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Math Help - Poisson process: quotient of two gamma r.v.'s

  1. #1
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    Poisson process: quotient of two gamma r.v.'s

    Let {N(t): t≥0} be a Poisson process of rate 1 and let S_r be the time to the rth point. Calculate the density function of X = S_2/S_6 using the Jacobian transformation method.

    To use the Jacobian method, I need to define another random variable Y that is a function of S_2 and S_6.
    X=S_2/S_6
    Y=???
    I know that S_2~gamma(2,1) and S_6~gamma(6,1), but I am having some trouble defining the other random variable Y, and the problem is that S_2 and S_6 are not independent.
    The Jacobian method expresses the joint density of X and Y in terms of S_2 and S_6, so I must first obtain the joint density of S_2 and S_6. But now it seems like that I have no way of obtaining the joint density of S_2 and S_6 because they are NOT independent random variables (S_2<S_6 always).
    So how can we solve this problem using the Jacobian method?

    Any help is much appreciated!
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  2. #2
    MHF Contributor matheagle's Avatar
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    Usually Y is either S_2 or S_6,
    S_6 might be easier
    AND you start with the joint distribution of S_2 and S_6
    You must find that first, and yes they are dependent.


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  3. #3
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    Quote Originally Posted by kingwinner View Post
    the problem is that S_2 and S_6 are not independent.
    But S_2 and S_6-S_2 are independent, and the second one has same distribution as S_4, so let's call it S'_4, so that \frac{S_2}{S_6}=\frac{S_2}{S_2+S'_4} and then you can do the usual method (like MathEagle said, Y may be chosen to be S_2 or S'_4).
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  4. #4
    MHF Contributor matheagle's Avatar
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    I was going to add that S6-S2 is indep of S2 since the time periods are separate, but I wanted kingswinner to figure that out.
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  5. #5
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    Quote Originally Posted by matheagle View Post
    I was going to add that S6-S2 is indep of S2 since the time periods are separate, but I wanted kingswinner to figure that out.
    Oh, sorry to spoil your pedagogy... I feel like we're quits now, since the same happened on an other thread in reverse (that was a couple of weeks ago)

    Best wishes,
    Laurent.
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  6. #6
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    Quote Originally Posted by Laurent View Post
    But S_2 and S_6-S_2 are independent, and the second one has same distribution as S_4, so let's call it S'_4, so that \frac{S_2}{S_6}=\frac{S_2}{S_2+S'_4} and then you can do the usual method (like MathEagle said, Y may be chosen to be S_2 or S'_4).
    Thanks matheagle and Laurent for the helpful comments!
    But I have some more questions:

    1) If I take Y to be (S2 + S4') which is the denominator, is that also OK? Will I get the correct answer?

    2) I don't quite follow your argument that (S6 - S2) would have gamma(4,1) distribution.
    What I have learned is the following...
    Theorem: if X1~gamma(r1,lambda) and X2~gamma(r2,lambda), with X1 and X2 INDEPENDENT, then the SUM (X1+X2)~gamma(r1+r2,lambda).

    But I am not too sure about a SUBTRACTION of gamma r.v.'s, particuarly when S6 and S2 are DEPENDENT?? Is the theorem above applicable?

    Thank you very much!
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  7. #7
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    Quote Originally Posted by kingwinner View Post
    2) I don't quite follow your argument that (S6 - S2) would have gamma(4,1) distribution.
    You should try to picture what S6-S2 is.

    (and no, you can't "substract", whatever that would mean)
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  8. #8
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    1) Actually, I am puzzled about this for a long time. The Jacobian method always requires TWO random variables. But we are almost always only given one, and somehow we have to "invent" another random variable, but a lot of times I have no idea how...
    Is there any general rule or hint about how this second random variable can be constructed?
    X=S2/(S2+S4')
    Y=???


    How do we know what to pick for Y, and how can we know whether it is going to work or not?
    Can we take Y=S4' ? How about Y=S2+S4'? Are there any other possible choice of Y that would work?



    2) I can picture S6-S2 as the time between the 2nd and 6th point, but I don't get why S6-S2 is necessarily gamma distributed.
    Suppose X1~gamma(6,1), X2~gamma(2,1), and X1 and X2 are NOT independent. Does this imply that X1-X2 ~gamma(6-2,1)=gamma(4,1) ? If so, how can we prove it? I haven't seen this theorem before...

    Thanks for any help!
    Last edited by kingwinner; December 12th 2009 at 02:45 AM.
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