# Joint Distribution of two dependent Bernoulli Random Variables for $\rho=1$

Say we have two random variables $$X\sim B(p_1),\ Y\sim B(p_2)$$ where $$B(p)=$$ Bernoulli with probability $$0\le p\le 1$$. I am interested in the case when the correlation $$\rho$$ of $$X,Y$$ tends to $$1$$.

If we set events $$A=\{X=1\}$$, $$B=\{Y=1\}$$, the conditional probability property gives us \begin{align}\tag{1} P(A\cap B) = P(A\mid B)\cdot P(B)\\ = P(B\mid A)\cdot P(A) \end{align} When $$\rho=1$$ we must have $$P(A\mid B)=P(B\mid A)=1$$ (since one event implies the other). Equation $$(1)$$ then yields $$P(A) = P(B)$$ Furthermore, if correlation is high (= tending to 1), then whenever event $$A$$ occurs then $$B$$ must also occur (and vice versa). So the probabilities of $$A,B$$ should be $$P(A)=P(B)=\max(p_1,p_2)\tag{2}$$ However, for any $$\rho=1-\epsilon$$ with $$\epsilon>0$$, we still have $$P(A)=p_1$$ and $$P(B)=p_2$$ as per definition.

So can $$(2)$$ even be correct? What am I missing?

Unfortunately, Joint distribution of dependent Bernoulli Random variables only discusses non-deterministic sequences, so it doesn't quite apply.

• "we still have $P(A)=p_1$ and $P(B)=p_2$". Try $p_1=0.2$ and $p_2=0.8$ and try to get a value of $\rho$ being 0.9. Can't do it. Given $p_1$ and $p_2$, there are certain values of $\rho$ that can't be obtained. So you can't fix $p_1$ and $p_2$ and take the limit of $\rho$ -> $1$. Hint: start with the definition of $\rho$.
– JimB
Jan 25, 2019 at 14:36

The apparent inconsistency is due to the fact that $$\rho$$ cannot take on all values between $$-1$$ and $$1$$ given specific values for $$p_1$$ and $$p_2$$.

Using the definition of a correlation coefficient

$$\rho={{Cov(x_1,x_2)} \over {\sqrt{Var(x_1)Var(x_2)}}}={{\text{Pr}(x_1=1,x_2=1)-p_1 p_2}\over{\sqrt{p_1(1-p_1)p_2(1-p_2)}}}$$

and noting that $$\text{Pr}(x_1=1,x_2=1)\leq \text{Min}(p_1,p_2)$$ we can find the maximum value of $$\rho$$ for all possible values of $$p_1$$ and $$p_2$$ (here using Mathematica):

sol = Maximize[{(P11 - p1 p2)/Sqrt[p1 (1 - p1) p2 (1 - p2)],
0 < p1 < 1 && 0 < p2 < 1 && 0 < P11 <= Min[p1, p2]}, P11][[1]]


$$\begin{array}{cc} \{ & \begin{array}{cc} \sqrt{\frac{\text{p1} (\text{p2}-1)}{(\text{p1}-1) \text{p2}}} & \left(0<\text{p2}<\frac{1}{2}\land 0<\text{p1}\leq \text{p2}\right)\lor \left(\frac{1}{2}\leq \text{p2}<1\land \text{p1}=\text{p2}\right)\lor \left(\frac{1}{2}\leq \text{p2}<1\land 0<\text{p1}<\text{p2}\right) \\ \sqrt{\frac{(\text{p1}-1) \text{p2}}{\text{p1} (\text{p2}-1)}} & \left(0<\text{p2}<\frac{1}{2}\land \text{p2}<\text{p1}<1\right)\lor \left(\frac{1}{2}\leq \text{p2}<1\land \text{p2}<\text{p1}<1\right) \\ -\infty & \text{True} \\ \end{array} \\ \end{array}$$

ContourPlot[sol, {p1, 0, 1}, {p2, 0, 1}, ContourLabels -> True]


In short, $$\rho=1$$ can only occur when $$p_1=p_2$$.