Question

Suppose that a continuous random variable 𝑋 has a probability density function

given by

𝑓(𝑥) = {

0 if 𝑥 ≤ 0

2𝑥

𝑅2

if 0 < 𝑥 ≤ 𝑅

0 if 𝑥 > 𝑅

Find the variance of 𝑋.

Answer 1

Let's find the mathematical expectation:

$M(X) = \int\limits_{ - \infty }^{ + \infty } {xf(x)dx} = \int\limits_0^R {x \cdot \frac{{2x}}{{{R^2}}}} dx = \left. {\frac{{2{x^3}}}{{3{R^2}}}} \right|_0^R = \frac{{2{R^3}}}{{3{R^2}}} = \frac{2}{3}R$

Then the variance is

$D(X) = \int\limits_{ - \infty }^{ + \infty } {{x^2}f(x)dx} - {M^2}(X) = \int\limits_0^R {{x^2} \cdot \frac{{2x}}{{{R^2}}}} dx - {\left( {\frac{2}{3}R} \right)^2} = \left. {\frac{{2{x^4}}}{{4{R^2}}}} \right|_0^R - \frac{4}{9}{R^2} = \frac{{{R^2}}}{2} - \frac{4}{9}{R^2} = \frac{1}{{18}}$

Answer: $D(X) = \frac{1}{{18}}$