How to prove $ P(A\cap B)= P(A) P(B) - P(\bar{A} )P(\bar{B})$

Question

I am given this question in my statistics course:

Let $A$ and $B$ be $2$ events such that $A\cup B = \Omega$. Prove that $ P(A\cap B)= P(A) P(B) - P(\bar{A} )P(\bar{B})$

Hint: Define the following two random variables and use covariance.

$\qquad\qquad\qquad X = \begin{cases} 1 & \text{if $A$ occurs} \\ 0 & \text{otherwise} \end{cases} \qquad\qquad Y = \begin{cases} 1 & \text{if $B$ occurs} \\ 0 & \text{otherwise} \end{cases} $

$\,$

I was able to prove it using this method:

$RHS =P(A)P(B) - (1-P(A))(1-P(B))$

$\space = P(A)P(B) - ( 1-P(B)-P(A)+ P(A)P(B))$

$\space = P(A) + P(B) - 1$

$\space= P(A) + P(B) - (P(A) + P(B) - P(A\cap B))$

$\space= LHS$

I was not, however, able to prove it using the hint (or covariance).

Any help would be much appreciated. Thank you!

Answer 1

Since $\Omega=A\cup B$:

$\mathsf{Cov}\left(\mathbf{1}_{A^{\complement}},\mathbf{1}_{B^{\complement}}\right)=P\left(A^{\complement}\cap B^{\complement}\right)-P\left(A^{\complement}\right)P\left(B^{\complement}\right)=-P\left(A^{\complement}\right)P\left(B^{\complement}\right)$

Also:

$\mathsf{Cov}\left(\mathbf{1}_{A^{\complement}},\mathbf{1}_{B^{\complement}}\right)=\mathsf{Cov}\left(1-\mathbf{1}_{A},1-\mathbf{1}_{B}\right)=\mathsf{Cov}\left(\mathbf{1}_{A},\mathbf{1}_{B}\right)=P\left(A\cap B\right)-P\left(A\right)P\left(B\right)$

IMV however it is not a very useful hint and is it better to work out straightforward.

Answer 2

Perhaps the hint asks you to do the following: $(1-X)(1-Y)=0$ since either $1-X$ or $1-Y$ is always $0$. This gives $XY=X+Y-1$. Taking expectations we get $P(A\cap B)=E(XY)=EX+EY-1=P(A)+P(B)-1$. Now $t+s-1=ts-(1-t)(1-s)$ for any two numbers $t, s\in [0,1]$. Hence we get $P(A\cap B)=P(A)P(B)+(1-P(A))(1-P(B))$ as required.

Not a great hint though!