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Math Help - Expected Value

  1. #1
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    Expected Value

    Hi,

    Can anyone please help with the question?

    Suppose E(U|X) = X^2. Suppose that E(X) = 0, Var(X) = 1 and E(X^3) = 0

    What is the E(UX)?

    Thanks alot for helping!
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  2. #2
    Moo
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    Hello,

    E[UX]=E[E[UX|X]]=E[XE[U|X]]=E[X^3]

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  3. #3
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    @ Moo - Thanks for your response.

    I do understand your solution.

    But I am not confortable with all the theorms/or rather properties of Expectation (in case of joint distributions). I basically get confused and at times rely on intuition.

    Will it be possible for you to reccomend me
    1. A source/book I can read to understand this better
    2. For e.g. Why is
    E(Y) = E(E(Y|X))
    And then how you got
    E[E[UX|X]]=E[XE[U|X]] - I understand it intuitively but not rigorously.

    Thanks
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  4. #4
    MHF Contributor matheagle's Avatar
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    E(Y) = E(E(Y|X))
    is easy to prove, just write the conditional expectation and integrate/sum a second time
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  5. #5
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    Thanks.

    But where I'm weak is a better understanding of Joint Distributions. I tried referring some online resources but couldn't get a very clear idea. And, I'm not attending any university so little hard for me to find out where can you read up on it. Thanks
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  6. #6
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    Thanks for your help!
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  7. #7
    MHF Contributor matheagle's Avatar
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    In the continuous setting....

    E(Y|X)=\int yf_{Y|X}(y)dy

    =\int y{f_{X,Y}(x,y)\over f_X(x)}dy

    now take the expectation wrt x...

    E(E(Y|X))=\int E(Y|X)f_X(x)dx

    and wave the Fubini wand...

    E(E(Y|X))=\int \int y{f_{X,Y}(x,y)\over f_X(x)}dyf_X(x)dx

    =\int \int yf_{X,Y}(x,y)dydx=E(Y)
    Last edited by matheagle; September 17th 2010 at 01:55 PM.
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  8. #8
    Moo
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    aman_cc,

    I have a very formal proof of the fact that E[E[X|Y]]=E[X] but it's kind of long, and very theoretical. You also need measure theory knowledge. Well I don't think it's a relevant one...

    The general idea is to say that in a probability space \Omega,\mathcal A,\mathbb P) and given a sigma-algebra \mathcal B, and X a random variable \in L^1(\Omega,\mathcal A,\mathbb P), then there exists a unique random variable Z=E[X|\mathcal B such that for any B\in \mathcal B,\int_B Z d\mathbb P=\int_B Xd\mathbb P.
    In particular, for B=\Omega, we get the desired equality.

    Note that when we write E[X|Y], it's in fact E[X|\sigma(Y), where \sigma(Y) is the sigma-algebra generated by Y. So it's in place of \mathcal B in the above paragraph.

    Also note that with the Aussie singer's method, matheagle, we have to assume that these random variables have a pdf.



    For the second point, the proof doesn't look that difficult, it's just long and painful... I don't know how to explain it with words...
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  9. #9
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    Thanks. I did read up about this proof in particular.

    If there is a good book at an intermediate level to read up on probabolity thoery and expectations, distribution (joint) etc please do refer that to me. I like books which start with clear definitions of concepts / axioms and then proceed to prove such theorms for there. You can consider me a novice at measure theory.
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  10. #10
    Moo
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    Hi,

    I know three books in English that look good. But I don't know if they're the best. My library isn't specialized in English books

    Amazon.com: Probability with Martingales (Cambridge Mathematical Textbooks) (9780521406055): David Williams: Books
    Amazon.com: Probability and Measure, 3rd Edition (9780471007104): Patrick Billingsley: Books
    Measure, Integral and Probability: Amazon.fr: Marek Capinski, Peter E. Kopp, Ekkehard Kopp: Livres en anglais

    well they're expensive, but I hope the library in your university has some of them !
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