# Expected Value of a Random Variable M

Question: After your complaint about their service, a representative of an insurance company promised to call you "between $$7$$ and $$9$$ this evening". Assume that this means that the time T of the call is uniformly distributed in the specified interval. Assume that you know in advance that the call will last exactly $$1$$ hour. From $$9$$ to $$9:30$$, there is a game show on $$TV$$ that you wanted to watch. Let $$M$$ be the amount of time of the show that you miss because of the call. Compute the expected value of $$M$$.

What I have understood is $$P(M | X < 8:00) = 0$$, i.e. probability that show will be missed is $$0$$ when call is received before $$8:00$$. If I consider time b/w $$8:00$$ to $$8:30$$, then expected value is $$(8:30-8:00)/2 = 15.$$ Is it the right way to proceed. I don't know what is actual answer.

• What you say in English is almost right. You just need to also consider the case that the phone call starts between 8:30-9:00. Note that what you write in maths is incorrect: you should say that given $X$ is the start time of the call, then $E(M|X<8:00)=0$. Dec 6, 2020 at 15:48
• @BenjaminWang thanks for correcting. After 8:30, getting show missed will become a sure event. i.e. $P(M | X > 8:30) = 1$. Right? Dec 6, 2020 at 16:34
• Again, the english is right, and the maths is off by a little. You should say that $P(M=30|X>8:30)=1$. Dec 6, 2020 at 16:50

If call arrives in the interval $$[8;8:30]$$ your M is uniform in $$[0;30]$$. This happens with probability $$p=\frac{1}{4}$$
If the call arrives after 8:30 you miss all your TV show. This happens with probability $$p=\frac{1}{4}$$
$$E(M)=15\frac{1}{4}+30\frac{1}{4}=11.25$$