Conditional pmf of X ∣ X + Y = n, with X,Y independent geometric r.v.Suppose, X,Y are independent geometric random variables with the same parameter p. We want to find the conditional probability function of X given that X + Y = n, where n &gt; 1.

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Conditional pmf of   X  ∣  X  +  Y  =  n, with X,Y independent geometric r.v.Suppose, X,Y are independent geometric random variables with the same parameter p. We want to find the conditional probability function of X given that   X  +  Y  =  n, where   n  &amp;gt;  1.

Lintynx · Accepted Answer

Step 1I assume that the geometric distribution starts from   k  =  1. Then for   1  ≤  k  ≤  n we have that                     P        (        X        =        k        ∣        X        +        Y        =        n        )                            =                              P            (            X            +            Y            =            n            ∣            X            =            k            )            P            (            X            =            k            )                                P            (            X            +            Y            =            n            )                          =                              P            (            Y            =            n            −            k            )            P            (            X            =            k            )                                P            (            X            +            Y            =            n            )                                              (1)                                  =                              p            (            1            −            p                          )                              n                −                k                −                1                                      p            (            1            −            p                          )                              k                −                1                                                          P            (            X            +            Y            =            n            )                          =                                            p              2                        (            1            −            p                          )                              n                −                2                                                          P            (            X            +            Y            =            n            )                              Step 2Now for any   n  ∈      N                      P        (        X        +        Y        =        n        )                            =                  ∑                      k            =            1                                n            −            1                          P        (        X        =        k        )        P        (        Y        =        n        −        k        )        =                  ∑                      k            =            1                                n            −            1                          p        (        1        −        p                  )                      k            −            1                          p        (        1        −        p                  )                      n            −            k            −            1                                                            =        (        n        −        1        )                  p          2                (        1        −        p                  )                      n            −            2                              so, substituting back in (1) gives  P  (  X  =  k  ∣  X  +  Y  =  n  )  =                    p        2            (      1      −      p              )                  n          −          2                            (      n      −      1      )              p        2            (      1      −      p              )                  n          −          2                      =      1          n      −      1      So   X  ∣  X  +  Y  =  n is uniformly distributed in   {  1  ,  2  ,  …  ,  n  −  1  }

Suppose, X,Y are independent geometric random variables with the same parameter p. We want to find the conditional probability function of X given that X+Y=n, where n>1.

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