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Multivariable Probability Distribution

  1. Apr 26, 2012 #1
    f(x,y) = ke^(-x-y), 0<x<∞ , 0<y<∞ , and x<y

    Find k to make this a PDF.

    So I set up: ∫(from 0 to ∞)∫(from 0 to ∞) [ke^(-x-y)]dxdy. Is this integral right?

    I also need to find the marginals of X and Y but my k has to be right first.
     
  2. jcsd
  3. Apr 26, 2012 #2
    I am not familiar with multi-variable, but if it is like single, you would need to set the integral = 1 and you could find your k.
     
  4. Apr 26, 2012 #3
    Yes, it's the same as single variable. I'm just unsure on my bounds of my integrals.
     
  5. Apr 26, 2012 #4
    [0,Infinity) seem reasonable.
     
  6. Apr 26, 2012 #5

    Ray Vickson

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    Your integration is over the two-dimensional set {(x,y}: 0 ≤ x < ∞, 0 ≤ y < ∞}, but you told us that f is defined only for x < y.

    RGV
     
  7. Apr 27, 2012 #6
    Oops! So it should be ∫from 0 to ∞ ∫ from 0 to x [f(x,y)dxdy] ?
     
  8. Apr 27, 2012 #7

    Ray Vickson

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    Yes, or the other way if you prefer (y from 0 to ∞ and x from 0 to y).

    RGV.
     
  9. Apr 27, 2012 #8
    What's the integral of the inner part evaluate to? I tried using integration by substitution but it's not working.
     
  10. Apr 27, 2012 #9

    Ray Vickson

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    Are you saying you don't know how to do the integral [itex] \int_0^x e^{-y} \, dy?[/itex] I hope you are joking.

    RGV
     
  11. Apr 28, 2012 #10
    I got that k=2, and checked it by seeing it integrated to 1.

    For the Marginal Distributions of X and Y, I got :
    f(x) = 2[e^(-x)-e^(-2x)] I [0,infinity) (x)
    f(y) = 2[e^(-y)-e^(-2y)] I [0,infinity) (y)

    Then, for the conditionals, I got:
    f(x|y) = e^(-x) / [1-e^(-y)] I[0,infinity) (x)
    f(y|x) = e^(-y) / [1-e^(-x)] I[0,infinity) (y)

    Can someone check those? Thanks!
     
  12. Apr 29, 2012 #11

    Ray Vickson

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    Your marginal for y is correct but your marginal for x is incorrect. Do NOT use the same f for three different functions. You already used f for f(x,y), so call the marginals something different, like g(x) and h(y), or fX(x) and fY(y).

    RGV
     
    Last edited: Apr 29, 2012
  13. Apr 29, 2012 #12
    Was my conditional of f(x|y) correct?

    Hmm is this the MD of x then?

    Integral from 0 to infinity [2e^(-x-y)dy]
    = -2(e^(-x-y)] from y=0 to y=infinity
    =2[1+e^(x)] I [0,infinity) (x)
     
    Last edited: Apr 29, 2012
  14. Apr 29, 2012 #13

    Ray Vickson

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    There is no point in answering questions about conditional densities until you get both marginal densities correct. The integral above is wrong.

    RGV
     
  15. Apr 29, 2012 #14
    Which part of f(x) = 2[e^(-x)-e^(-2x)] I [0,infinity) (x) is wrong?
     
  16. Apr 29, 2012 #15

    Ray Vickson

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    All of it. You seem to have claimed
    [tex] \int_x^{\infty} e^{-x-y}\, dy = e^{-x} - e^{-2x}.[/tex]
    but this is absolutely not correct.

    RGV
     
  17. Apr 29, 2012 #16
    Let's start over.

    I originally had for the marginal of X:
    ∫ from 0 to x of [2e^(-x-y)]dy
    =-2[e^(-x-y)], evaluated from y=0 to y=x
    = 2[e^(-x)-e^(-2x)] I[0,∞) (x)

    To check this, I integrated this and it came out to 1, so I'm not sure why this is wrong.
     
  18. Apr 29, 2012 #17

    Ray Vickson

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    It is wrong because you integrated y from 0 to x instead of from x to infinity.

    RGV
     
  19. Apr 29, 2012 #18
    Ok, hopefully here's the final check:

    fx(x) =
    ∫ from x to infinity [2e^(-x-y)]dy
    =-2[e^(-x-y)] from y=x to y=∞
    = 2e^(-2x) I[0,∞) (x)

    fy(y) = 2[e^(-y)-e^(-2y)] I [0,infinity) (y)

    f(x|y) = 2e^(-x-y) / 2[e^(-y)-e^(-2y)]
    = [e^(-x)] / (1-e^(-y)) I[0,∞) (x)

    f(y|x) = 2e^(-x-y) / 2e^(-2x)
    = e^(-y) / e^(-x) I[0,∞) (y)
     
  20. Apr 30, 2012 #19

    Ray Vickson

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    These look OK now, except you should get rid of those annoying I[0,∞)(x), etc., symbols. Just say in words that x and y are >= 0.

    RGV
     
  21. Apr 30, 2012 #20
    Thanks a bunch!

    I, too, hate the annoying indicator functions, but my professor seems the need to "require" us to use them.
     
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