Problem source

The random variable $X$ takes  only the values $x_1$ and $x_2$  (where $ x_1 \not= x_2 $),  and the random variable $Y$ takes only  the values $y_1$ and $y_2$  (where $y_1 \not= y_2$). 
Their joint distribution is given by  
$$ 
\P ( X = x_1 , Y = y_1 ) = a \ ; \ \  
\P ( X = x_1 , Y = y_2 ) = q - a \ ; \ \  
\P ( X = x_2 , Y = y_1 ) = p - a \ . 
$$  
Show that if $\E(X Y) = \E(X)\E(Y)$ then  
$$  
(a - p q ) ( x_1 - x_2 ) ( y_1 - y_2 ) = 0 . 
$$ 
Hence show that two random variables each taking only two distinct values are independent if  
$\E(X Y) = \E(X) \E(Y)$. 
 
Give a  joint distribution for two random variables  $A$ and $B$, each taking the three values $- 1$, $0$ and $1$  with probability ${1 \over 3}$,  which have $\E(A B) = \E( A)\E (B)$, but which are not independent.

Solution source

\begin{align*}
\mathbb{P}(X = x_1) &= a + q - a = q \\
\mathbb{P}(X = x_2) &= 1 - q \\
\mathbb{P}(Y = y_1) & = a + p - a = p \\
\mathbb{P}(Y = y_2) & = 1 - p
\end{align*}

\begin{align*}
\mathbb{E}(X)\mathbb{E}(Y) &= \l qx_1 + (1-q)x_2 \r \l p y_1 + (1-p)y_2\r  \\
&= qpx_1y_1 + q(1-p)x_1y_2 + (1-q)px_2y_1 + (1-q)(1-p)x_2y_2 \\
\mathbb{E}(XY) &= ax_1y_1 + (q-a)x_1y_2 + (p-a)x_2y_1 + (1 + a - p - q)x_2y_2 &= 
\end{align*}

Therefore $\mathbb{E}(XY) - \mathbb{E}(X)\mathbb{E}(Y)$ is a degree 2 polynomial in the $x_i, y_i$. 

If $x_1 = x_2$ then we have:

\begin{align*}
\mathbb{E}(X)\mathbb{E}(Y) &=x_1 \l p y_1 + (1-p)y_2\r  \\
\mathbb{E}(XY) &= x_1(ay_1 + (q-a)y_2 + (p-a)y_1 + (1 + a - p - q)y_2) \\
&= x_1 (py_1 + (1-p)y_2)
\end{align*}

Therefore $x_1 - x_2$ is a root and by symmetry $y_1 - y_2$ is a root. Therefore it remains to check the coefficient of $x_1y_1$ which is $a - pq$ to complete the factorisation.

For any two random variables taking two distinct values, we can find $a, q, p$ satisfying the relations above. We also note that $X$ and $Y$ are independent if $\mathbb{P}(X = x_i, Y = y_i) = \mathbb{P}(X = x_i)\mathbb{P}(Y = y_i)$. Since $x_1 \neq x_2$ and $y_1 \neq y_2$ and $\E(A B) = \E( A)\E (B) \Rightarrow a = pq$. But if $a = pq$, we have  $\mathbb{P}(X = x_1, Y = y_1) = \mathbb{P}(X = x_1)\mathbb{P}(Y = y_1)$ and all the other relations drop out similarly.

Consider
\begin{align*}
\mathbb{P}(A = -1, B = 1) &= \frac{1}{6} \\
\mathbb{P}(A = -1, B = -1) &= \frac{1}{6} \\
\mathbb{P}(A = 0, B = 0) &= \frac{1}{3} \\
\mathbb{P}(A = 1, B = -1) &= \frac{1}{6} \\
\mathbb{P}(A = -1, B = -1) &= \frac{1}{6}
\end{align*}