Characterization of the geometric distributionX,Y are i.i.d. random variables with mean μ, and taking values in {0,1,2,...}.Suppose for all m ≥ 0, P ( X = k | X + Y = m ) = 1 m + 1 , k = 0 , 1 , . . . m. Find the distribution of X in terms of μ.

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Characterization of the geometric distributionX,Y are i.i.d. random variables with mean   μ, and taking values in {0,1,2,...}.Suppose for all   m  ≥  0,   P  (  X  =  k      |    X  +  Y  =  m  )  =      1          m      +      1      ,   k  =  0  ,  1  ,  .  .  .  m. Find the distribution of X in terms of   μ.

Erika Brady · Accepted Answer

Step 1We have   P  (  X  =  k      |    X  +  Y  =  m  )  =            P      (      X      =      k            &amp;amp;            Y      =      m      −      k      )              P      (      X      +      Y      =      m      )        =                    p                  k                            p                  m          −          k                                    ∑                  j          =          0                          m                            p                  j                            p                  m          −          j                      =      1          m      +      1      for all   0  ≤  k  ≤  m  +  1. In particular, we have       p          0            p          m        =      p          1            p          m      −      1        =      p          2            p          m      −      2        =  ⋯  =      p          m            p          0        .This gives             p              1                    p              0              =            p              m                    p              m        −        1             for all   m  ≥  1, so that       p          m            /        p          m      −      1       is a constant for all   m  ≥  1 and this characterizes X as a geometric distribution.Step 2It seems like we may use generating function as well. If we put   f  (  x  )  =      ∑          k      ≥      0            p          k            x          k        , then the identity actually gives a differential equation   f  (  x      )          2        =      p          0        (  f  (  x  )  +  x      f    ′    (  x  )  )  =      p          0        (  x  f  (  x  )      )    ′   with   f  (  1  )  =  1 and       f    ′    (  1  )  =  μ. I think this will also show that X,Y follows geometric distribution.

X,Y are i.i.d. random variables with mean mu, and taking values in {0,1,2,...}.Suppose for all m ge 0, P(X=k|X+Y=m)=1/(m+1), k=0,1,...m. Find the distribution of X in terms of mu.

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