Question

Let the $pdf$ of X be given by $f(x)=0.25exp(-0.25x),x>0$

Show that $E(X)$ =4 and $Var(X)$ =16.

Answer 1

We have given that,

$f(x) = 0.25e^{-0.25x},x>0$

We have to find the value of E(X),

We know,

$E(X) = \int_{0}^{\infin} x0.25e^{-0.25x}dx$

$= 0.25\int_{0}^{\infin} xe^{-0.25x} dx$

$= 0.25[-4xe^{-0.25x}-16e^{-0.25x}]_{0}^{\infin}$

$= 0.25 \times 16$

$= 4$

Hence, $E(X) = 4$

$Var(X) = E(X^2)-{E(X)}^2$

$E(X^2) = 0.25 \int_{0}^{\infin} x^2e^{-0.25x}d$ x

$= 0.25[-4x^2e^{-0.25x}-32xe^{-0.25x}-128e^{-0.25x}]_{0}^{\infin}$

$= 0.25 \times 128$

$= 32$

Hence, $Var(X) = 32-16 = 16$