What is the probability of maximum of two iid geometric random variable?- Let X,Y be independent geometric random variables, where both are having same parameter (p).- Let Z = m a x ( X , Y ).I would like to find P ( Z = i ) for some real values of i.As we know for K = m i n ( X , Y ), K is geometric distributed with parameter ( 2 p − p 2 ). Does Z also geometric distributed ?.

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What is the probability of maximum of two iid geometric random variable?- Let X,Y be independent geometric random variables, where both are having same parameter (p).- Let   Z  =  m  a  x  (  X  ,  Y  ).I would like to find   P  (  Z  =  i  ) for some real values of i.As we know for   K  =  m  i  n  (  X  ,  Y  ), K is geometric distributed with parameter   (  2  p  −      p    2    ). Does Z also geometric distributed ?.

AtticaPlotowvi · Accepted Answer

Step 1Since   Pr  [  Z  ≤  i  ]  =  (  1  −  (  1  −  p      )    i        )    2    , then   Pr  [  Z  =  i  ]  =  (  1  −  (  1  −  p      )    i        )    2    −  (  1  −  (  1  −  p      )          i      −      1            )    2    =  (  2  −  (  2  −  p  )  (  1  −  p      )          i      −      1        )  (  1  −  p      )          i      −      1        p  , where   i  ∈  {  1  ,  2  ,  3  ,  …  }  . his cannot be written as a geometric distribution because Z is not geometrically distributed.Step 2To prove that the PMF for Z is not geometric, note that for any geometric distribution, we must have             Pr      [      Z      =      i      +      1      ]              Pr      [      Z      =      i      ]        =  1  −  p  , which is constant with respect to i. But in this case, we can easily see that for   p  =  1      /    2, the PMF of Z is  Pr  [  Z  =  i  ∣  p  =  1      /    2  ]  =                    2                  i          +          1                    −      3              4      i        , hence             Pr      [      Z      =      3      ∣      p      =      1              /            2      ]              Pr      [      Z      =      2      ∣      p      =      1              /            2      ]        =            13              /            64              5              /            16        =      13    20    , but             Pr      [      Z      =      2      ∣      p      =      1              /            2      ]              P      r      [      Z      =      1      ∣      p      =      1              /            2      ]        =            5              /            16              1              /            4        =      5    4    .If Z were geometric, these ratios would be equal.

bucstar11n0h · Accepted Answer

Step 1To show Z is not geometrically distributed, it suffices to show that there exists no   q  ∈  (  0  ,  1  ] such that   (  1  −  (  1  −  p      )    i        )    2    =  1  −  (  1  −  q      )    i   for each positive integer i, since the right-hand side is the CDF of a geometric random variable with generic parameter q.Step 2Plug in   i  =  1 to deduce       p    2    =  q. Thus   (  1  −  (  1  −  p      )    i        )    2    =  1  −  (  1  −      p    2        )    i  . If   i  =  2, this reads   (  1  −  (  1  −  p      )    2        )    2    −  (  1  −  (  1  −      p    2        )    2    )  =  0, which only has positive solution   p  =  1. We can check that if   p  =  1,   Z  ∼  Geometric  (  1  ), but of course this edge case is uninteresting.

Let X,Y be independent geometric random variables, where both are having same parameter (p). Let Z=max(X,Y). I would like to find P(Z=i) for some real values of i. As we know for K=min(X,Y), K is geometric distributed with parameter (2p-p^2) Does Z also geometric distributed?

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