Question:
Suppose that X is a random variable taking on only the values 0 and 1 and that Y is a random variable taking on only the values −1,0,1. It is known that Var(X) = 1/4, Var(Y ) = 1/2 and E[Y ] = 0. In addition, it is known that P(X = 0|Y = 0) = 0. Find Cov(X,Y ).
I can't seem to figure this question out. I have done pretty much everything that comes to mind, but always reached a dead end. The only thing I've conclusively managed to do is find out the probability mass function of both X and Y.
Let us draw joint p.m.f. as a table
\begin{array}{lr|c|c|c|c} % \hline & Y=b & -1 & \, 0\, & \, 1 \,& \mathbb P(X=a) \\[-1mm] X=a & & & & & \\ \hline 0 & & x & y & z & x+y+z=p \\ \hline 1 & & u & v & w & u+v+w=1-p \\ \hline \mathbb P(Y=b)& & x+u & y+v & z+w & 1 \\ \end{array} and start from $\mathbb P(X=0\mid Y=0)$. Can you find $y=\mathbb P(X=0,Y=0)$ from here?
Next, $Var(X)=1/4=p-p^2$, Find $p$ from here.
$\mathbb E[Y]=0$ implies both $x+u=z+w$ and $$Var(Y)=\mathbb E[Y^2]=(-1)^2(x+u)+1^2(z+w)=x+u+z+w=\frac12.$$ Then $y+v=\frac12$ too. So $v=\frac12$ and $x+u=z+w=\frac14$.
\begin{array}{lr|c|c|c|c} % \hline & Y=b & -1 & \, 0\, & \, 1 \,& \mathbb P(X=a) \\[-1mm] X=a & & & & & \\ \hline 0 & & & 0 & & \frac12 \\ \hline 1 & & & \frac12 & & \frac12 \\ \hline \mathbb P(Y=b)& & \frac14 & \frac12 & \frac14 & 1 \\ \end{array} You can fill the rest cells in the table.