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Probability Review - Expectations

  1. Feb 5, 2012 #1
    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

    [tex]\displaystyle\int_{0}^{1}\int_{0}^{a}b\,db\,da + \displaystyle\int_{0}^{1} \int_{a}^{1}a\,db\,da[/tex]

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

    [tex]=1/6+3/6-2/6=1/3[/tex]
     
  2. jcsd
  3. Feb 6, 2012 #2

    vela

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    Looks good to me.
     
  4. Feb 7, 2012 #3
    Thanks. I also need to find [tex]E(|A-B|)[/tex] and [tex]E((A+B)^2)[/tex]

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

    Can somebody give me a hint for: [tex]E(|A-B|)[/tex]
     
    Last edited: Feb 7, 2012
  5. Feb 7, 2012 #4

    vela

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    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.
     
  6. Feb 7, 2012 #5

    Ray Vickson

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

    RGV
     
  7. Feb 7, 2012 #6
    Oh yes, I forgot to square [itex]A[/itex] and [itex]B[/itex]. For this problem they are independent (guess I should have mentioned that). So:
    [tex]E((A+B)^2)=E(A^2)+E(B^2)+2E(A)B(A)[/tex]
     
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