A doctor has made two appointments, one at 15:00 and one at 15:30. The duration time of each examination is a random exponential variable with mean of 30 minutes. Moreover the durations are independent. We assume that the patients be punctual. a) Find the waiting time distribution b) Find the expected value of this distribution. c) If we assume that the first patient arrives in a time uniformly distributed between 14:50 and 15:10, how change the answers?
a) The waiting time distribution is $F_T(t)=\left\{\begin{matrix} 0;t<0\\ 1-e^{-1},t=0\\ 1-e^{-(\frac{t}{30}+1)},t>0\end{matrix}\right.$
b) $\mathbb{E}[T]=\frac{30}{e}$
c) Now the problem. I obtain $F_T(t)=\left\{\begin{matrix} 0;t<0\\ 1-\frac{1}{2}[e^{-\frac{2}{3}}+e^{-\frac{4}{3}}],t=0\\ 1-\frac{1}{2}[e^{-\frac{2}{3}}+e^{-\frac{4}{3}}]e^{-\frac{t}{30}},t>0\end{matrix}\right.$ but the solution is $F_T(t)=\left\{\begin{matrix} 0;t<0\\ 1-\frac{3}{2}[e^{-\frac{2}{3}}-e^{-\frac{4}{3}}],t=0\\ 1-\frac{3}{2}[e^{-\frac{2}{3}}-e^{-\frac{4}{3}}]e^{-\frac{t}{30}},t>0\end{matrix}\right.$. I don't understand where I wrong.
Could you please help me? Thanks in advance!
Say the arrival time of the first patient is $A$ minutes after 15:30 (thus $A<0$ always) and the exam time is $E$ minutes. You want the probability distribution of $\max \{ 0,A+E \}$. Begin with just the probability distribution of $A+E$. You have
$$F_{A+E}(t) =\int_{-\infty}^t \int_{-\infty}^{t-x} f_A(x) f_E(y) dy dx \\ = \frac{1}{600} \int_{-\infty}^t \int_{-\infty}^{t-x} 1_{[-40,-20]}(x) 1_{[0,\infty)}(y) e^{-y/30} dy dx \\ = \frac{1}{600} \int_{(-\infty,t] \cap [-40,-20]} \int_{(-\infty,t-x] \cap [0,\infty)} e^{-y/30} dy dx.$$
Now you need to work out these intersections. In $(-\infty,t] \cap [-40,-20]$, the left endpoint is always $-40$ but the right endpoint might be $-20$ or $t$ depending on whether $t>-20$ or not; moreover the intersection is empty if $t<-40$, so that case has zero probability.
For the other case, the left endpoint is always $0$ and the right endpoint is always $t-x$. The interval would be empty if $t-x$ were negative but this isn't in the outer domain of integration.
So you'll define $F_{A+E}(t)$ in three pieces: for $t<-40$, for $-40 \leq t<-20$, and for $t \geq -20$.
Finally, to take the maximum, you can simply set $F_T(t)=\begin{cases} 0 & t<0 \\ F_{A+E}(t) & t \geq 0 \end{cases}$; this corresponds to moving all the probability that $A+E$ took on some negative value into the event $T=0$, which makes sense in the model.
I find that when I finish the problem I get the $3/2$; my guess is that you did not properly handle the case work.