# ISI MStat PSB 2009 Problem 3 | Gamma is not abNormal

This is a very simple but beautiful sample problem from ISI MStat PSB 2009 Problem 3. It is based on recognizing density function and then using CLT. Try it !

## Problem- ISI MStat PSB 2009 Problem 3

Using and appropriate probability distribution or otherwise show that,

$$\lim\limits_{x\to\infty}\int^n_0 \frac{exp(-x)x^{n-1}}{(n-1)!}\,dx =\frac{1}{2}$$.

### Prerequisites

Gamma Distribution

Central Limit Theorem

Normal Distribution

## Solution :

Here all we need is to recognize the structure of the integrand. Look, that here, the integrand is integrated over the non-negative real numbers. Now, event though here it is not mentioned explicitly that $$x$$ is a random variable, we can assume $$x$$ to be some value taken by a random variable $$X$$. After all we can find randomness anywhere and everywhere !!

Now observe that the integrand has a structure which is very identical to the density function of gamma random variable with parameters $$1$$ ande $$n$$. So, if we assume that $$X$$ is a $$Gamma(1, n)$$, then our limiting integral transforms to,

$$\lim\limits_{x\to\infty}P(X \le n)$$.

Now, we know that if $$X \sim Gamma(1,n)$$, then its mean and variance both are $$n$$.

So, as $$n \uparrow \infty$$, $$\frac{X-n}{\sqrt{n}} \to N(0,1)$$, by Central Limit Theorem.

Hence, $$\lim\limits_{x\to\infty}P(X \le n)=\lim\limits_{x\to\infty}P(\frac{X-n}{\sqrt{n}} \le 0)=\lim\limits_{x\to\infty}\Phi (0)=\frac{1}{2}$$. [ here $$\Phi(z)$$ is the cdf of Normal at $$z$$.]

Hence proved !!

## Food For Thought

Can, you do the proof under the "Otherwise" condition !!

Give it a try !!