Question

1. 20 points Let X be a random variable with the following probability density function: f(x)--e+1" with ? > 0, ? > 0, constants x > ?, (a) 5 points Find the value of constant c that makes f(x) a valid probability mass function. (b) 5 points Find the cumulative distribution function (CDF) of X.

Answer 1

1)

a)

For f(x) to be probability function, we should have $\int f(x)dx=1$ .

$\int_{\alpha}^{\infty}\frac{c}{x^{\theta+1}}dx=1\Rightarrow c\int_{\alpha}^{\infty}x^{-\theta-1 }dx=1\Rightarrow c[\frac{x^{-\theta }}{-\theta}]_{\alpha}^{\infty}=1$

$\Rightarrow \frac{c}{-\theta}[\frac{1}{x^{\theta}}]_{\alpha}^{\infty}=1\Rightarrow \frac{c}{-\theta}[0-\frac{1}{\alpha^{\theta}}]=1\Rightarrow c=\theta \alpha^{\theta}$

$f(x)=\frac{\theta \alpha^{\theta}}{x^{\theta+1}},x>\alpha$

b)

The cummulative distribution function of X is given by F(x) = P(X<x)

$F(x)=\int_{\alpha}^{x}\frac{\theta \alpha^{\theta}}{x^{\theta+1}}dx=\theta \alpha^{\theta}\int_{\alpha}^{x}x^{-\theta-1}dx=\theta \alpha^{\theta}[\frac{x^{-\theta}}{-\theta}]_{\alpha}^{x}=-\alpha^{\theta}[\frac{1}{x^{\theta}}]_{\alpha}^{x}=-\alpha^{\theta}[\frac{1}{x^{\theta}}-\frac{1}{\alpha^{\theta}}]=1-(\frac{\alpha}{x})^{\theta}$

c)

$P(\beta <X <\lambda \beta)=\int_{\beta}^{\lambda \beta}\frac{\theta \alpha^{\theta}}{x^{\theta+1}}dx$

$=\theta \alpha^{\theta}\int_{\beta}^{\lambda \beta}x^{-\theta-1}dx=\theta \alpha^{\theta}[\frac{x^{-\theta}}{-\theta}]_{\beta}^{\lambda \beta}=-\alpha^{\theta}[\frac{1}{x^{\theta}}]_{\beta}^{\lambda \beta}=-\alpha^{\theta}[\frac{1}{(\lambda \beta)^{\theta}}-\frac{1}{\beta^{\theta}}]=(\frac{\alpha}{\beta})^{\theta}[1-\frac{1}{\lambda}]$

d)

$E(X)=\int_{\alpha}^{\infty}xf(x)dx=\int_{\alpha}^{\infty}x\frac{\theta \alpha^{\theta}}{x^{\theta+1}}dx=\theta \alpha^{\theta}\int_{\alpha}^{\infty}x^{-\theta}dx=\theta \alpha^{\theta}[\frac{x^{-\theta+1}}{-\theta+1}]_{\alpha}^{\infty}=(\frac{\theta}{-\theta +1})\alpha^{\theta}[\frac{1}{x^{\theta-1}}]_{\alpha}^{\infty}=(\frac{\theta}{-\theta +1})\alpha^{\theta}[0-\frac{1}{\alpha^{\theta-1}}]=\frac{\theta}{\theta-1}\alpha$