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I'm having trouble determining the conditional probability of the sum of two independent Poisson Point Processes, $X, Y$, with parameters $\lambda,\mu$ respectively. If $X+Y=W$, I would like to find:

$$P(W_t=w|Y_s=y)$$

where $s<t$. I tried using Bayes' Rule because I thought the following form might be simpler to solve:

$$\rightarrow P(X_t+Y_t=w|Y_s=y)=\frac{P(Y_s=y|X_t+Y_t=w)\cdot P(X_t+Y_t=w)}{P(Y_s=y)}$$

The right term in the numerator and the term in the denominator are simple enough to calculate. The left term in the numerator is:

$$\rightarrow \frac{P(Y_s=y, X_t+Y_t=w)}{P(X_t+Y_t=w)}$$

The denominator here cancels with the right term in our original numerator, so we have:

$$\rightarrow \frac{P(Y_s=y, X_t+Y_t=w)}{P(Y_s=y)}$$

It is here that I am stuck, as I don't know how to separate the numerator here into two separate, calculatable probabilities. I was thinking I could do some simplification like this:

$$\rightarrow \frac{P(Y_s=y, X_t=w-y)}{P(Y_s=y)}=\frac{P(Y_s=y)P(X_t=w-y)}{P(Y_s=y)}=P(X_t=w-y)$$

But I'm not sure how, if at all, this is justifiable. Any suggestions would be sincerely appreciated. Cheers.

scoopfaze
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    If you let $A=X_t, y=Y_s, B=Y_t-Y_s$ then you get $P(W_t=w \mid Y_s=y) =P(A+B=w-y)$ – Henry May 30 '20 at 01:21
  • @Henry I arrived at this conclusion as well, but I wasn't certain what good it was. I thought I should search for some expression which allows me to either use $\lambda_\mu$ to solve, or each of them individually. If I am to do what you suggest, is it really just as simple as plugging the sum into the expression: $\frac{(\lambda t)^n}{n!}\times e^{-\lambda t}$? – scoopfaze May 30 '20 at 01:28
  • I would have thought $W_t-y$ has a Poisson distribution with parameter $\lambda t + \mu (t-s)$ – Henry May 30 '20 at 15:21
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    Using independence, realize that $P(Y_s=y,X_t+Y_t=w)=P(Y_s=y)P(X_t+Y_t-Y_s=w-y)$. Further, note that $X_t$ and $Y_t-Y_s$ are independent too.I hope this is enough for you to go on. – CharlieCornell May 31 '20 at 02:27

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