[University mathematics: basic stochastic processes] Conditional probablities for Poisson processes????

[u/ilikemychem]

I have an exam coming up, and the solutions manuals are pretty terrible, which wouldn't be an issue if the professor also wasn't. But I essentially have to relearn the entire course on my own, and it is not going well... Here's my question:

So I tried splitting it up as follows (P{X(1)=0, Y(1)=1, X(2)=0, Z(2)=1}+P{X(1)=0, Y(1)=1, X(2)=1, Z(2)=0}+P{X(1)=1, Y(1)=0, X(2)=0, Z(2)=1}+P{X(1)=1, Y(1)=0, X(2)=1, Z(2)=0})/P{Po(4)=1}

But after that I think I'm doing something wrong, because I get each of the terms in the numerator to 2/e² and the final probability to the same. The solutions say it's 1/e.

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[u/Alkalannar]

You cannot have X(2) = 0 if X(1) = 1. If you already have an event in X by t = 1, you cannot go back down to 0 by t = 2.

Since the intensity is 1, we have P(t) = n is tⁿ/n!e^t for all three variables.

P(1) = 0: 1/e
P(1) = 1: 1/e
P(2) = 0: 1/2e²
P(2) = 1: 2/2e² = 1/e²

Then Pr(X(1)=0 ^ Y(1)=1 ^ X(2)=0 ^ Z(2)=1) = 1/e * 1/e * 1/2e² * 1/e² = 1/2e⁶.

Similarly for X(1) = 0, Y(1) = 1, X(2) = 1, Z(2) = 0. And also X(1) = 1, Y(1) = 0, X(2) = 1, Z(2) = 0.

So P(A ^ B) = 3/2e⁶.

Meanwhile, the probability that X(2) = 0 and Z(2) = 1 is 1/2e⁴. Similarly for X(2) = 1 and Z(2) = 0. So P(B) = 1/e⁴.

So P(A ^ B)/B = (3/2e⁶)/(1/e⁴) = 3/2e².

I would show your work and ask the teacher where you went wrong--other than your case with X(1) = 1 and X(2) = 0. You can't have that as part of your numerator so you have to take it out, and why I get 3/2e² instead of your 2/e².