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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]
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