Expected value of two person meeting with another one - exponential distribution

Question

A manager set a time for two of his employees to meet with him at the same time, unfortunately he can see only one person at the same time and so one of them must wait. the time of each meeting follows an exponential distribution with mean = 30 minutes ($\lambda = 1/30$). We want to calculate the Expected Value of the time-distance between first man entrance to second man exit if (for example if first man enters at 11:00 and exits at 11:10 and second man enters at 11:10 and exits at 12:00, time-distance will be 60 minutes. i.e. 12:00 -11:00)

a) we know that person 1 arrives on time and person b arrives 5 minutes later.

b) we know that person 1 arrives on time and b arrives X minutes later and X has a exponential distribution with mean = 5 minutes.

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For a) I think we have two conditions. First person meeting ends before 5 minutes and first person meeting ends after 5 minutes. the probability that it takes less than 5 minutes is $1- e^ { - \frac{1}{30} * 5}= 0.1535 $ and in this condition, the expected value is $5 + 30$ (5 because person b arrives 5 minutes late). the probability that it takes more than 5 minutes is 0.8465 and so expected value in this condition is $\int_5^\infty x\frac{1}{30} e^{- \frac{1}{30} x} + 30 = 59.6269$ and the final answer will be $0.1535 \times 35 + 0.8465 \times 59.6269 = 55.8466$

I don't know any idea to solve part b.

I want to know whether my answer for part a is correct and how to solve part b

Answer 1

HINT + debugging the OP answer.

First of all, your answer of $55.8$ is definitely wrong because it is $< 60$. Just the two meetings alone would take an average of $60$ (even without any extra "idle" time).

The key to both problems is to calculate this quantity: $E[\max(A,B)]$ where

• $A =$ meeting length for person 1,

• $B =$ arrival time for person 2,

then the second meeting starts at time $C=\max(A,B)$ and run for an average of $30$ more minutes, so the overall answer is simply $E[\max(A,B)] + 30$.

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For part (a), we have $B=5$ and $E[C]$ can be calculated easily this way:

• $E[C] = E[C \mid A > 5] P(A > 5) + E[C \mid 5 > A] P(5 > A)$ by law of total probability

• $E[C \mid A > 5] = E[A \mid A > 5] = 30+5 $ by the memoryless property

  • In your OP, you miscalculated this $E[A \mid A > 5]$ as $\int_5^\infty x\frac{1}{30} e^{- \frac{1}{30} x} dx = 29.6269$. That calculation is wrong because you forgot to rescale by the condition probability ${1 \over P(A > 5)}$. If you had done that, you would have found ${29.6269 \over 0.8465} = 35$. But honestly, using the memoryless property of $A$ is much faster.

• $E[C \mid 5 > A] = 5$

• You can calculate $P(A>5)$ easily.

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For part (b) we have $B \sim Exp(1/5)$ and $A \sim Exp(1/30)$. The problem is that $C =\max(A,B)$ is no longer an exponential variable.

In particular, if you use the law of total probability, you will need to evaluate e.g. $E[A \mid A > B]$ and there is no easy way to do this. E.g., you cannot argue from memoryless property that this equals $E[A] + E[B] = 35$.

Instead of using the law of total prob, you can calculate $E[C]$ from first principles. Its CDF is:

$$P(C \le c) = P(\max(A,B) \le c) = P(A \le c) P(B \le c)$$

and as you can see, the RHS above is not of the exponential CDF form. Then you can calculate $E[C]$ based on the CDF and some integration. Alternatively, there is a super nice trick - see this answer.