PDF of a Continuous Random Variable Proof

Question

$V$ is the speed uniformly distributed between $[40, 100]$. With a distance of $200$ km and uniform linear motion, the pdf of the time traveled was given as: $$ f_T(t) =\begin{cases} \frac{10}{3} \frac{1}{t^2} & 2\leq t\leq 5\\ \end{cases} $$

What I understand is that since the speed (V) has uniform linear motion then: $V = \frac{Distance}{Time}$. I tried to solve for the pdf of $t$. Where I used $f(t) = \frac{d}{V} = \frac{10}3t$ $$ f_T(t) = \int_2^5 \,\frac{10}3t\,dt$$ But all this results to $\frac{10}3 \ln(\frac{5}2)$

What is the correct $f(t)$ to use in this case? I feel I am missing something simple.

Answer 1

One approach you can try:

• $V$ is uniformly distributed on $[40,100]$
• For $40 \le v \le 100$ find the cdf $F_V(v)=P(V \le v)$ in terms of $v$
• $T=\frac{200}{V}$ which is a decreasing function
• So for $2 \le t \le 5$ find an expression for $P(T \ge t)$ in terms of $t$
• Then find the cdf $F_T(t)= P(T \le t)$ in terms of $t$
• Then find the density $f_T(t)$

If you do each step correctly, you should get $F_V(v)=\frac{v-40}{60}$ and $P(T \ge t)= \frac{10}{3t}-\frac23$ and $F_T(t)=\frac53 -\frac{10}{3t}$ and $f_T(t)=\frac{10}{3t^2}$ on the interval of interest