Marcos

Posted on Thursday, 13 November, 2003 - 02:32 pm:

Can we express E(AB) in terms of E(A) and E(B) when A and B aren't independant?
So far I've managed to show (although I dunno if they're correct):
E(A + B) = E(A) + E(B), for any A and B
E(AB) = E(A)E(B), for independant A and B

Basically I'm trying to express Var(A + B) in terms of Var(A), Var(B), E(A) and/or E(B) only. I've gotten as far as showing that:
Var(A + B) = Var(A) + Var(B) + 2[E(AB) - E(A)E(B)]

[This expression reduces to Var(A + B) = Var(A) + Var(B) for independant events]

Thanks in advance,
Marcos

P.S. I'm talking about discrete random variables although I suspect these formulas should be applicable for continuous ones aswell.

David Loeffler

Posted on Thursday, 13 November, 2003 - 03:00 pm:

There is a quantity called the covariance, which is defined by

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

. Note that

Cov (X, X) = Var (X)

, and

Cov (X, Y) = 0

X

Y

are independent (but the converse isn't true.); so the covariance in some sense measures the strength of the relationship between

X

and

Y

. We can then define the correlation coefficient

ρ = Cov (X, Y) / \sqrt{Var (X) Var (Y)}

. Exercise:

- 1 \leq ρ \leq 1

David

Tristan Marshall

Posted on Thursday, 13 November, 2003 - 03:17 pm:

To emphasise to what David's just said, you've found that Var(A+B) = Var(A) + Var(B) + 2Cov(A,B), and this is as far as you can go unless you know more about A and B.