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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 !

Using and appropriate probability distribution or otherwise show that,

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

Gamma Distribution

Central Limit Theorem

Normal Distribution

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 !!

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

Give it a try !!

Content

[hide]

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 !

Using and appropriate probability distribution or otherwise show that,

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

Gamma Distribution

Central Limit Theorem

Normal Distribution

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 !!

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

Give it a try !!

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