Let $X$ and $Y$ be two discrete random variables with respective probability distributions $\Bbb P(X=x_1)=p_1,\Bbb P(X=x_2) = 1 – p_1$ and $\Bbb P(Y=y_1) = p_2, \Bbb P(Y=y_2)=1-p_2$. Show that if they are uncorrelated then they are independent.
Since $X$ and $Y$ are uncorrelated we find that $\Bbb E(XY) = \Bbb E(X) \Bbb E(Y)$. So $$\Bbb E(XY) = [x_1p_1 + x_2 (1-p_1)][y_1p_2 + y_2 (1-p_2)].$$
So $\Bbb E(XY) = x_1 y_1 p_1p_2 + x_1 y_2 p_1 (1- p_2) + x_2 y_1 p_2(1-p_1) + x_2y_2 (1-p_1)(1-p_2)$. On the other hand $\Bbb E(XY) = x_1 y_1 \Bbb P(XY=x_1y_1) + x_1 y_2 \Bbb P(XY=x_1y_2) + x_2 y_1 \Bbb P(XY=x_2y_1) + x_2 y_2 \Bbb P(XY=x_2y_2)$.
From these two relations how can I conclude that $X$ and $Y$ are independent? Please help me in this regard.
Thank you very much.
Best Answer
First let us do it for $X', Y'$ such that $x_1=y_1=1$ and $x_2=y_2=0$.
Then: $$0=\mathsf{Cov}(X',Y')=\mathbb EX'Y'-\mathbb EX'\mathbb EY'=P(X'=1\wedge Y'=1)-P(X'=1)P(Y'=1)$$
This shows that the events $\{X'=1\}$ and $\{Y'=1\}$ are independent and since both random variables take a.s. at most two distinct values this implies that $X'$ and $Y'$ are independent.
For this note that $P(A\cap B)=P(A)P(B)$ implies that: $$P(A^{\complement}\cap B)=P(B)-P(A\cap B)=P(B)-P(A)P(B)=(1-P(A))P(B)=P(A^{\complement})P(B)$$
In general if $X'$ and $Y'$ are independent then so are $aX'+b$ and $cY'+d$.
From this we conclude that also $X=(x_1-x_2)X'+x_2$ and $Y=(y_1-y_2)Y'+y_2$ are independent.