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Homework Help: Expected value distribution

  1. Nov 3, 2012 #1
    1. The problem statement, all variables and given/known data
    (X,Y) has joint density f_X,Y ( (x,y) = (3/16)(x+2y) for 0<y<x , 0< x<2

    Evaluate E(Z) where Z = (3X+4Y)/(X+2Y)

    2. Relevant equations

    3. The attempt at a solution
    Getting the marginal densities
    f_X (x) =(3/8)x^2 for 0<x<2

    f_Y (y) = (3/2)+(3/4)y for 0<y<2

    Would I find the new distribution of 3X , 4Y, and 2Y then do the ratio distribution to solve for the distribution of Z? If this is correct, is there a shorter way?
    Last edited: Nov 3, 2012
  2. jcsd
  3. Nov 3, 2012 #2

    now use fx/z=f(x,z)/f(z) and integration over suitable rigions in the xy plane.
    recall that E[x/z]=integral(xfx/z)
  4. Nov 3, 2012 #3

    Ray Vickson

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    You don't need to find marginal distributions (which would be of no help anyway) and there is no need to find the distribution of Z (although doing so would not be harmful). Just use the so-called "theorem (or law) of the unconcious statistician"; use Google if you have never heard of this.

  5. Nov 3, 2012 #4
    I tought the symbol / denotes conditional variable,but maybe you ment fraction.
  6. Nov 3, 2012 #5
    Oh yeah! Using the marginals would lose information.

    So E[Z]=integral 0 to 2 integral 0 to x (3x+4y)/(x+2y) (3/16)(x+2y) dy dx = integral 0 to 2 (3/16)(3x+4y) dx = 5/2 Am I using the Unconscious Statistician theorem correctly? Thanks for all the help! <3
    Last edited: Nov 3, 2012
  7. Nov 3, 2012 #6

    Ray Vickson

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    Well, you should write (3x+4y)/(x+2y), not (3x+4y/x+ 2y) --- because this last means
    [tex] 3x + \frac{4y}{x} + 2y.[/tex]
    Also, when I do the y-integral I do not get 3x^2(3x+1)/8. This last integrates to 11/2 as x goes from 0 to 2, so I don't know how you get 5/2. (The 5/2 is correct; your computation is not.)

  8. Nov 3, 2012 #7
    Thanks again! I typed the function incorrectly, but wrote it down correctly.
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