Proof of $ \text{Poisson}(\lambda p) $-arrivals. We have a queue where people pass out of it with $ \text{Poisson}(\lambda) $ and they come in with probability $ p $.
I understand that the arrivals follow $ \text{Poisson}(\lambda p) $, but how can I prove it?
 A: Let X be the number of arrivals.
\begin{eqnarray*}
&&P(X=r)=\sum\limits_{k=r}^{\infty}\dfrac{\lambda ^k}{k!}\exp{(-\lambda)}C_k^rp^r(1-p)^{k-r}\\
&&=\sum\limits_{k=r}^{\infty}\exp{(-\lambda)}\dfrac{\lambda ^k k!}{r!(k-r)!k!}p^r(1-p)^{k-r}\\
&&=\dfrac{\lambda^r p^r}{r!}\sum\limits_{k=r}^{\infty}\exp{(-\lambda)}\dfrac{\lambda ^{k-r} }{(k-r)!}(1-p)^{k-r}\\
&&=\dfrac{\lambda^r p^r}{r!}\exp{(-\lambda)}\exp{(\lambda(1-p))}=\dfrac{(\lambda p)^r}{r!}e^{-\lambda p}
\end{eqnarray*}
So $X\sim P(\lambda p)$
A: Here is a cool proof I thought about.
I am going to split the original process into two parts, one that can arise with probability p and the other with probability 1-p. Here in your example, the original process is passing out, with proba p going in, proba 1-p not going in. 
Now what we want to prove is how the splitted process is still a poisson process, and this will give us the proof that your process is indeed poisson with parameter $\lambda p$.
Also, since it is the case that interested me, I prove it for poisson proceses, not poisson random variables (replace $\lambda$ by $\lambda t $.

We have the poisson process
$$X(t) = X_1(t) + X_2(t) $$
First we calculate the joint probability 
$$P[X_1(t) = k, X_2(t) = m] = \sum_{n=0}^{\infty} P[X_1(t) = k, X_2(t)
> = m \mid X(t) = n]P[X(t) = n]$$
Note that
$$P[X_1(t) = k, X_2(t) = m \mid X(t) = n] = 0 \:\:\: \text{when}\:\: n
> \neq k+m$$
Now
$$P[X_1(t) = k, X_2(t) = m] = P[X_1(t) = k, X_2(t) = m \mid X(t) =
> k+m]P[X(t) = k+m]$$
$$= P[X_1(t) = k, X_2(t) = m \mid X(t) = k+m]e^{-\lambda
> t}\frac{(\lambda t)^{k+m}}{(k+m)!}$$
Now, given that $k+m$ events occurred, since each event has
  probability $p$ of being a type $1$ event and probability $1-p$ of
  being a type $2$ event, it follows that
$$P[X_1(t) = k, X_2(t) = m \mid X(t) = k+m] = \binom{k+m}{k}
> p^k(1-p)^m$$
Thus, 
$$P[X_1(t) = k, X_2(t) = m] = \binom{k+m}{k} p^k(1-p)^m e^{-\lambda
> t}\frac{(\lambda t)^{k+m}}{(k+m)!}$$
$$= \frac{(k+m)!}{k!m!} p^k(1-p)^m e^{-\lambda t}\frac{(\lambda
> t)^{k+m}}{(k+m)!}$$
$$= e^{-\lambda p t} \frac{(\lambda p t)^k}{k!} e^{-\lambda(1-p) t}
> \frac{[\lambda (1 - p)t]^m}{m!} \:\:\:\:\:\:\:\:\:(1)$$
Then 
$$P(X_1 = k) = \sum_{m=1}^{\infty} P[X_1(t) = k, X_2(t) = m]$$
$$=e^{-\lambda p t} \frac{(\lambda p t)^k}{k!} e^{-\lambda(1-p) t}
> \sum_{m=1}^{\infty} \frac{[\lambda (1 - p)t]^m}{m!}$$
$$= e^{-\lambda p t} \frac{(\lambda p t)^k}{k!} e^{-\lambda(1-p) t}
> e^{\lambda(1-p) t}$$
$$=e^{-\lambda p t} \frac{(\lambda p t)^k}{k!} \:\:\:\:\:\: (2)$$
which indicates that $X_1(t)$ is a poisson process with rate $\lambda
> p$.
Similarly, we can obtain
$$P(X_2(t) = m) = \sum_{k=1}^{\infty} P[X_1(t) = k, X_2(t) = m]$$
$$= e^{-\lambda(1-p) t} \frac{[\lambda (1 - p)t]^m}{m!} \:\:\:\:\:\:
> (3)$$
and so $X_2(t)$ is a poisson process with rate $\lambda (1-p)$.  
Finally, from equations $(1)$, $(2)$ and $(3)$
$$P[X_1(t) = k, X_2(t) = m] = P[X_1(t) = k]P[X_2(t) = m]$$
Hence, $X_1(t)$ and $X_2(t)$ are independent.

Then you have the answer to your problem ! I am not sure that there is a simplier way to prove it, other than intuition.
The original part of the proof comes from here : Splitting Poisson process formal proof and the reward of the work should be attributed to JKnecht. I wanted to use his proof for another interesting similar case, which is yours, because in the end, it is the same problem.
