Hi just a quick question, I have in my notes that:

$\displaystyle Cov(X,Y) = Var(X)$ for any $\displaystyle X$

Is this true? Wouldn't this mean that you could just work out the variance of $\displaystyle X$ and not bother with $\displaystyle Cov(X,Y)$?

Cheers.

Edit just realised that it was in fact :

$\displaystyle Cov(X,X) = Var(X)$ for any $\displaystyle X$, oops!