroup$ OH I GOT IT. Thank you very much for the concept. ^^ $\endgroup$
    – MusMeong
    Jan 29, 2016 at 15:31

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