Question

Let the probability density function of a random variable, X, be given as:

$f_x\left(x\right)=\frac{3}{2}{e}^{-3x}u\left(x\right)+a{e}^{4x}u(-x)$

where u(x) is the unit step function.

Then the value of 'a' and Prob $X\le 0,$ respectively are

• A. $2,\frac{1}{2}$
• B. $4,\frac{1}{2}$
• C. $2,\frac{1}{4}$
• D. $4,\frac{1}{4}$

Answer 1

The correct option is A $2,\frac{1}{2}$

$f_x\left(x\right)=\frac{3}{2}{e}^{-3x}u\left(x\right)+a{e}^{4x}u(-x)$

We know that,

${\int }_{-\infty }^{\infty }{f}_x\left(x\right)dx=1$

$\therefore {\int }_{-\infty }^{0}a{e}^{4x}dx+{\int }_0^{\infty }\frac{3}{2}{e}^{-3x}dx=1$

$\because u\left(x\right)=(\begin{array}{cc}1,& x\ge 0\\ 0,& x<0\end{array}$

$\&u(-x)=(\begin{array}{cc}1,& x\le 0\\ 0,& x>0\end{array}$

$\Rightarrow {\left[\frac{a{e}^{4x}}{4}\right]}_{-\infty }^0+\frac{3}{2}{\left[\frac{e^{-3x}}{-3}\right]}_0^{\infty }=1$

$\Rightarrow \frac{a}{4}+\frac{1}{2}=1$

$\Rightarrow a=4\times \frac{1}{2}=2$

$Prob\left(x\le 0\right)={\int }_{-\infty }^0{f}_x\left(x\right)dx={\int }_{-\infty }^{0}a{e}^{4x}dx$

$={\int }_{-\infty }^{0}2e^{4x}dx=2{\left[\frac{e^{4x}}{4}\right]}_{-\infty }^0$

$=2\times \frac{1}{4}=\frac{1}{2}$