Waiting after finishing a single queue

Question

I'm a little confused by a conditional expectation question:

You have two exponentially distributed random variables, and you need to compute an expectation that looks like $$ E[T_{1}|T_{2} > T_{1}] $$

For example, you and a friend are in tandem $M/M/1$ queues, so the sojourn times will be exponential rvs. What the expected time you will have to wait given that you finish first.

I know that for an exponential rv $$ E[X|X>x] = x + E[X] $$

because of the memoryless property, but I can't see how to relate it to the above problem.

Answer 1

You could find the conditional joint pdf, and integrate...

Another way: consider two independent Poisson processes with rates $\lambda_1$ and $\lambda_2$. These could be realized by taking a combined Poisson process with rate $\lambda_1 + \lambda_2$, where each occurrence is assigned (independently of everything else) to process $1$ with probability $\lambda_1/(\lambda_1 + \lambda_2)$, otherwise to process $2$.
It should then be clear that the conditional distribution of the time of the first occurrence of process $1$, given that this happens before the first occurrence of process $2$ (i.e. that the first occurrence of the combined process is assigned to process $1$), is the same as the distribution of the first occurrence of the combined process. In particular, the conditional expectation is $1/(\lambda_1 + \lambda_2)$.

Answer 2

You can work out what $E[T_1|T_1<x]$ is. This should be $g(x) = \dfrac{1}{\lambda} - \dfrac{xe^{-\lambda x}}{1-e^{-\lambda x}}$

Then you should compute $\int^\infty_0 g(x)f(x)\text{dx}$ where $f$ is the distribution of $T_2$. I have looked at it, it doesnt look like an integral you can easily do.