# Estimate Gamma distribution using central limit theorem

The random variable X has a Gamma distribution with parameters $\alpha$ = na, and $\beta$ = b , $a, b > 0$, and n a natural number different from zero. What is the probability that X is strictly greater than $(n+1)ab$ for n big enough ?

i start with: $$X_i > (n+1)ab$$ so: $$X_i -nab > ab$$

now divide by standard deviation of Gamma distribution:$$\frac{X_i -nab}{\sqrt {nab^2} } > \frac{a}{\sqrt {na}}$$

and now take the sum over all $X_i$'s: $$\sum_{i=0}^n \frac{X_i -nab}{\sqrt {nab^2} } > \frac{na}{\sqrt {na}}$$

and for last i multiply $\frac{1}{\sqrt n}$ with the term above: $$\frac{1}{\sqrt n}\sum_{i=0}^n \frac{X_i -nab}{\sqrt {nab^2} } > \frac{na}{\sqrt {a}n}$$

or equivalently:$$\sum_{i=0}^n \frac{X_i -nab}{\sqrt {a} nb } > \frac{a}{\sqrt {a}}$$

so i get exactly the structure of central limit theorem by manipulating the fact that $X_i > (n+1)ab$ that says it is equal to standard normal distribution the left hand side is now standard normal distributed , but clearly i make a mistake some where but i don't see it! Can someone please explain why my method doesn't work ?

• "and now take the sum over all X's" Sorry but what does that mean? – Did Sep 2 '17 at 7:37
• @Did i edited the question up a bit , by adding indices 'i' so now it make sense to take sum over all $X_i's$ – Sam Farjamirad Sep 2 '17 at 8:08
• Except that $$X_i-nab>ab$$ with $X_i$ Gamma$(a,b)$, does not interest you... – Did Sep 2 '17 at 8:11

## 1 Answer

I don't quite understand your question, but I'll give it a shot.

Let $X\sim\Gamma(na,b)$, $n\in\mathbb{N}_{>1};a,b\in\mathbb{R}_{>0}$. Now define iid $X_i\sim\Gamma(a,b)$, so that $$X=\sum_{i=1}^n X_i$$ First, let $\mu={a}{b}$ and $\sigma^2=ab^2$ (assuming shape-scale, not shape-rate, parametrization). Then, by the central limit theorem, $$Z = \frac{X - n\mu}{\sigma\sqrt{n}}$$ converges in distribution to the standard normal, i.e. $$Z=\frac{X - nab}{b\sqrt{an}} \xrightarrow[\;\;\;]{\text{d}} \mathcal{N}\left(0,1\right)$$ However, it appears you are interested in the fact that the distribution of $X/n$ is close to $\mathcal{N}(\mu,\sigma^2/n)$ for large enough $n$ (verbatim from Wiki). Please correct me if I'm misunderstanding.

So, your original question asks for $$p=P(X>(n+1)ab)$$ Now, for large $n$, we assume $X\sim\mathcal{N}(n\mu,n\sigma^2)$ and so we get: $$p=P\left( \frac{X - nab}{b\sqrt{an}} > \frac{(n+1)ab-nab}{b\sqrt{an}} \right) \approx P\left( Z > \frac{\sqrt{a}}{\sqrt{n}} \right) = 1 - \Phi\left(\sqrt{\frac{a}{n}}\,\right)$$ where $\Phi$ is the normal cdf.

• Thanks, you got it right , but why you didn't multiply Z (third equation to the last by $\frac {1}{\sqrt n}$) and i think 'a' should be over n instead of b at the last equation, at the end. am i right ? – Sam Farjamirad Sep 2 '17 at 7:07
• FYI, the assertion that $$\frac{1}{n}X \xrightarrow[\;\;\;]{\text{d}} \mathcal{N}\left(\mu,\frac{\sigma^2}{n}\right)$$ is meaningless. Every convergence statement $$x_n\to x$$ requires that the limit $x$ does not depend on $n$. Same remark regarding $$X \xrightarrow[\;\;\;]{\text{d}} \mathcal{N}\left(n\mu,n{\sigma^2}\right)$$ – Did Sep 2 '17 at 7:35
• Finally, the approach you suggest, once corrected, does not lead to the limit $$1 - \Phi\left(\sqrt{\frac{a}{b}}\,\right)$$ but to the limit \frac12$– Did Sep 2 '17 at 7:40 • @upvoter Please explain your vote. – Did Sep 2 '17 at 7:40 • Dear @Did, thanks for your comments. I agree that the "convergence" notation is misleading. All I wanted to say was that$X$approximately follows that distribution for large enough finite$n$, and did not mean a limit to infinity. What notation do you suggest? Or am I strictly incorrect in substance as well as notation? As for the final limit being$1/2$, I think we are understanding the question differently (and indeed it is not optimally worded). The OP, I think, assumes a large but finite$n$, hence it does not disappear in the final expression. (Note: I will correct the$b$to an$n\$) – user3658307 Sep 2 '17 at 16:42