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Let $\mathbf{X}$ be Binomial point process in $W = [0, 6] \times [0, 4]$ with $n$ points. Let $A_1 = [0, 2] \times [0, 4]$, $A_2 = [0, 6] \times [0, 2]$, and $A_3 = [2, 6] × [2, 4]$. I want to find $E[N_{\mathbf{X}}(A_1) N_{\mathbf{X}}(A_2)]$, where $N_{\mathbf{X}}(A_1)$ is the number of points from $\mathbf{X}$ in $A_1$ and $N_{\mathbf{X}}(A_2)$ is the number of points from $\mathbf{X}$ in $A_2$. It is clear that $N_{\mathbf{X}}(A_1)$ and $N_{\mathbf{X}}(A_2)$ are dependent, but we can see from $A_1 = [0, 2] \times [0, 4]$ and $A_2 = [0, 6] \times [0, 2]$ that they're also not disjoint. If these were disjoint, then I would know how to calculate $E[N_{\mathbf{X}}(A) N_{\mathbf{X}}(B)]$ using the multinomial distribution, but I'm not sure how to calculate this.

I calculated that $E[N_{\mathbf{X}}(A_1)] = \dfrac{8}{24} = \dfrac{1}{3}$ and $E[N_{\mathbf{X}}(A_2)] = \dfrac{12}{24} = \dfrac{1}{2}$.

I then try to calculate $E[N_{\mathbf{X}}(A_1) N_{\mathbf{X}}(A_2)]$ using conditional expectation:

$$ \begin{align*} E[N_{\mathbf{X}}(A_1) N_{\mathbf{X}}(A_2)] &= E[N_{\mathbf{X}}(A_1)]E[N_{\mathbf{X}}(A_2) | N_{\mathbf{X}}(A_1)] \\ &= \left( n \cdot p_1 \right) \left( (n - r) \cdot p_2 \right) \ \ \ \text{(where $n - r$ is the number of points in $A_2$ that are not in $A_1$.)} \\ &= n^2 p_1 p_2 \cdot (1 - r/n) \\ &= n^2 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \left( 1 - \frac{r}{n} \right) \\ &= \frac{n}{6} \left( n - r \right). \end{align*} $$

But I'm not sure if this is correct.

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  • $\begingroup$ I think your formula is not correct, it should be $E[N_X(A_1)N_X(A_2)] = E[N_X(A_1) E[N_X(A_2)|N_X(A_1)]]$ $\endgroup$
    – D F
    Aug 15, 2023 at 14:25

1 Answer 1

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If you know how to compute the expectation for two disjoint sets, then you can define $A_{1, 1} = [0, 2] \times [2, 4]$ and $A_{1, 2} = [0, 2] \times [0, 2]$, and $A_{2, 1} = [2, 6] \times [0, 2]$. Notice that $A_{1, 1}, A_{1, 2}, A_{2, 1}$ are disjoint. Now, clearly $$N_X(A_1) = N_X(A_{1, 1}) + N_{X}(A_{1, 2}),$$ and $$N_X(A_2) = N_{X}(A_{1, 2}) + N_X(A_{2, 1}).$$ Hence, $$E[N_X(A_1)N_X(A_2)] = E[N_X(A_{1, 1})N_X(A_{1, 2})] + E[N_X(A_{1, 1})N_X(A_{2, 1})] + E[N_X(A_{1, 2})^2] + E[N_X(A_{1, 2})N_X(A_{2, 1})].$$ So you have three terms of disjoint products that you know how to compute, and the squared term can also be easily computed.

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  • $\begingroup$ The sets you mention are not disjoint. For example $A_{1,1} \cap A_{1,2} = [0,2] \times \{2 \}$. $\endgroup$ Aug 15, 2023 at 19:28

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