Results 1 to 3 of 3

Thread: Gamma/Poisson Distribution- Probability

  1. #1
    Member
    Joined
    Aug 2007
    Posts
    239

    Gamma/Poisson Distribution- Probability

    Let $\displaystyle Y $ be a gamma random variable with parameters $\displaystyle (s, \alpha) $. That is, its density is $\displaystyle f_{y}(y) = Ce^{- \alpha y}y^{s-1}, y > 0 $, where $\displaystyle C $ is a constant that does not depend on $\displaystyle y $. Suppose also that the conditional distribution of $\displaystyle X $ given that $\displaystyle Y = y $ is Poisson with mean $\displaystyle y $. That is $\displaystyle P \{X = i| Y = y \} = e^{-y}y^{i}/i!, \ i \geq 0 $. Show that the conditional distribution of $\displaystyle Y $ given that $\displaystyle X = i $ is the gamma distribution with parameters $\displaystyle (s+i, \alpha + 1) $.

    So the problem is to find $\displaystyle P \{Y = y | X = i \} $ and show that it is the gamma distribution. Now $\displaystyle P \{Y =y |X =i \} = \frac{P \{Y =y, \ X = i \}}{P \{X = i\}} $. So I am guessing that the $\displaystyle \frac{1}{P \{Y = y \}} $ and the $\displaystyle \frac{1}{P \{X = i \}} $ are the factors that affect whether the distribution is Poisson or Gamma.

    PS: Is R the best stats software in general? Is Jmp good?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    using Baye's theorem we get:

    $\displaystyle
    P\{ Y = y|X = i\} = \frac{{P\{ X = i|Y = y\} f_y (y)}}{{\int\limits_0^\infty {P\{ X = i|Y = y\} f_y (y)dy} }}
    $

    $\displaystyle
    P\{ Y = y|X = i\} = \frac{{Ce^{ - (\alpha + 1)y} \frac{{y^{s + i - 1} }}{{i!}}}}{{C\int\limits_0^\infty {e^{ - (\alpha + 1)y} \frac{{y^{s + i - 1} }}{{i!}}dy} }} = \frac{{e^{ - (\alpha + 1)y} y^{s + i - 1} }}{{\int\limits_0^\infty {e^{ - (\alpha + 1)y} y^{s + i - 1} dy} }} =
    $

    $\displaystyle
    = \frac{{e^{ - (\alpha + 1)y} y^{s + i - 1} }}{{\frac{{(s + i - 1)!}}{{(\alpha + 1)^{s + i} }}}} = \frac{{(\alpha + 1)^{s + i} }}{{(s + i - 1)!}}e^{ - (\alpha + 1)y} y^{s + i - 1} $

    which is exactly the gamma distribution with the parameters $\displaystyle (s + i,\alpha + 1)$
    Last edited by Peritus; Jan 20th 2008 at 12:21 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2007
    Posts
    239
    Ah I knew it. Whenever you are given $\displaystyle P(A|B) $ and you want to find $\displaystyle P(B|A) $ just write it as $\displaystyle P(B|A) = \frac{P(B)P(A|B)}{P(A)} $.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Poisson Distribution Probability
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: Oct 17th 2011, 02:55 PM
  2. Poisson Probability Distribution
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: Mar 4th 2011, 10:18 AM
  3. Probability Poisson Distribution
    Posted in the Statistics Forum
    Replies: 1
    Last Post: May 4th 2010, 01:56 AM
  4. Fourier transform of Gamma function and Poisson distribution
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: Apr 30th 2009, 12:20 PM
  5. poisson probability distribution
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Mar 9th 2008, 09:33 PM

Search Tags


/mathhelpforum @mathhelpforum