Question

Examples for Continuous rv 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

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