Exercise
May 04'23
Answer
Solution: E
First note R = 10/T . Then
[[math]]
F_R(r) = \operatorname{P}[R ≤ r] = \operatorname{P}[\frac{10}{T} \leq r ] = \operatorname{P}[T \geq \frac{10}{r}] = 1 - F_T(\frac{10}{r}).
[[/math]]
Differentiating with respect to
[[math]]
\begin{align*}
r f_R(r) = F^{'}_R(r) = d/dr (1 - F_T(\frac{10}{r})) = -(\frac{d}{dt}F_T(t)) (\frac{-10}{r^2}) \\
\frac{d}{dt}F_T(t) = f_T(t) = \frac{1}{4}
\end{align*}
[[/math]]
since [math]T[/math] is uniformly distributed on [8,12]. Therefore
[[math]]
f_R(r) = \frac{-1}{4} (\frac{-10}{r^2}) = \frac{5}{2r^2}.
[[/math]]