The Cov(X,Y) on a square dart board

[BookMark440]

A point (X, Y) is picked at random, uniformly from the square with corners at (0,0), (1,0), (0,1) and (1,1). Compute Cov{X, Y}.

I think of this as darts thrown at a unit square dart board.

Cov(X,Y) = E(XY) - E(X)E(Y).

I compute that E(X)=E(Y)= 0.5 using the integral xf(x).

But I cannot figure out how to approach computing E(XY). Or is there a better strategy for solving this?

THANKS!

 

[tiny-tim]

Welcome to PF!

I compute that E(X)=E(Y)= 0.5 using the integral xf(x).

But I cannot figure out how to approach computing E(XY).

Hi BookMark440! Welcome to PF!

(I'm not sure exactly what you mean by f(x))

If you got E(X) from ∫x, why can't you get E(XY) from ∫∫xy ?

 

[BookMark440]

Thanks. I did that and ended up getting zero for an answer, which always makes me think I made an error. In this case, I guess it just verifies that X,Y are independent.

 

[KingNothing]

A covariance of zero is very common in textbook examples.