An identity involving geometric distributionLet ( Ω , A , P ) be probability space and X : Ω → N 0 be a geometrically distributed random variable with parameter p ∈ ( 0 , 1 ) X ∽ Geo ( p ) . Show that P ( X = m + n | X ≥ m ) = P ( X = n ) , for all m , n ∈ N 0 .

Question

An identity involving geometric distributionLet   (  Ω  ,      A    ,  P  ) be probability space and   X  :  Ω  →            N      0       be a geometrically distributed random variable with parameter   p  ∈  (  0  ,  1  )   X  ∽      Geo          (      p      )      . Show that   P  (  X  =  m  +  n         |    X  ≥  m  )  =  P  (  X  =  n  )  ,       for all           m      ,      n      ∈                        N          0                      .

Aedan Hatfield · Accepted Answer

Step 1That propertie is known in probability theory as memorylessness.The probability mass function for a geometric random variable X is   f  (  x  )  =  p  (  1  −  p      )    x  . The probability that X is greater than or equal to x is   P  (  X  ≥  x  )  =  (  1  −  p      )    x  .Step 2So the conditional probability of interest is   P  (  X  =  m  +  n      |    X  ≥  m  )  =            P      (      X      =      m      +      n      ,      X      ≥      m      )              P      (      X      ≥      m      )        =            P      (      X      =      m      +      n      )              P      (      X      ≥      m      )        =            p      (      1      −      p              )                  (          m          +          n          )                            (      1      −      p              )        m              =  p  (  1  −  p      )          n        =  P  (  X  ≥  n  ), which proves the memoryless property.

Let (Omega, mathfrakA, P) be probability space and X:Omega rightarrow mathbbN_0 be a geometrically distributed random variable with parameter p in (0,1) X∽Geo(p). Show that P(X=m+n | X ge m)=P(X=n), for all m, n in mathbbN_0.

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