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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)

Sagot :

Use the definitions of expectation and variance.

  • Expectation

[tex]E(X) = \displaystyle \int_{-\infty}^\infty x f_X(x) \, dx = \frac14 \int_0^\infty x e^{-x/4} \, dx[/tex]

Integrate by parts,

[tex]\displaystyle \int_a^b u \, dv = uv \bigg|_a^b - \int_a^b v \, du[/tex]

with

[tex]u = x \implies du = dx \\\\ dv = e^{-x/4} \, dx \implies v = -4 e^{-x/4}[/tex]

Then

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

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

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

  • Variance

[tex]V(X) = E\bigg((X - E(X))^2\bigg) = E(X^2) - E(X)^2[/tex]

Compute the so-called second moment.

[tex]E(X^2) = \displaystyle \int_{-\infty}^\infty x^2 f_X(x)\, dx = \frac14 \int_0^\infty x^2 e^{-x/4} \, dx[/tex]

Integrate by parts, with

[tex]u = x^2 \implies du = 2x \, dx \\\\ dv = e^{-x/4} \, dx \implies v = -4 e^{-x/4}[/tex]

Then

[tex]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)[/tex]

[tex]E(X^2) = 8 E(X) = 32[/tex]

and the variance is

[tex]V(X) = 32 - 4^2 = \boxed{16}[/tex]