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Exercise
May 01'23
Answer
Solution: C
The pdf of x is given by [math]f(x) = \frac{k}{(1+x)^4}, \, 0 \lt x \lt \infty [/math]. To find [math]k[/math], note
[[math]]
1 = \int_0^{\infty} \frac{k}{(1+x)^4} dx = - \frac{k}{3} \frac{1}{(1+x)^3} \Big |_{0}^{\infty} = \frac{k}{3}
[[/math]]
which implies that [math]k = 3 [/math]. It then follows that
[[math]]
\operatorname{E}[x] = \int_0^{\infty} \frac{3x}{(1+x)^4} dx [[/math]]
and substituting [math]u = 1+x, du = dx [/math], we see
[[math]]
\begin{align*}
\operatorname{E}[x] = \int_1^{\infty} \frac{3(u-1)}{u^4} du &= 3 \int_{1}^{\infty} (u^{-3} - u^{-4}) du \\ &= 3\left[\frac{u^{-2}}{-2} - \frac{u^{-3}}{-3}\right]_0^{\infty} \\ &= 3\left[\frac{1}{2}-\frac{1}{3}\right] = 3/2 - 1 = 1/2.
\end{align*}
[[/math]]