Waiting after finishing a single queue 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.
 A: 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.
A: 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 $1/(\lambda_1 + \lambda_2)$.
