# (Using conditional expectation to calculate) expected value of the product of two dependent random variables

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.

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

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