Proof of $ \text{Poisson}(\lambda p) $-arrivals.

Question

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?

Answer 1

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 $X\sim P(\lambda p)$

Answer 2

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.