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.
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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