An SMS message is sent through a communication channel with multiple servers, starting from 0. In each second, the message can be transmitted to the next server with a probability P, or remain on the server and wait.
Let $Xi$ be the random variable which gives the number of seconds needed for the message to be received by the $i$-th server. Find the distributions of $x1$ and $x2$
I think this is poisson distribution but I've no idea where or how to start. Any help?
$X_1$ = the number of seconds it takes to move from server 0 to 1. As we are the counting Bernoulli trials until a success, this will have a Geometric distribution, $X_1\sim\mathcal{Geo}_1(p)$.
$$X_1\sim\mathcal{Geo}_1(p) \quad\iff\quad \mathsf P(X_1=x_1)= (1-p)^{x_1-1}p$$
$X_2-X_1$ = the number of seconds it takes to move from server 1 to 2. This will also have a geometric distribution, iid to $X_1$.
Thus we have the join pmf: $\;P_{X_1,X_2}(x_1,x_2)= \mathsf P(X_1{=}x_1, X_2{=}x_2)\\ \qquad = \mathsf P(X_1{=}x_1)\;\mathsf P(X_2{-}X_1{=}x_2{-}x_1) \\ \qquad \ddots \text{(can you complete?)}$
You can use this joint distribution to evaluate the marginal for $X_2$.
However we can also assert that $X_2$ will have a negative binomial distribution as it is the count of trials until the second success. Thus to obtain $X_2=x_2$ we require 1 success and $x_2-2$ failures in some order, followed by a final success.
$$X_2\sim\mathcal{NB}(2, p) \quad\iff\quad \mathsf P(X_2{=}x_2) = \dbinom{n}{x_2-2}(1-p)^{x_2-2}p^2$$