Question

The probability density functions of two independent random variables X and Y are given by,
Where a, b are positive real constants and u(•) represents the unit step function. The probability density functibn of the random variable Z = X + Y will be
$f_x\left(x\right)=a{e}^{-ax}u\left(x\right)and{f}_y\left(y\right)=b{e}^{-by}u\left(y\right)$

• A. $\frac{ab}{(b-a)}[e^{-az}-e^{-bz}]u\left(z\right)$
• B. $\frac{ab}{(a+b)}[e^{-az}-e^{-bz}]u\left(z\right)$
• C. $\frac{ab}{(b-a)}\left[e^{-(a+b)z}\right]u\left(z\right)$
• D. $\frac{ab}{(b-a)}[e^{-bz}-e^{-az}]u\left(z\right)$

Answer 1

The correct option is A $\frac{ab}{(b-a)}[e^{-az}-e^{-bz}]u\left(z\right)$

$f_z\left(z\right)=f_x\left(z\right)\ast f_y\left(z\right)$
$f_x\left(z\right)=a{e}^{-az}u\left(z\right)$
$f_y\left(z\right)=b{e}^{-bc}u\left(z\right)$

$L\left[f_x\right(z\left)\right]=\frac{a}{s+a}andL\left[f_y\right(z\left)\right]=\frac{b}{s+b}$

$f_z\left(z\right)$ = $L^{-1}\left[\frac{ab}{(s+a)(s+b)}\right]$
= $L^{-1}${ $\frac{ab}{b-a}[\frac{1}{(s+a)}-\frac{1}{(s+b)}]$}

= $\frac{ab}{(b-a)}[e^{-az}-e^{-bz}]u\left(z\right)$