What can we say about Covariance?

  • Context: Graduate 
  • Thread starter Thread starter Karnage1993
  • Start date Start date
  • Tags Tags
    Covariance
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
Karnage1993
Messages
131
Reaction score
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?
 
Physics news on Phys.org
[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.

Karnage1993 said:
My question is, what can we conclude if we have that the covariance of two random variables... is equal to the expectation of the product of those two random variables?

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].