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Math Help - Expectation with 3 variables?

  1. #1
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    Expectation with 3 variables?

    I have this question:



    I know I need to find E[X1], E[X2] and E[X3] and then fiddle a bit to get the answers, but when I integrate do I just treat the two variables I am not interested in as constants?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Expectation with 3 variables?

    By definition the expected value of an g(x_{1},x_{2},x_{3}) is...

    E(G)= \int_{1}^{3} \int_1}^{2} \int_{0}^{1} g(x_{1},x_{2},x_{3})\ \frac{x_{1}\ x_{2}\ x_{3}}{3}\ d x_{1}\ d x_{2}\ d x_{3} (1)



    Marry Christmas from Serbia

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    Last edited by chisigma; December 16th 2011 at 11:05 AM.
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  3. #3
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    Re: Expectation with 3 variables?

    Hello,

    To be more precise on what chisigma wrote above, you'd be looking for the expected value of g(X_1,X_2,X_3) (in general, capital letters point to random variables, whereas small letters point to variables).

    But here, it's much better to calculate the first 2 moments of X_1,X_2,X_3, then say that since the joint pdf can be written as functions of only x1, x2 or x3, the three variables are independent. And hence this simplifies quite a lot the calculations, especially for the last 2 expectations.

    You'd be able to treat the other rv's as constants if you're considering a conditional expectation with respect to some given rv or sigma-algebra. But given the kind of problem you're working on, it's not something you've studied yet, maybe soon
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  4. #4
    MHF Contributor matheagle's Avatar
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    Re: Expectation with 3 variables?

    the marginal densities are obvious by inspection.
    You need not integrate out the other variables.

    Clearly the density of X1 is cX1 where c\int_1^3 xdx=1
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    Re: Expectation with 3 variables?

    Okay understood, thanks guys.
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  6. #6
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    Re: Expectation with 3 variables?

    IT's useful,thanks
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  7. #7
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    Re: Expectation with 3 variables?

    Just an update.

    I am using chisigmas method for this question as a means of practice..

    for the first one E[3X1+5X2+4X3] is there a shorthand alternative other than calculating the first 2 moments? As opposed to integrating:

    E[3X1+5X2+4X3] * (X1X2X3)/3

    ?
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  8. #8
    MHF Contributor matheagle's Avatar
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    Re: Expectation with 3 variables?

    the density of X3 by inspection, no integration, is  2X_3 on (0,1)
    that's a beta distribution, the mean is alpha/(apha+beta)=2/3
    you can do the same with the other two random variables, even though they aren't beta's

    Then  E(4X_3)=4(2/3)
    Last edited by matheagle; January 5th 2012 at 12:10 PM.
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  9. #9
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    Re: Expectation with 3 variables?

    Quote Originally Posted by matheagle View Post
    the density of X3 by inspection, no integration, is  2X_3 on (0,1)
    that's a beta distribution, the mean is alpha/(apha+beta)=2/3
    you can do the same with the other two random variables, even thougt they aren't beta's

    Then  E(4X_3)=4(2/3)
    Ah it makes perfect sense now, thanks for that.
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