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Probability function of two random variables, another non-convergent integral

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data

    The joint probability density function of the random variable (X, Y) is given by:

    [tex] f(x,y) = \frac{2x}{y^2} \text{where} \; 0 \leq x\leq 1 \; \text{and} \; y\geq 1 [/tex]
    and 0 elsewhere.

    Find the probability density function of the folowing random variable:

    U=X+Y


    2. Relevant equations

    I'm not sure of this equation, (have I mentioned my book should be burned?) but I think ultimately I'm looking for
    [tex] \int_{-\infty }^{\infty } f(u-v,v) \mathrm{d} v [/tex]

    3. The attempt at a solution

    [tex] U=X+Y [/tex]
    [tex] x= u-v [/tex]
    [tex] y=v [/tex]

    the jacobian of the transformation is 1, so we can leave out further mention of it in equations.

    My integral is [tex] \int_a^b \frac{2u-v}{v^2} \mathrm{d} v [/tex]
    I'm not sure about the bounds. Since v=y, should the bounds on v equal the bounds on y? That is, from 1 to infinity?

    I already tried that and got a non-convergent integral (the sad story of my day today) :(
    [tex] \int_1^{\infty } \frac{2u-v}{v^2} \mathrm{d} v = [/tex]
    [tex] \int_1^{\infty } \frac{2u}{v^2} \mathrm{d} v - \int_1^{\infty } \frac{1}{v} \mathrm{d} v = [/tex]
    [tex] 2u \int_1^{\infty } v^{-2} \mathrm{d} v - \int_1^{\infty } \frac{1}{v} \mathrm{d} v = [/tex]
    [tex] -2u(\frac{1}{v} |_1^\infty ) - \mathrm{ln} v |_1^{\infty } [/tex]
    And since [itex] \mathrm{ln} (\infty ) = \infty [/itex], part of my problem says [itex] 0-\infty [/itex] Which is not good...
     
    Last edited: Nov 26, 2011
  2. jcsd
  3. Nov 26, 2011 #2

    vela

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    Two ways to solve the problem:

    1. Find FZ[z] = P(Z ≤ z) and then fZ(z) = FZ'(z).
    2. Find f(z,v) and integrate out v to find the marginal probability density fZ(z).

    For either case, I suggest you sketch the domain of f(x,y) in the xy-plane as well as the line x+y=c.
     
  4. Nov 26, 2011 #3
    I changed Z to U because I just noticed my formulas use the letter U. I don't think that changed anything other than improving my weak grasp on the problem.

    I don't really understand your first suggestion, Vela.

    As for the second suggestion, I think that's what I'm already doing, isn't it? Maybe I don't understand though. I'm really shaky on this change of variable thing.

    As for the sketch, I sketched the domain of f(x,y) and shaded in a region above y=1 and between x=0 and x=1. As for x+y= c, I didn't know how to graph that since I don't know what c is.
     
  5. Nov 26, 2011 #4

    I like Serena

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    Your integral should be written as:
    [tex]\int\limits_{\stackrel{0 \le u-v \le 1}{v \ge 1}} \frac{2(u-v)}{v^2} \mathrm{d} v[/tex]
    This is due to the the definition of f(x,y) where it is non-zero.
    Where that is exactly depends on the value of u.

    This not so easy, and you'll have to divide it into different cases.
    For this you need to draw your domain with v and (u-v) on the axes (hence vela's suggestion to replace u-v by z).

    First you should consider the case that u=1 and u=2, and after that the 3 cases that u is less, in between, or more.
     
  6. Nov 26, 2011 #5

    vela

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    I was just noting two different ways you can solve the problem. The first method is bit easier to grasp intuitively, at least to me. As you noted, you're taking the second approach. Either way, it should work out.
    The c is a constant. When you integrate f(x,y) along the line x+y=c — or equivalently, the line u=c in the uv-plane, you're calculating the probability that U=c.

    As ILS noted, you're going to have to consider different cases depending on the value of U to figure out what limits you should be integrating over.
     
  7. Nov 26, 2011 #6
    Why isn't it as simple as: since v=y, and y is greater than or equal to 1, v is greater than or equal to 1?
     
  8. Nov 26, 2011 #7

    I like Serena

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    But it is!

    So you get:
    [tex]\def\d{\textrm{ d}}
    \int_{-\infty}^\infty f(u-v,v) \d v
    = \int_{1}^\infty f(u-v,v) \d v[/tex]

    It's only that the next step is something like:
    [tex]\int_{1}^\infty f(u-v,v) \d v
    = \int_{1}^\infty {2(u-v) \over v^2} \cdot (1 \textrm{ if }(0 \le u-v \le 1), 0 \textrm{ otherwise}) \d v[/tex]
     
  9. Nov 26, 2011 #8
    but u is defined as x+v, and x=u-v, so, I can't nail down u. it's slippery like soap!
     
  10. Nov 26, 2011 #9
    I don't even know what I'm supposed to do :(
     
  11. Nov 26, 2011 #10

    I like Serena

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    Consider 3 cases: u<1, 1<u<2, u>2.

    Start with u<1.
    What is the range of (u-v) if v>1?
     
  12. Nov 26, 2011 #11
    0 to -infinity
     
  13. Nov 26, 2011 #12

    I like Serena

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    So....?

    What does that tell you about fU(u) if u<1?
     
  14. Nov 26, 2011 #13
    What's f sub u of u?
     
  15. Nov 26, 2011 #14

    I like Serena

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    I made it up.

    I had to think of something, since you didn't make something up yourself.
    It's based on this sentence in your problem:
     
  16. Nov 26, 2011 #15
    Maybe that's what the book is calling h(u). h(u) is found by the integral I'm trying to solve. I still don't see how the bounds of u come into play since the integral is in terms of v and i can't solve it.
     
  17. Nov 26, 2011 #16

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    f(u-v,v)=0 if (u-v) and v are not both in the proper ranges as defined for f(x,y).

    For u<1, (u-v) is not in range, so f(u-v,v)=0, therefore h(u)=0.
     
  18. Nov 26, 2011 #17

    I like Serena

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    You already have:
    [tex]\def\d{\textrm{ d}}
    h(u)= \int_{1}^\infty f(u-v,v) \d v[/tex]

    Perhaps you should do a substitution with x=u-v (with respect to integration of v).
     
  19. Nov 26, 2011 #18
    Okay, but for larger values of v, u-v isn't in range for u between 1 and 2, and for any fixed u greater than 2, there are v's for which it isn't in range again. So, I dont get it still.
     
  20. Nov 26, 2011 #19
    I don't know what you mean. This might be one of those times when you take pity on me and give me more info than usual, since I'm jus too lost to see what you're pointing to.
     
  21. Nov 27, 2011 #20

    I like Serena

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    Okay. I'll give you the next step.


    We have:
    [tex]\def\d{\textrm{ d}}
    h(u)= \int_{1}^\infty f(u-v,v) \d v[/tex]

    Substituting x=u-v, we get:
    [tex]h(u) = \int_{-\infty}^{u-1} f(x, u-x) \d x[/tex]

    Now we have 3 cases: u-1<0, 0<u-1<1, u-1>1.
    Note that the y-parameter of f, which is (u-x) is always greater than 1.


    Case 1: u-1<0 (or u<1)

    This means that within the bounds of the integral x<0 and so f(x,u-x)=0, so the integral is zero too.

    So h(u)=0 if u<1.



    Case 2: 0<u-1<1 (or 1<u<2)

    This means that within the bounds of the integral for x such that 0<x<u-1, we have f(x,u-x)=2x/(u-x)^2, so the integral is:
    [tex]h(u) = \int_{0}^{u-1} f(x, u-x) \d x = \int_{0}^{u-1} {2x \over (u-x)^2} \d x[/tex]

    I leave it to you to evaluate this.



    Case 3: u-1>1 (or u>2)

    This means that within the bounds of the integral for x such that 0<x<1, we have f(x,u-x)=2x/(u-x)^2, so the integral is:
    [tex]h(u) = \int_{0}^{1} f(x, u-x) \d x = \int_{0}^{1} {2x \over (u-x)^2} \d x[/tex]

    Again I leave it to you to evaluate this.
     
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