Question

Examples for Continuous rv X

1. Given the pdf:
f(x) {1/4 e^-1/4x} x>0
0, otherwise
Find: a. E(X) and b. Var(X)

Answer 1

Use the definitions of expectation and variance.

• Expectation

`E(X) = \displaystyle \int_(-\infty)^\infty x f_X(x) \, dx = \frac14 \int_0^\infty x e^(-x/4) \, dx`

Integrate by parts,

`\displaystyle \int_a^b u \, dv = uv \bigg|_a^b - \int_a^b v \, du`

with

`u = x \implies du = dx \\\\ dv = e^(-x/4) \, dx \implies v = -4 e^(-x/4)`

Then

`E(X) = \displaystyle \frac14 \left(\left(-4x e^(-x/4)\right)\bigg|_0^\infty + 4 \int_0^\infty e^(-x/4) \, dx\right)`

`E(X) = \displaystyle \int_0^\infty e^(-x/4) \, dx = \boxed{4}`

(since the integral of the PDF is 1, and this integral is 4 times that)

• Variance

`V(X) = E\bigg((X - E(X))^2\bigg) = E(X^2) - E(X)^2`

Compute the so-called second moment.

`E(X^2) = \displaystyle \int_(-\infty)^\infty x^2 f_X(x)\, dx = \frac14 \int_0^\infty x^2 e^(-x/4) \, dx`

Integrate by parts, with

`u = x^2 \implies du = 2x \, dx \\\\ dv = e^(-x/4) \, dx \implies v = -4 e^(-x/4)`

Then

`E(X^2) = \displaystyle \frac14 \left(\left(-4x^2 e^(-x/4)\right)\bigg|_0^\infty + 8 \int_0^\infty x e^(-x/4) \, dx\right)`

`E(X^2) = 8 E(X) = 32`

and the variance is

`V(X) = 32 - 4^2 = \boxed{16}`