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Poisson and Gamma Distributions

  1. Nov 3, 2012 #1
    Let [itex]Y|X[/itex] be a Poisson(X), and X be [itex]Gamma(\alpha, \beta)[/itex]. Find E(X|Y)...

    Since Y|X is Poisson(X), we have [itex]f(Y|X)= \frac{m^x e^{-m}}{x!}[/itex]...

    Since X is [itex]Gamma(\alpha, \beta)[/itex], we have [itex]f(x)= \frac{x^{\alpha-1} e^{-x/B}}{\Gamma(\alpha) \beta^{\alpha}}[/itex]...

    Since [itex]f(Y|X) = \frac{f(x,y)}{f(x)}[/itex] ====> [itex]f(x,y) = \frac{f(Y|X)}{f(x)}[/itex]...

    So now we have...

    [itex]f(x,y) = \frac{(\frac{m^x e^{-m}}{x!})}{\frac{x^{\alpha-1} e^{-x/B}}{\Gamma(\alpha) \beta^{\alpha}}}[/itex]

    So now I have to find f(y), but I'm sort of confused...because the poisson distribution is discrete while the gamma distribution is continuous...so do I need to handle them seperately?

    Thanks in advance.
     
  2. jcsd
  3. Nov 3, 2012 #2

    Ray Vickson

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    What is P{Y = k} for k = 0,1,2,... ? What is P{x < X < x+dx|Y=k} for k = 0,1,2, ... ?

    RGV
     
  4. Nov 4, 2012 #3
    When you say P(Y=k), do you mean P(Y=k|X)? (since we are told that it is poisson when it is conditioned by X...) However, if you mean that it is not P(Y=k|X)...actually that was what I was trying to compute by integrating out x from the continuous part...and then looking at the discrete part (but I was confused, because I wasn't sure if I could handle them seperately).

    P(Y=k) = [itex]\sum \frac{m^k e^{-m}}{k!}[/itex]

    and...

    [itex] P(x < X < x+dx|Y=k) = \frac{P(x < X < x+dx) ∩ P(Y=k)}{P(Y=k)}[/itex]
     
  5. Nov 4, 2012 #4

    Ray Vickson

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    Where do you get the formula
    [tex] P(Y=k) = \sum \frac{m^k e^{-m}}{k!}?[/tex] What is m? what are you summing over?

    The way I read the problem, we have
    [tex] P\{ Y = k | X = x \} = \frac{x^k e^{-x}}{k!}.[/tex]
    That is *exactly* what the problem says!

    Now, of course, you need to find P{Y = k}, an unconditional probability.

    RGV
     
  6. Nov 4, 2012 #5
    The reason that I wrote summation is because this is the cumulative distribution...not the probability mass function. So for continuous cases, the cdf is the integral of the pdf, and for discrete cases the cdf is the summation of the pmf...and what you're summing over depends on the question. Isn't that correct?
     
  7. Nov 4, 2012 #6

    Ray Vickson

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    OK, fine: you want to find the cdf of something. However, your expression is still lacking in two ways: (i) what is the summation index?; and (ii) what are the summation limits? Just writing Ʃ does not tell anyone whether you are summing over k for fixed m or over m for fixed k, and what the summation limits are.

    Anyway, nothing in this question asks about the cdf (unless there are parts of the question you have not written in your first post).

    At this point I am finished with this thread.

    RGV
     
  8. Nov 5, 2012 #7
    Actually, I wasn't trying to find the cdf at first. But when you asked me to find P{Y = k} and P{x < X < x+dx|Y=k}...P(anything) can be found using the cdf.

    Anyways, I'll probably ask my professor about this...
     
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