Geometric Distribution - ProbabilityHow we can proof that if X ∼ G e o ( p )Than: P r ( X &gt; a ) = ( 1 − p ) a How can we proof it formally? There is an example for that to understand it intuitively?

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Geometric Distribution - ProbabilityHow we can proof that if   X  ∼  G  e  o  (  p  )Than:   P  r  (  X  &amp;gt;  a  )  =  (  1  −  p      )    a  How can we proof it formally? There is an example for that to understand it intuitively?

Zackary Hatfield · Accepted Answer

Step 1To proof it formally you should know the formula for the partial sum of a geometric series. The pmf of X is   P  (  X  =  a  )  =  p  (  1  −  p      )          a      −      1        =  p      q          a      −      1          ∀        a  =  1  ,  2  ,  …And the cdf is   P  (  X  ≤  a  )  =  p      ∑          i      =      1        a        q          i      −      1          ∀        a  =  1  ,  2  ,  …Step 2For a discrete random variable X we have   P  (  X  &amp;gt;  a  )  =  1  −  P  (  X  ≤  a  ).Thus   P  (  X  &amp;gt;  a  )  =  1  −  p      ∑          i      =      1        a        q          i      −      1        =  1  −      p    q        ∑          i      =      1        a        q          i      The sum is a partial sum of a geometric series:       ∑          i      =      1        a        q          i        =  q  ⋅                    q        a            −      1              q      −      1      . Use this and the fact that   p  =  1  −  q to obtain   P  (  X  &amp;gt;  a  )  =  (  1  −  p      )    a

Simone Watts · Accepted Answer

Step 1The geometric distribution gives the probability that the first occurrence of success requires a independent trials, each having a success probability of p. Then the probability that the ath trial is the first success is   P  (  X  =  a  )  =  (  1  −  p      )          a      −      1        p  , by independence.Step 2Thus,                    P                  (          X          &amp;gt;          a          )                                    =                  ∑                      k            =            a            +            1                    ∞                P        (        X        =        k        )                                          =                  ∑                      k            =            a            +            1                    ∞                (        1        −        p                  )                      k            −            1                          p                                          =                  ∑                      k            =            1                    ∞                (        1        −        p                  )                      k            −            1                          p                 −                  ∑                      k            =            1                    a                (        1        −        p                  )                      k            −            1                          p                                          =                  p                      1            −            (            1            −            p            )                          −                              p                          (              1              −              (              1              −              p                              )                a                            )                                            1            −            (            1            −            p            )                          =        (        1        −        p                  )          a                .

How we can proof that if X~Geo(p). Than: Pr(X>a)=(1-p)^a. How can we proof it formally? There is an example for that to understand it intuitively?

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