I'm looking at the following question
The solution presented is
I'm confused at how the arrived at the expected arrival of $\frac{1}{\mu+\lambda}$ in the case where the first arrival was from the second process. Should it not be $\frac{2}{\lambda}$ since $\mathbb{E}(T_1+T_2)=\frac{2}{\lambda}$? Could someone fill in the details?
(The solution is written in a very confusing way. First, the author talks about three Poisson processes, and then about two. I think the idea is correct though.)
Say you have two independent Poisson processes, one with rate $\lambda$ and one with rate $\mu$. As the author says, we are interested in the expected time at which either the first process has had two arrivals or the second process has had one arrival.
If the first arrival is from the second process, you know that (i) $S \leq T_1$, and (ii) $Z = S$. So what you want in this case is $$ E[Z|S \leq T_1] = E[S | S \leq T_1], $$ where $S$ is exponential with rate $\mu$ and $T_1$ is exponential with rate $\lambda$. Before this problem, you will probably have proven that the distribution of $S$ conditional on $S \leq T_1$ (i.e. the distribution of $\min(S,T_1)$ conditional on $S = \min(S,T_1)$) is exponential with rate $\mu+\lambda$. In other words, $$ E[Z|S \leq T_1] = E[S|S \leq T_1] = 1/(\mu+\lambda). $$
Edit: Proof that if $X$ is exponential with rate $\mu$ and $Y$ is exponential with rate $\lambda$, then $$ X | X \leq Y \sim {\rm Exp}(\mu+\lambda). $$ Take any $z>0$. To show the above, it suffices to show that $$ P(X \leq z | X \leq Y) = 1-e^{-(\mu+\lambda)z}, $$ since this is the CDF of the exponential distribution with rate $\mu+\lambda$.
First, we know that $P(X \leq Y) = \mu/(\mu+\lambda)$. Therefore, $$ P(X \leq z | X \leq Y) = \frac{P(X \leq z, X \leq Y)}{P(X \leq Y)} = \frac{\mu+\lambda}{\mu} P(X \leq z, X \leq Y). \quad (\ast) $$ To calculate $P(X \leq z, X \leq Y)$, use that they are independent, so their joint pdf is $f(x,y) = \mu \lambda e^{-\mu x} e^{-\lambda y}$. Therefore, \begin{align*} P(X \leq z, X \leq Y) &= \int_0^z \int_x^\infty \mu \lambda e^{-\mu x}e^{-\lambda y} dx dy\\ &= \int_0^z \mu e^{-\mu x} \big(\int_x^\infty \lambda e^{-\lambda y} dy\big) dx \\ &= \int_0^z \mu e^{-\mu x} \big[-e^{-\lambda y}\big]_x^\infty dx \\ &= \int_0^z \mu e^{-\mu x} e^{-\lambda x} dx \\ &= \mu \int_0^z e^{-(\mu+\lambda)x} dx \\ &= \mu \big[-1/(\mu+\lambda) \cdot e^{-(\mu+\lambda)x}\big]_0^z \\ &= \mu/(\mu+\lambda) \cdot (1-e^{(\mu+\lambda)z}). \end{align*} Combine with $(\ast)$ to get $$ P(X \leq z | X\leq Y) = 1-e^{-(\mu+\lambda)z}, $$ as desired. Note that you can also write this result as $$ P(\min(X,Y) \leq z | X = \min(X,Y)) = 1-e^{-(\mu+\lambda)z}, $$ which is how you may have seen it before. You already know that $\min(X,Y)$ has the exponential distribution with rate $\mu+\lambda$, but this says that even if you condition on $X$ being the smaller of the two, the distribution of the minimum does not change.