Derivation of a conditional probability mass function which involves Geometric and Gamma random variablesLet N be a geometric random variable with parameter p. Suppose that the conditional distribution of X given that N = n is the gamma distribution with parameters n and λ . Find the conditional probability mass function of N given that X = x ..

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Derivation of a conditional probability mass function which involves Geometric and Gamma random variablesLet N be a geometric random variable with parameter p. Suppose that the conditional distribution of X given that   N  =  n is the gamma distribution with parameters n and   λ  . Find the conditional probability mass function of N given that   X  =  x  ..

gortepap6yb · Accepted Answer

Step 1The infinite series in the denominator is missing a factor of             (      1      −      p      )              n      −      1        p. Basically you need to add up the value from the numerator over all possible n.Step 2After including this factor, you can cancel out the p in the numerator, and the series will sum to       e          λ              (        1        −        p        )            x      , which is what you need.

fabler107 · Accepted Answer

Step 1It is much more convenient to work with the kernels of each distribution, rather than the density and PMF.Our hierarchical model is   N  ∼  Geometric  ⁡  (  p  )  ,    X  ∣  N  ∼  Gamma  ⁡  (  N  ,  λ  )  ,    Pr  [  N  =  n  ]  =  (  1  −  p      )          n      −      1        p  ,    n  ∈  {  1  ,  2  ,  …  }  ,        f          X      ∣      N        (  x  )  =                    λ        N                    x                  N          −          1                            e                  −          λ          x                            Γ      (      N      )        ,    x  &amp;gt;  0.Then as you noted, Bayes&#039; rule gives   Pr  [  N  =  n  ∣  X  =  x  ]  =                    f                  X          ∣          N                    (      x      )      Pr      [      N      =      n      ]              Pr      [      X      =      x      ]        .Step 2But we don&#039;t care about the denominator   Pr  [  X  =  x  ], because it does not depend on n. All that is needed is to observe that the kernel of the LHS must equal the product of the kernels on the RHS, up to some constant of proportionality with respect to n:  Pr  [  N  =  n  ∣  X  =  x  ]  ∝                    λ        n                    x                  n          −          1                            (      n      −      1      )      !        (  1  −  p      )          n      −      1        ∝            (      λ      x      (      1      −      p      )              )                  n          −          1                            (      n      −      1      )      !        ,where we have removed every factor that does not depend on n. In the second step, we removed a constant factor of   λ. This gives us the kernel of a Poisson distribution with rate       λ    ∗    =  λ  x  (  1  −  p  ); that is to say, the posterior distribution of   N  −  1 given the observation   X  =  x is Poisson with rate       λ    ∗  .The location transformation is required because we selected a parametrization of the prior geometric distribution of N that had support on the strictly positive integers, hence the posterior cannot have   N  =  0. Thus   N  −  1  ∣  X is Poisson, rather than N∣X.

Let N be a geometric random variable with parameter p. Suppose that the conditional distribution of X given that N=n is the gamma distribution with parameters n and lambda. Find the conditional probability mass function of N given that X=x.

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