I'm trying to show that as $\alpha$ tends to 0, the gamma distribution $$\Gamma(\lambda,\alpha),$$ is properly standardised, tends to the standard normal distribution. I have figured out that the moment generating function for the gamma distribution is $$\left(\frac{\lambda}{\lambda-t}\right)^\alpha.$$ Also, I've worked out that the mean and variance of a gamma random variable is $$\frac{\alpha}{\lambda}$$ and $$\frac{\alpha}{\lambda^2}$$ respectively.

However, I am not sure how to proceed further. I tried by defining $$Z=\frac{X-\frac{\alpha}{\lambda}}{\frac{\alpha^0.5}{\lambda}}$$ and using the fact that $M_{_Z}(t) = e^{bt}M_{X}(at)$

However, I can't show that $$M_{_Z}(t)=e^{t^2/2}$$ which is the moment generating function of a standard normal random variable. Is this the correct way to proceed?

  • 1
    $\begingroup$ It seems you're using the notation $t$ for two distinct quantities. Take a close look, for example, at your second display equation. $\endgroup$ – cardinal Jun 22 '11 at 22:28
  • $\begingroup$ The displayed function is not the mgf of anything I recognize. It is a really really bad idea to use $t$ as a parameter for the distribution and simultaneously as the variable of the mgf! If you write down the right mgf, scale it which is easy, the calculation will be a straightforward limit. $\endgroup$ – André Nicolas Jun 22 '11 at 22:46
  • $\begingroup$ Two comments: you wrote $\alpha\to0$ but everybody below (and I) assumed you meant $\alpha\to+\infty$ instead. // The CLT argument in 6312's answer works for integer valued shapes (true, it can then be extended to non integer shapes but this is messy) while the MGF argument in Robert's answer works directly for all real valued shapes (despite the choice of notation $k$, which might seem to indicate that the shape should be an integer, this is not so). $\endgroup$ – Did Jun 23 '11 at 6:07

First of all, you seem to be using $t$ for two different purposes: a parameter of the Gamma distribution and the variable in the moment generating function. These should be completely different. The Gamma distribution with shape parameter $k$ and rate parameter $r$ has mean $\mu = k/r$, variance $\sigma^2 = k/r^2$, and moment generating function $M_X(t) = \left(\frac{r}{r-t}\right)^k$. The limit you should be taking is $k \to \infty$ with $r$ fixed. The MGF of the scaled and translated variable $Y = (X-\mu)/\sigma$ is then $M_Y(t) = \left( 1 - \frac{t}{\sqrt{k}}\right)^{-k} e^{-\sqrt{k} t}$. I suggest taking logarithms, and using a degree-2 Taylor expansion of $\ln(1 - t/\sqrt{k})$.

  • $\begingroup$ @cardinal and @user6312: Thanks for pointing out the confusion in the notation. I realized I mixed up the parameter t in the gamma distribution with that for the moment-generating function. I have edited the parameters to make the notation clear. $\endgroup$ – JP83 Jun 23 '11 at 3:55

Your post starts by saying that you want to show that the Gamma distribution, when scaled properly, tends to the standard normal. If you need to prove this by using an explicit moment generating function argument, then you should follow the procedure described by Robert Israel.

If, however, you can use standard facts, then you will know that a Gamma is a sum of independent exponentials with the same parameter, and then all one needs to do is to quote the Central Limit Theorem. Indeed, if a moment generating function argument is to be used, the general argument is really not very different from the specific argument for the Gamma distribution.

  • $\begingroup$ @Israel, thanks for the heads-up. I managed to prove the convergence to the standard normal based on your advice after clearing up the notation. Thanks a lot everyone! $\endgroup$ – JP83 Jun 23 '11 at 3:58

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