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What can we say about Covariance?

  1. Nov 1, 2013 #1
    I'm working on a problem that wants me to show that $$Cov(X,Y) = 0$$ and I am up to the point where I simplified it down to $$Cov(X,Y) = E(XY)$$. In other words, $$E(X)E(Y) = 0$$ to make the above true. My question is, what can we conclude if we have that the covariance of two random variables (not independent) is equal to the expectation of the product of those two random variables?
     
  2. jcsd
  3. Nov 1, 2013 #2
    [itex]Cov(X,Y) = \mathbb E(XY) - (\mathbb EX)(\mathbb EY) = \mathbb E[(X-\mathbb EX)(Y-\mathbb EY)][/itex]. Either of these two equivalent expressions could be the definition of covariance, depending who you ask.

    So really, all you can conclude is that at least one of [itex]X[/itex] or [itex]Y[/itex] is mean-zero. There's no information there whatsoever about how [itex]X,Y[/itex] are/aren't related.

    ~~~~~

    What does it mean to say [itex]Cov(X,Y)=0[/itex], on an intuitive level?? (This condition is sometimes said as "[itex]X,Y[/itex] are orthogonal", which has a geometric meaning behind it.) It means that, on average variation in [itex]X[/itex] tells us nothing about variation in [itex]Y[/itex].
     
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