# Homework Help: Probability Review - Expectations

1. Feb 5, 2012

### spitz

1. The problem statement, all variables and given/known data

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

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

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

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

2. Feb 6, 2012

### vela

Staff Emeritus
Looks good to me.

3. Feb 7, 2012

### 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: $$E(|A-B|)$$

Last edited: Feb 7, 2012
4. Feb 7, 2012

### vela

Staff Emeritus
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.

5. Feb 7, 2012

### Ray Vickson

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

6. Feb 7, 2012

### spitz

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