Q. 5.14 - A First Course in Probability

Question

Suppose that the cumulative distribution function of the random variable $X$ is given by

$F\left(x\right)=1-e^{-x^2}x>0$

Evaluate $\text{ (a) }P[X>2];\text{ (b) }P[1<X<3)$; (c) the hazard rate function of $F_i\text{ (d) }E\left[X\right]\text{; (e) }\mathrm{Var}\left(X\right)$.

Hint: For parts $\left(d\right)$ and $\left(e\right)$, you might want to make use of the results of Theoretical Exercise  $5.5$.

Step-by-Step Solution

Answer

(a) The cumulative distribution function of  $P(X>2)=e^{-4}$

(b) The cumulative distribution function of random variables is $P(1<X<3)=e^{-1}-e^{-9}$

(c) The cumulative distribution function of hazard rate function  is ${\mu }_X\left(x\right)=2x$

(d)  $EX=\frac{\sqrt{\pi }}{2}$

(e) $\mathrm{Var}\left(X\right)=\frac{3\pi }{4}$

Step :1 The cumulative distribution of the function P [ X &#62; 2 ] (part a)

The cumulative distribution of the function $P[X>2]$

$$\begin{gathered} P(X>2)=1-P(X\le 2) \\ =1-F_X\left(2\right) \\ =1-\left(1-e^{-2^2}\right) \\ P(X>2)=e^{-4} \end{gathered}$$

Step :2 The cumulative distribution of the function P [ 1 &#60; X &#60; 3 ) (part b)

$$\begin{gathered} P(1<X<3)=P(X<3)-P(X<1) \\ =F_X\left(3\right)-F_X\left(1\right) \\ =e^{-1}-e^{-9} \\ P(1<X<3)=e^{-1} e^{-9} \end{gathered}$$

Step :3 The hazard rate of function (part c)

We have that, 

$$\begin{gathered} f_X\left(x\right)=\frac{d}{dx} \\ {F}_X\left(x\right)=\frac{d}{dx}\left(1-e^{-x^2}\right) \\ =2x{e}^{-x^2} \\ {F}_X\left(X\right)=2x{e}^{-x^2} \end{gathered}$$

Here the hazard function 

$$\begin{gathered} {\mu }_X\left(x\right)=\frac{f_X\left(x\right)}{1-F_X\left(x\right)} \\ =\frac{2x{e}^{-x^2}}{1-\left(1-e^{-x^2}\right)} \\ {\mu }_X\left(X\right)=2x \end{gathered}$$

Step :4 Value of E [ X ] (part d)

$$\begin{gathered} EX={\int }_{\mathrm{ℝ}} x{f}_X\left(x\right)dx \\ ={\int }_0^{\mathrm{\infty }} x\cdot 2x{e}^{-x^2}dx \\ ={\int }_0^{\mathrm{\infty }} 2x^2{e}^{-x^2}dx \end{gathered}$$

$u=x\text{ and }dv=2x{e}^{-x^2}dx\text{. Thus }du=dx\text{ and }v=-e^{-x^2}$

$$\begin{gathered} {\int }_0^{\mathrm{\infty }} 2x^2{e}^{-x^2}dx \\ =-{\left.x\cdot e^{-x^2}\right|}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }} e^{-x^2}dx \\ =\frac{\sqrt{\pi }}{2} \end{gathered}$$

Step : 5 variable Var &#8289; ( X )

In order to calculate the variance, let's find the second moment. We ve got 

$$\begin{gathered} E\left(X^2\right)={\int }_0^{\mathrm{\infty }} x^2{f}_X\left(x\right)dx \\ ={\int }_0^{\mathrm{\infty }} 2x^3{e}^{-x^2}dx \end{gathered}$$

$u=x^2\text{ and }dv=2x{e}^{-x^2}dx\text{. Thus }du=2xdx\text{ and }v=-e^{-x^2}$

$$\begin{gathered} {\int }_0^{\mathrm{\infty }} 2x^3{e}^{-x^2}dx=-{\left.x^2\cdot e^{-x^2}\right|}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }} 2x{e}^{-x^2}dx \\ ={\left.e^{-x^2}\right|}_0^{\mathrm{\infty }} \\ =1 \end{gathered}$$

Hence, the variance is adequate to

$$\begin{gathered} \mathrm{Var}\left(X\right)=E\left(X^2\right)-E(X{)}^2 \\ =1-\frac{\pi }{4} \\ =\frac{3\pi }{4} \end{gathered}$$