Probability mass function of sum of two independent geometric random variablesHow could it be proved that the probability mass function of X + Y, where X and Y are independent random variables each geometrically distributed with parameter p; i.e. p X ( n ) = p Y ( n ) = { p ( 1 − p ) n − 1 n = 1 , 2 , . . . 0 o t h e r w i s e equas to P X + Y ( n ) = ( n − 1 ) p 2 ( 1 − p ) n − 2

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Probability mass function of sum of two independent geometric random variablesHow could it be proved that the probability mass function of   X  +  Y, where X and Y are independent random variables each geometrically distributed with parameter p; i.e.       p    X    (  n  )  =      p    Y    (  n  )  =      {                            p          (          1          −          p                      )                          n              −              1                                                n          =          1          ,          2          ,          .          .          .                                      0                          o          t          h          e          r          w          i          s          e                         equas to             P              X        +        Y              (    n    )    =          (      n      −      1      )                   p      2        (    1    −    p          )              n        −        2

taguetzbo · Accepted Answer

Step 1Since   X  ,  Y  ≥  1, the summation should run over   k  =  1  ,  2  ,  …  ,  n  −  1.Step 2Using this your convolution becomes                    P        (        X        +        Y        =        n        )                            =                              ∑                      k            =            1                                n            −            1                                    p          2                (        1        −        p                  )                      n            −            2                                                            =                              p          2                (        1        −        p                  )                      n            −            2                                    ∑                      k            =            1                                n            −            1                          1                                          =                              p          2                (        1        −        p                  )                      n            −            2                          (        n        −        1        )        .

Ruby Briggs · Accepted Answer

Step 1A geometric random variable is the count of Bernouli trial until a success. We measure the probability of obtaining   n  −  1 failures and then 1 success.      P    (  X  =  n  )  =  (  1  −  p      )          n      −      1        p    :  n  ∈  {  1  ,  2  ,  …  }The sum of two such is the count of Bernouli trials until the second success. We measure the probability of obtaining 1 success and   n  −  2 failures, in any arrangement of those   n  −  1 trials, followed by the second success.      P    (  X  +  Y  =  n  )  =  (  n  −  1  )  (  1  −  p      )          n      −      2            p    2      :  n  ∈  {  2  ,  3  …  }Step 2This may also be counted by summing                               P                (        X        +        Y        =        n        )                            =                  ∑                      k            =            1                                n            −            1                                    P                (        X        =        k        ,        Y        =        n        −        k        )                    note the range                                          =                  ∑                      k            =            1                                n            −            1                                    P                (        X        =        k        )                  P                (        Y        =        n        −        k        )                    by independence                                          =                  ∑                      k            =            1                                n            −            1                          (        1        −        p                  )                      k            −            1                          p        ⋅        (        1        −        p                  )                      n            −            k            −            1                          p                                          =        (        1        −        p                  )                      n            −            2                                    p          2                          ∑                      k            =            1                                n            −            1                          1                                          =        (        n        −        1        )        (        1        −        p                  )                      n            −            2                                    p          2                    Since   X  +  Y must equal n and neither can be less than 1, then neither can be more than   n  −  1. Hence this the range of X values we must sum over.

How could it be proved that the probability mass function of X + Y, where X and Y are independent random variables each geometrically distributed with parameter p

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