Results 1 to 3 of 3

Math Help - Gamma/Poisson Distribution- Probability

  1. #1
    Member
    Joined
    Aug 2007
    Posts
    239

    Gamma/Poisson Distribution- Probability

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

    So the problem is to find  P \{Y = y | X = i \} and show that it is the gamma distribution. Now  P \{Y =y |X =i \} = \frac{P \{Y =y, \ X = i \}}{P \{X = i\}} . So I am guessing that the  \frac{1}{P \{Y = y \}} and the  \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:

    <br />
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} }}<br />

    <br />
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} }} = <br />

    <br />
 = \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 (s + i,\alpha  + 1)
    Last edited by Peritus; January 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  P(A|B) and you want to find  P(B|A) just write it as  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: October 17th 2011, 02:55 PM
  2. Poisson Probability Distribution
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: March 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: April 30th 2009, 12:20 PM
  5. poisson probability distribution
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: March 9th 2008, 09:33 PM

Search Tags


/mathhelpforum @mathhelpforum