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Expected Values . E(X^2(3Y-1))

  1. Jul 30, 2011 #1
    Expected Values..... E(X^2(3Y-1))

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

    What is the Expected Value of E(X2(3Y-1))

    2. Relevant equations

    Properties of Expected Values.
    E(X+c) = E(X) + c
    E(X+Y) = E(X) + E(Y)


    3. The attempt at a solution

    I tried googling for a few of the properties I could use but I'm not sure which ones to use, or maybe I haven't found the required property?
    I have been given the data and worked out the marginal probability mass functions so I can work out E(X), E(Y), and E(XY)

    But I was wondering if I could decompose the problem quickly using properties of expected values, instead of having to do all the math involved manually calculating the expected value?
    If so, which properties should I take a look at in order to achieve this? Thanks!
     
  2. jcsd
  3. Jul 30, 2011 #2
    Re: Expected Values..... E(X^2(3Y-1))

    Ok I gave it a crack using basic expansion, lemmy know if im close :)

    E(X2(3Y-1))
    E(3X2Y-X2)
    3E(X2Y)-E(X2)

    Then from there its a fairly simple process of taking the integrals of the various functions. In my examples case the marginal prob dens functions are only defined for certain values so all I gotta do is E = 3sum(X2.y(x)) - sum(X2.y(x))and badabing.

    Yay or nay? :D
     
  4. Jul 30, 2011 #3

    Ray Vickson

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    Homework Helper

    Re: Expected Values..... E(X^2(3Y-1))

    You need to work out E(X^2*Y). If X and Y are independent, this would be E(X^2)*E(Y), but it they are correlated, there is no simple expression. By the way, you say you have worked out the marginal probability mass functions so can work out E(XY). This statement (the "so can ..." part) is false if X and Y are not independent (or, at least, uncorrelated), because in that case you need the full bivariate probability mass function f(x,y) in order to work out E(XY).

    RGV
     
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