Answer to Question #176727 in Statistics and Probability for Eryl

Question #176727

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

given by

𝑓(π‘₯) = {

0 if π‘₯ ≀ 0

2π‘₯

𝑅2

if 0 < π‘₯ ≀ 𝑅

0 if π‘₯ > 𝑅

Find the variance of 𝑋.


1
Expert's answer
2021-03-31T07:16:37-0400

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}}"


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