# Finding the conditional distribution of Poisson random variables [duplicate]

Question

Let $$X$$ and $$Y$$ be the number of accidents which will occur at each of two intersections over the next year. Suppose that $$X$$ and $$Y$$ are independent Poisson random variables , with means $$a$$ and $$b$$ respectively. Find the conditional distribution of the number of accidents which will occur at the first intersection over the next year, given the total number of accidents.

My working

Let $$W = X + Y$$

$$\implies f_W(w) = \frac {(a + b)^w e^{-(a + b)}} {w!}$$

$$f_{X \mid W}(x \mid w) = \frac {f_{X, W}(x, w)} {f_W(w)}$$

This is where I am stuck. I believe that, in order to find the conditional distribution of $$X$$ on $$W$$, I need the joint distribution of $$X$$ and $$W$$ - I know of no other way to approach the problem. However, I am not sure if I am given enough information to find this joint distribution. In particular, it is obvious that $$X$$ and $$W$$ are not independent. How should I continue? Any intuitive explanations will be greatly appreciated :)

Just note that for $$a,b\in \Bbb Z$$ \begin{align*} F_{X,W}(a,b)&=\Pr [X\leqslant a, X+Y\leqslant b]\\ &=\sum_{\{(t,s)\in \mathbb{Z}^2:t\leqslant a,t+s\leqslant b\}}f_{X,Y}(t,s)\\ &=\sum_{(t,s)\in\mathbb{Z}^2}\mathbf{1}_{(-\infty ,a]}(t)\mathbf{1}_{(-\infty ,b]}(t+s)f_{X,Y }(t,s)\\ &=\sum_{(t,s)\in\mathbb{Z}^2}\mathbf{1}_{(-\infty ,a]}(t)\mathbf{1}_{(-\infty ,b-t]}(s)f_{X,Y}(t,s)\\ &=\sum_{t=-\infty }^a\sum_{s=-\infty }^{b-t}f_{X,Y}(t,s) \end{align*} and $$f_{X,W}(a,b)=\nabla_a\nabla _bF_{X,W}(a,b)=\nabla_b\nabla_a F_{X,W}(a,b)$$ where $$\nabla_a g(a,b):=g(a,b)-g(a-1,b)$$ for any function $$g$$, and similarly for $$\nabla_b g(a,b)=g(a,b)-g(a,b-1)$$, hence you find that

$$f_{X,W}(a,b)=\nabla_b\sum_{s=-\infty }^{b-a} f_{X,Y}(a,s)=f_{X,Y}(a,b-a)=f_X(a)f_Y(b-a)$$

if there is no weird mistake somewhere. I'm sure you can finish from here.

• Thank you for your answer! Sorry but I just wanted to check, are the $a$ and $b$ you used in the first equality the same as the $a$ and $b$ used in the question, or are they just arbitrary letters you introduced? – Ethan Mark May 4 at 16:01
• @EthanMark they are arbitrary letters that I introduced here, I didn't noticed that you used the same letters in your question – Masacroso May 4 at 16:04
• I see. Also, is this like a known proof/identity or something like that? – Ethan Mark May 4 at 16:21
• Where do the densities and integration come from? These are discrete random variables. – StubbornAtom May 4 at 16:21
• @Stubborn LOL, I didn't noticed, I assumed they where absolutely continuous random variables. Well I think we can just use PMFs instead of densities and sums instead of integrals. then derivatives must be changed by backward differences also. – Masacroso May 4 at 16:38