Probability help please?

Question

X (a random variable) has a double exponential distribution with parameter p>0 if its density is:

f_X(x) = 1/2 * e^(-p*|x|) for all x.

Prove that the variance of X = 2/(p^2).

I guess I'm having trouble performing the integration. If anyone could show how to do it step by step, it would be very helpful:

Answer 1

Isn't f = (p/2) e^(-p*|x|) ? Regardless, you have

integral from -∞ to ∞ of x² e^(-p*|x|) dx

which by symmetry is twice

integral from 0 to ∞ of x² e^(-p*x) dx

(we can drop the absolute value sign because x is positive throughout the interval of integration).

It requires two applications of integration by parts. The general formula is

∫ u dv = u*v - ∫ v du

or equivalently

∫ u (dv/dx) dx = u*v - ∫ v (du/dx) dx

First usage:

u = x², so du = 2xdx
dv = e^(-p*x) dx, so v = (-1/p) e^(-p*x)

and the integral becomes

(x²) (-1/p) e^(-p*x) evaluated at 0 and ∞
- integral from 0 to ∞ of (2x)(-1/p) e^(-p*x) dx

The evaluated term is zero at both limits, so we're left with

(2/p) integral from 0 to ∞ of x e^(-p*x) dx

Integrate by parts again:

u = x, so du = dx
dv = e^(-p*x) dx, so v = (-1/p) e^(-p*x)

The integral becomes

(2/p) [ (x)((-1/p) e^(-p*x) evaluated at 0 and ∞
- integral from 0 to ∞ of (-1/p) e^(-p*x) dx]

I'll leave the rest to you.