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I have a setup with a homogeneous Poisson Point Process (PPP) of intensity $\lambda$ in $W \subseteq \mathbb{R}^d$ and a set $A \subseteq W$. I'm looking for the expected value of points in set $A$, given the knowledge that at point $x\notin A, x \in W$ is exactly one point which is part of the PPP.

I konw that the expected number of points in two disjoint sets $A_1,A_2$ depends only on the volume/measure of the respective set. But how does a fixed point (which is part of the PPP) influence that expected value?

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Your question states that the fixed point $x$ is not in $A$. In that case it doesn't change anything. The number of points in $A$ follows a Poisson distribution with mean $\lambda |A|$, where $|A|$ is the volume of $A$.

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  • $\begingroup$ This was also my first guess, but I'm not sure about it. Assuming e.g. that the expected number of points in $W$ is $\lambda \vert W\vert = 1$, then, shouldn't it make a difference when I say that I already know that outside of $A$ but in $W$ is definitively 1 point? If I now claim that I have knowledge about 10 point in $W$ but outside $A$, the poisson process still has to achieve on average 1 point in $W$ which is then almost impossible. $\endgroup$
    – user93427
    Commented Oct 29, 2015 at 8:16
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    $\begingroup$ This is irrelevant for what happens in $A$. That is exactly the independence property of the Poisson process. You can also think about the fact that you can simulate the process on $W$ by independently simulating on $A$ and $W \setminus A$, so no matter what you do in $W \setminus A$ it doesn't have an effect on what happens in $A$. $\endgroup$
    – Ege Rubak
    Commented Oct 29, 2015 at 15:17
  • $\begingroup$ Your example of observing 10 points outside $A$ is not a problem. The event that you have 10 points in $W \setminus A$ is highly unlikely with your assumption $\lambda |W|=1$, but you condition on this unlikely event and the fact that something unlikely happened elsewhere doesn't change the chances of what happens in $A$ when it is a Poisson process. It's like observing "red" 100 times on the roulette: Extremely unlikely, but if it has happened it doesn't change the chances in the next game (assuming it really is a fair roulette). $\endgroup$
    – Ege Rubak
    Commented Oct 29, 2015 at 15:23

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