Expectation of a couple's date duration.

Question

I encountered an question to the solution to a question in my probability textbook (Problem 4.23. Introduction to Probability 2nd edition, by Dimitri P. Bertsekas and John N. Tsitsiklis)

The question is copied as follows:

Pat and Nat are dating, and all of their dates are scheduled to start at 9PM. Nat always arrives promptly at 9PM. Pat is highly disorganized and arrives at a time that is uniformly distributed between 8PM and 10PM. Let X be the time in hours between 8PM and the time when Pat arrives. If Pat arrives before 9PM., their date will last exactly 3 hours. If Pat arrives after 9PM., their date will last for a time that is uniformly distributed between 0 and 3-X hours. The date starts at the time they meet. Nat gets irritated when Pat is late and will end the relationship after second date on which Pat is late by more than 45 minutes. All dates are independent of any other dates.

I am stuck with the second question (What is the expected duration of any particular date?) for the part 0 < D < 3-X, where D is the random variable for the duration of date.

I applied the formula for uniform random variable and got $$ f_{D|X}(x)=\frac{1}{3-X}$$ However, when I look at the expectation in the solution, which is E[D|X] = $$\frac{3-X}{2}$$ and I know there is no way I would get this answer.

What have I done wrong?

Answer 1

If $Y$ is uniformly distributed on $(0,\Delta)$ then the mean of $Y$ is $\frac 1{\Delta} \int_0^{\Delta} xdx=\frac {\Delta} 2$.

Answer 2

Here is a graphical representation of fD|X(t):

And we have:

$$E[D|X]=\int_{1}^{2}tf_{D|X}(x)dt=\int_{1}^{2}t(3-X)dt$$ $$=\frac{1}{2} (1)^{2}(3-X)=\frac{3-X}{2}$$