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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 :)

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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.

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  • $\begingroup$ 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? $\endgroup$ – Ethan Mark May 4 at 16:01
  • $\begingroup$ @EthanMark they are arbitrary letters that I introduced here, I didn't noticed that you used the same letters in your question $\endgroup$ – Masacroso May 4 at 16:04
  • $\begingroup$ I see. Also, is this like a known proof/identity or something like that? $\endgroup$ – Ethan Mark May 4 at 16:21
  • $\begingroup$ Where do the densities and integration come from? These are discrete random variables. $\endgroup$ – StubbornAtom May 4 at 16:21
  • $\begingroup$ @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. $\endgroup$ – Masacroso May 4 at 16:38

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