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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 !!


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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 !!


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


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