Probability Review - Expectations

[spitz]

Homework Statement

I'm trying to review basic probability; haven't looked at it in a couple of years. Am I on the right track here?

A and B are independent random variables, uniform distribution on [0,1]. Find: E(min(A,B))

2. The attempt at a solution

\displaystyle\int_{0}^{1}\int_{0}^{a}b\,db\,da + \displaystyle\int_{0}^{1} \int_{a}^{1}a\,db\,da

=\displaystyle\int_{0}^{1}\frac{a^2}{2}\,da+\int_{0}^{1}a-a^2\,da

=1/6+3/6-2/6=1/3

 

[vela]

Looks good to me.

 

[spitz]

Thanks. I also need to find E(|A-B|) and E((A+B)^2)

For the second one: E((A+B)^2)=E(A)+2E(A)E(B)+E(B) and so on ...

Can somebody give me a hint for: E(|A-B|)

 

[vela]

You'll need to break the integral up into two regions again, A<B and B>A. In one region, |A-B| = A-B, and in the other, |A-B| = B-A.

 

[Ray Vickson]

Thanks. I also need to find E(|A-B|) and E((A+B)^2)

For the second one: E((A+B)^2)=E(A)+2E(A)E(B)+E(B) and so on ...

Can somebody give me a hint for: E(|A-B|)

The claim E((A+B)^2)=E(A)+2E(A)E(B)+E(B) is false. For general bivariate (A,B) the correct result is E (A+B)^2 = E(A^2) + E(B^2) + 2E(AB). If A and B happen to be independent (or, at least, uncorrelated) then we have E(AB) = E(A) \cdot E(B), but for general (A,B) this fails. More generally, if A has variance \sigma_A^2, B has variance \sigma_B^2 and the pair (A,B) has covariance \sigma_{AB}, then
E(A+B)^2 = \mbox{Var}(A+B) + (EA + EB)^2 = \sigma_A^2 + \sigma_B^2 + 2 \sigma_{AB} + (EA + EB)^2.

RGV

 

[spitz]

Oh yes, I forgot to square A and B. For this problem they are independent (guess I should have mentioned that). So:
E((A+B)^2)=E(A^2)+E(B^2)+2E(A)B(A)