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# 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 !!

## Subscribe to Cheenta at Youtube

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 !!

## Subscribe to Cheenta at Youtube

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