Please help me with the following. For a long time I am trying to understand the part of the example:
The part I don't understand I marked by the red line. It has no sense to me, why the sum is for $k\geq i$. To be more specific. Take for example $X_n=1$ for some $n$ and ask about $P(Y_n=1|X_n=1)$. In this particular case at most $1$ customer can be served, so why the sum take into account all posibilities $k\geq 1$?
I think the idea is that the Poisson departure process do not directly correspond to the actual customer departure process, because the latter depends on the arrival process. You can think of it as letting a Poisson process run in the background, and whenever there is at least one customer in the queue, the next Poisson departure actually causes an actual customer departure.
Going back to your question, let there be $i$ customers, but assume the arrival process is much slower on average than the service process. Then the next customer will take a long time to arrive, during which there might be an arbitrarily large number of Poisson "departures" (I put this in quotes because there is no actual departure). The summation from $i$ to $\infty$ counts all these events. Note that the actual customer departures are denoted by $Y_{n}$, that is why the probability is zero when $j>i$.