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Finding P(X1<X2<X3) and P(X1=X2<X3)

  1. Nov 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Let f(x1, x2, x3) = e-(x1+x2+x3), 0<x1,2,3<infinity, zero elsewhere be a joint pdf of X1, X2, X3.

    Compute P(X1< X2< X3) and P(X1= X2< X3)

    Determine the joint mgf.

    3. The attempt at a solution

    P(X1< X2< X3) = triple integral of e-(x1+x2+x3) dx2dx1dx3 as x2 goes from x1 to x3, x1 goes from 0 to x2 and x3 goes from x2 to infinity.

    When I solve this integral I get 0 for an answer. The back of the book suggests the answer is 1/6. Clearly, I am setting up the problem wrong, but I don't know where my mistake is. Similarly, I'm having trouble setting up the second part of the question as well.

    Joint mgf = E(et1x1+t2x2+t3x3) but this integral is not easy by hand. Is there another way to do it?
     
  2. jcsd
  3. Nov 25, 2009 #2
    Yeah, your integration limits are incorrect. For P(x_1<x_2<x_3), start from x_1. Say it goes from zero to infinity. Then x_2 has to go from x_1 to infinity, and likewise x_3 goes from x_2 to infinity.

    What's hard about solving the MGF?

    [tex] E(e^{t(x_1+x_2+x_3)}) = \int dx_1 dx_2 dx_3 f(x_1,x_2,x_3) e^{t(x_1+x_2+x_3)} = \int dx_1 dx_2 dx_3 \exp [(t-1)(x_1+x_2+x_3)] [/tex]

    I'm sure you know how to integrate an exponential function.
     
  4. Nov 25, 2009 #3
    Oh! I was doing it with three different t's ... t1 t2 and t3 which made it a million times harder. That makes sense. Thanks so much!
     
  5. Nov 25, 2009 #4
    I guess you do need multiple t's. It doesn't make it any harder though.
     
  6. Nov 25, 2009 #5

    statdad

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    No, you don't need multiple ts.

    [tex]
    \phi(t)_{X_1+X_2+X_3} = \idotsint e^{t(x_1+x_2+x_3)} f(x_1,x_2,x_3)\,dx_1dx_2dx_3
    [/tex]

    Aren't your variables independent exponentials?
     
  7. Nov 25, 2009 #6
    Yep, finally got the answer. Thank you everybody for your help!
     
  8. Sep 22, 2011 #7
    What about P(X1=X2<X3)?
     
  9. Sep 22, 2011 #8

    Mute

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    The random variables are drawn from a continuous distribution, and are independent. Do you know what the probability is that a continuous random variable will be exactly some specified number?
     
  10. Sep 22, 2011 #9
    zero, it must be an interval... So then would it be P(X1=X2<X3) = P(X1<X3)*P(X2<X3) or even P(0<X3)?
     
  11. Sep 22, 2011 #10

    Ray Vickson

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    You are correct that this is the joint mgf (with three different t's). If you had t1 = t2 = t3 = t, as some others have suggested, you would be finding the mgf of the single random variable X = X1 + X2 + X3. Anyway, the integral is *very easy* by hand.

    RGV
     
  12. Oct 30, 2012 #11
    I have a question about the mgf...

    So this is the integral that we need to integrate, right?

    [itex]\int_0^{\infty} \int_0^{\infty} \int_0^{\infty} e^{x_1(t_1-1)} e^{x_2(t_2-1)} e^{x_3(t_3-1)} dx_1 dx_2 dx_3 [/itex]

    But I'm having some trouble with this, because this is what I get after integrating...

    [itex]\frac{1}{t_1-1}e^{x_1(t_1-1)} |_0^{\infty} \frac{1}{t_2-1}e^{x_2(t_2-1)} |_0^{\infty} \frac{1}{t_3-1}e^{x_3(t_3-1)} |_0^{\infty}[/itex]...but if we plug in infinity to [itex]\frac{1}{t_1-1}e^{\infty(t_1-1)}[/itex]...don't we get infinity, since e increases forever?
     
  13. Nov 1, 2012 #12
    That depends on whether t-1 is positive or negative.
     
  14. Nov 1, 2012 #13
    Thanks a lot...
     
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