Distribution of the number of outcomes from a Poisson distribution with two splits

Question

I expect guests for my birthday party. Their number follows a Poisson distribution with some mean $\lambda$. With probability $p$ any arriving guest will be left-handed and with probability $1-p$, right-handed. Left-handed guests bring a present with probability $\alpha$, right-handed guests bring a present with probability $\beta$. How do I derive the distribution of the number of presents I will receive? Will it also be Poisson?

Answer 1

If you are familiar with the coloring lemma for poisson point processes, you can see this problem in the following way: for each point of the poisson process, you are assigning one of the possible colors $\{(L, 0), (L, 1), (R, 0), (R, 1)\}$ ($1$ is representing the guest brought a gift) with probabilities respectively $p(1 - \alpha)$, $p\alpha$, $(1-p)(1 - \beta)$ and $(1-p)\beta$. Actually, you are interested in all points with color being either $(L, 1)$ or $(R, 1)$. So, the distribution of number of gifts is poisson with with mean $\lambda (p\alpha + (1-p)\beta)$.

Answer 2

This is a problem of classification of a Poisson random variable. I can give you a hint about the distribution of the number of left-handed and right-handed people. Let $L$ denote the number of left-handed people that will come to your birthday party and $R$ denote the number of the right-handed. What do you know about $L$ and $R$? Of course, $L+R$ is the total number of people who will come to your party and let us denote it as $N$. Knowing this, you can condition on $N$ and use the Law of Total Probability to get the joint distribution of $L$ and $R$. Then sum $L$ from $0$ to $\infty$ to get the marginal distribution of $L$ and similarly get the marginal distribution of $R$.

You can find that $L$ follows a Poisson distribution with parameter $p\lambda$ and $R$ follows a Poisson distribution with parameter $(1-p)\lambda$ and these two variables are actually independent!

As for the number of gifts you will receive, denote $L_0$ as the number of presents you receive from the left-handed, so that $L_0|L=l$ follows the distribution of $Binomial(l,\alpha)$ and use the Law of total Total Probability again.

I've been busy with some exams recently. Otherwise, I will present you with the full answer. Besides, you can turn to Introduction to Probability Models by Shelton M. Rose on Page 116 Example 3.23 for the solution of a similar question.