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Probability - Find PDF

  1. Mar 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Let the random variable X have the continuous probability density function f(x). find the probability density function of Y= F(x)


    2. Relevant equations

    I *think* we are meant to use something like [itex] h(y) = f(g^{-1}(y)) * ((dg^{-1}(y)) /dy)[/itex]

    (That is as good as I know how to format right now - I hope that's readable).


    But I'm given just f(x) and I'm being asked for F(x). And yet it's not asking for a CDF, it's asking for a PDF. So I'm really confused on what's being asked for here.

    (Unless this is another one of my book typos) Does the question even make sense?

    -Dave K
     
  2. jcsd
  3. Mar 5, 2013 #2

    Ray Vickson

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    Find the cdf of Y as ##F_Y(y) = P\{ Y \leq y \} = P\{ F(X) \leq y \},## then find the density of Y (the pdf) by differentiation. This method applies even when the formula you use fails (for example, because of intervals or points where f(x) = 0 but inside the range of X). Doing a bit of on-line research, or looking into a book on simulation, may be very helpful.
     
  4. Mar 7, 2013 #3
    So am I to integrate first for the CDF, and then differentiate back down again?

    I think it's perhaps so trivial that I'm getting confused. I even have the "back of book" answer but I don't know the reasoning.
     
  5. Mar 7, 2013 #4

    Ray Vickson

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    That is why I suggested you do a bit of on-line research or look into a book on simulation. Specifically, look at 'methods of generating random samples' or 'random variable generation' or something similar.
     
  6. Mar 7, 2013 #5
    I've looked everywhere online. But I'll try that.
     
  7. Mar 7, 2013 #6

    vela

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    Is Y=F(x) exactly how it was written in your book? If so, it looks like a typo or two. You may want to ask your professor for clarification.

    Perhaps you're supposed to find the cdf F(x) and then define the new random variable Y=F(X) (a function of X, not x) and find its density function.
     
  8. Mar 8, 2013 #7
    Yes, and the book is full of typos. And the professor gets lost in the middle of class. and..and.. you get the idea.

    I'm not sure. I'm really confused and frustrated. I have something written down and I'm handing it in.

    Thanks for the help everyone.

    -Dave K
     
  9. Mar 8, 2013 #8
    Dang David, next time try running it by me. I already had that class so I should be able to work it out if I can see the problem in the book. :)
     
  10. Mar 8, 2013 #9

    haruspex

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    No, I think Ray's interpretation is right. Yes, it should be Y=F(X), not F(x), but I doubt there's any relationship between f and F.
     
  11. Mar 8, 2013 #10

    vela

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    But it doesn't sound like F(X) was specified, which makes it impossible to find fY(y).
     
  12. Mar 8, 2013 #11

    Ray Vickson

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    No. Finding ##f_Y(y)## is "easy" (once you see it!); this is one of the important standard results that are used all the time in Monte-Carlo simulation.
     
  13. Mar 8, 2013 #12

    vela

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    I don't see it. You're given random variable X with probability density fX(x). Then you're asked to find fY(y) where Y=F(X). Haruspex has suggested F(X) bears no relation to fX(x), so it's completely arbitrary. You have virtually no information about Y other than it's a function of X.
     
  14. Mar 8, 2013 #13

    haruspex

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    The answer will involve F'.
     
  15. Mar 8, 2013 #14

    Ray Vickson

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    I will be happy to elucidate as soon as I am convinced the OP has already handed in his solution.
     
  16. Mar 8, 2013 #15

    Ray Vickson

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    In my previous responses I have been assuming all along that F is the cdf of X (just because of the notation you have used). If that is not what you meant then all my previous responses are irrelevant.
     
  17. Mar 9, 2013 #16
    I can attest to the fact that the assignment has been handed in. OP is on vacation now so you probably won't hear from him for a week.
     
  18. Mar 10, 2013 #17

    Ray Vickson

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    OK, so if X has pdf f(x) and cdf F(x), then Y = F(X) is uniformly distributed on (0,1). This is true even if X does not have a density (such as for discrete random variables, or mixed random variables that are partly continuous and partly discrete); the only thing that matters is that F(x) is a well-defined function (defined so as to be right-continuous). See, eg.,
    http://en.wikipedia.org/wiki/Inverse_transform_sampling
    for illustration and proof.

    This method is of fundamental importance in Monte-Carlo simulation.

    Note: the OP did not actually *say* that F was the cdf of X, but I just assumed he meant that. If not, then F is a more-or-less arbitrary function and there will be no simple, general result.
     
  19. Mar 15, 2013 #18
    Oh, the whole assignment was a mess, as you know. I'd be bugging you for every question.

    I am dropping the class anyway. My first drop, ever. I took some time on spring break to think about it. (I'm still there, just at a hotel in the middle of nowhere, PA). My final act was to get together with the class (we are down to about 8 people) and we agreed on some solutions to the assignment.

    Thanks everyone.

    -Dave K
     
  20. Mar 15, 2013 #19

    HallsofIvy

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    In post 17, Ray Vickson is assuming that F(x) is the integral of f(x). Is that true? My first thought was that they were unrelated functions.

    In that case, [tex]P(y< Y)= P(F(x)< Y= \int_{-\infty}^Y F(t)f(t)dt[/tex] and the PDF is the derivative of that, with respect to Y.
     
  21. Mar 15, 2013 #20
    The question was word for word as found in the book. There was no further information and I don't know what we were supposed to assume.
     
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