Question

Let the pdf of X be given by

f(x) = 0.25 exp((0.25x), x > 0

Show that E(X) = 4 and V ar(X) = 16

Answer 1

E(X)=$\int xf(x)dx$

=$\int _0^\infin 0.25e^{0.25x}.x dx$

=$\int _0^\infin 0.25xe^{0.25x} dx$

apply integration by parts

$=0.25\left(4e^{0.25x}x-\int \:4e^{0.25x}dx\right)$

$=0.25\left(4e^{0.25x}x-16e^{0.25x}\right)]_0^\infin$

$=0.25((0-0)-(0-16))$

=4

Var(X)

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

$E(X^2)=\int x^2f(x)dx$

$=\int _0^\infin 0.25e^{0.25x}.x^2 dx$

$=\int _0^\infin 0.25x^2e^{0.25x} dx$

apply integration by parts

$=0.25\left(4e^{0.25x}x^2-\int \:8e^{0.25x}xdx\right)$

$=0.25\left(4e^{0.25x}x^2-8\left(4e^{0.25x}x-16e^{0.25x}\right)\right)]_0^\infin$

=0.25((0-0)-(0-8(0-16)))

=0.25*128

=32

$Var(x)=32-4^2$

=16