We are given a gamma random $X\sim\text{Gamma}(\alpha,\beta)$ where $\beta=\alpha/\mu$. Then $\mathbb{E}[X]=\mu$ and the pdf of $X$ is \begin{equation} f_{X}(x)=\frac{(\alpha/\mu)^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\alpha x/\mu}. \end{equation}

What is $\lim_{\alpha\to\infty}f_{X}(x)$?

By central limit theorem, I suspect the solution is \begin{equation} \lim_{\alpha\to\infty}f_{X}(x)=\delta(x-\mu). \end{equation}

  • $\begingroup$ Look at the variance as a function of $\alpha$ and $\mu$. $\endgroup$ – heropup Dec 19 '17 at 16:00
  • $\begingroup$ @heropup Right. $\mathrm{var}[X]=\mu^{2}/\alpha$ so as $\alpha\to\infty$ we end up with a r.v. with $\mathbb{E}[X]=\mu$ and $\mathrm{var}[X]=0$ which means $X\sim\delta(x-\mu)$. I was more interested in solving the limit directly, i.e. computing $\lim_{\alpha\to\infty}f_{X}(x)$. $\endgroup$ – Aaron Hendrickson Dec 19 '17 at 16:04

\begin{equation} f_{X}(x)=\frac{(\alpha/\mu)^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\alpha x/\mu}\Rightarrow \ln f_X(x)=\alpha\ln(\alpha/\mu)-\ln\Gamma(\alpha)+(\alpha-1)\ln x-\frac{\alpha x}{\mu}\ . \end{equation} Use Stirling's approximation for the Gamma function $$ \ln\Gamma(\alpha)\sim (\alpha-1/2)\ln\alpha-\alpha+\ln(2\pi)/2 $$ to conclude that $$ \ln f_X(x)\sim \alpha(1+\ln (x/\mu)-x/\mu)+(1/2)\ln\alpha+\ldots $$ Therefore, all depends on the combination $1+\ln(x/\mu)-x/\mu$. If it is $=0$ (which happens when $x=\mu$), then $\ln f_X(x)\sim (1/2)\ln\alpha\to\infty$ as $\alpha\to\infty$. If it is not equal to zero, then we know it is always negative (let's call this coefficient $-|\gamma|$), therefore $f_X(x)\sim e^{-a |\gamma|}\to 0$. So indeed $f_X(x)=\delta(x-\mu)$ (which is $=\infty$ if $x=\mu$ and $0$ otherwise).


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