An equilateral triangle has one vertex at the origin of $Oxy$ plane, one vertex at $(1;0)$, and one in the 1st quadrant. Find $cov(X;Y)$

Question

An equilateral triangle has one vertex at the origin of $Oxy$ plane, one vertex at $(1;0)$, and one in the 1st quadrant. Suppose you choose one of these three vertices uniformly at random (i.e. each vertex has the same chance to be chosen). Let $(X;Y)$ be the point you chose. Find $cov(X; Y ).$

I know that $Cov(X;Y) = E[(X-E[X])(Y - E[Y])]$ but I'm having trouble applying this to this particular problem.

Answer 1

It may be easier to use the fact that the covariance is $E(XY)-E(X)E(Y)$.

First we compute $E(X)$. The random variable $X$ is equally likely to be $0$, $1/2$, and $1$. By symmetry we have $E(X)=1/2$.

Now we compute $E(Y)$. The $y$-coordinate is $0$ with probability $2/3$, and $\sqrt{3}/2$ with probability $1/3$. Thus $E(Y)=\sqrt{3}/6$.

Finally, we compute $E(XY)$. We have $XY=0$ unless we have chosen the point $(1/2,\sqrt{3}/2)$. So $E(XY)=\sqrt{3}/12$.

Now put the pieces together. All this work for nothing!