independence and uncorrelation proof Suppose $X$ and $Y$ are two indicator random variables (an indicator random variable for the event $A$ is an r.v. receiving $1$ if $A$ occurs and $0$ otherwise). Show that $X$ and $Y$ are independent if and only if $\text{Cov}(X,Y)=0.$
Also, can we claim $X,Y$ are independent if $E[XY] = E[X] E[Y]$ ?
 A: If we have independence, and the appropriate expectations exist, the covariance is always $0$. So we need only show that in the case of indicator random variables, the converse holds. Of course it does not hold in general. For a simple example, let $X=-1$, $0$, and $1$, each with probability $1/3$, and let $Y=|X|$. The $X$ and $Y$ are uncorrelated, but not independent. 
We want to prove that for any $x$ and $y$, the probability that $X=x$ and $Y=y$ is equal to $\Pr(X=x)\Pr(Y=y)$. Luckily, there are only $4$ cases to check, since $x$ and $y$ can only take on the values $0$ and $1$. And it will turn out that there really is only one case to check. 
Let $\Pr(X=1)=p_X$ and $\Pr(Y=1)=p_Y$. Then $E(X)=p_X$ and $E(Y)=p_Y$. 
The expectation $E(XY)$ of $XY$ is the probability that $XY=1$. This is the probability that $X=1$ and $Y=1$, since that's the only way the product can be $1$.
If the covariance is $0$, then $E(XY)=E(X)E(Y)$, and therefore the probability that $X=1$ and $Y=1$ is $p_Xp_Y$. This takes care of the case $x=1$, $y=1$. 
Next we take care of the case $x=1$, $y=0$ by showing that the probability that $X=1$ and $Y=0$ is $p_X(1-p_Y)$. The probability that $X=1$ and $Y=0$, plus the probability that $X=1$ and $Y=1$ is just $p_X$. So by the previous calculation, the probability that $X=1$ and $Y=0$ is $p_X-p_Xp_Y$. This is $p_X(1-p_Y)$, as desired. 
Now dealing with the last two cases is easy. The case $x=0$, $y=1$ is dealt with just like $x=1$, $y=0$, and the case $x=0$, $y=0$ in almost the same way. 
