Geometric sum of geometric random variablesI am looking to find the probability mass function of Y = ∑ i = 1 N X i where X i ∼ Geometric ( a ) and N ∼ Geometric ( b ). I attempted to do this by finding the probability generating function of Y and comparing it to known probability generating functions to take advantage of the uniqueness property. (In my searches online, it sounds like I should find that Y ∼ Geometric ( a b ).)

Question

Geometric sum of geometric random variablesI am looking to find the probability mass function of   Y  =      ∑          i      =      1        N        X    i   where       X    i    ∼      Geometric    (  a  ) and   N  ∼      Geometric    (  b  ). I attempted to do this by finding the probability generating function of Y and comparing it to known probability generating functions to take advantage of the uniqueness property. (In my searches online, it sounds like I should find that   Y  ∼      Geometric    (  a  b  ).)

Cody Petty · Accepted Answer

Step 1Assume that                     (1)                              P                [                  X          i                =        k        ]        =        p        (        1        −        p                  )                      k            −            1                          ,                k        =        1        ,        2        ,        …            and       P    [  N  =  n  ]  =  r  (  1  −  r      )          n      −      1       for   r  =  1  ,  2  ,  …. Then:                               P                [        Y        =        m        ]                            =                              ∑                      n            ≥            1                                    P                [        N        =        n        ]        ⋅                  P                [                  X          1                +        …        +                  X          n                =        m        ]                            (2)                                  =                              r                      1            −            r                                    ∑                      n            ≥            1                          (        1        −        r                  )          n                                      (                          p                              1                −                p                                      )                    n                (        1        −        p                  )          m                ⋅        r        (        m        ,        n        )            Where:                     r        (        m        ,        n        )                            =                    #        {        (                  a          1                ,        …        ,                  a          n                )        :                  a          i                ∈                              N                                &amp;gt;            0                          ,                  a          1                +        …        +                  a          n                =        m        }                                          =                    [                  x          m                ]                              (            x            +                          x              2                        +                          x              3                        +            …            )                    n                                    (3)                                  =                    [                  x          m                ]                              x            n                                (            1            −            x                          )              n                                      =                                            (                                                          m              −              1                                      n              −              1                                                          )                                          Step 2Hence:                               P                [        Y        =        m        ]                            =                    r        p        (        1        −        p                  )                      m            −            1                                    ∑                      n            ≥            1                                                              (                                                          m              −              1                                      n              −              1                                                          )                                                            (                                          p                −                p                r                                            1                −                p                                      )                                n            −            1                                                            =                    r        p        (        1        −        p                  )                      m            −            1                                                (            1            +                                          p                −                p                r                                            1                −                p                                      )                                m            −            1                                              (4)                                  =                    p        r        (        1        −        p        r                  )                      m            −            1                              proving your claim for geometric distributions supported on 1,2,…. There is little to change in order to deal with geometric distributions supported on 0,1,…, too.

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