CDF and Survival Function of Geometric DistributionI am having trouble understanding the intuition behind the CDF and survival probability of a geometric distribution on both {0,1,…} and on {1,2,3,…}.I know that a geometric starting from 0 is the number of failures before the first success and the geometric starting at 1 is the number of failures including the first success.Can someone please explain the intuition behind the CDF and survival functions and explain what the formula is?I think that the part confusing me is: Why is the following not true? p ( X &gt; n ) = ( 1 − p ) n + 1 + ( 1 − p ) n + 2 + . . . for a geometric starting at 0.

Question

CDF and Survival Function of Geometric DistributionI am having trouble understanding the intuition behind the CDF and survival probability of a geometric distribution on both {0,1,…} and on {1,2,3,…}.I know that a geometric starting from 0 is the number of failures before the first success and the geometric starting at 1 is the number of failures including the first success.Can someone please explain the intuition behind the CDF and survival functions and explain what the formula is?I think that the part confusing me is: Why is the following not true?  p      (    X    &amp;gt;    n    )    =            (      1      −      p      )              n      +      1        +            (      1      −      p      )              n      +      2        +  .  .  . for a geometric starting at 0.

Alison Mcgrath · Accepted Answer

Step 1If X is the number of failures before the arrival of first success then the event   {  X  &amp;gt;  n  } is the same as the event that the first   n  +  1 trials are not a success, so the probability on that is   (  1  −  p      )          n      +      1      .Then consequently:      F    X    (  n  )  =  P  (  X  ≤  n  )  =  1  −  P  (  X  &amp;gt;  n  )  =  1  −  (  1  −  p      )          n      +      1      Similarly we find:      F    X    (  n  )  =  P  (  X  ≤  n  )  =  1  −  P  (  X  &amp;gt;  n  )  =  1  −  (  1  −  p      )          n       if X denotes the number of trials need to arrive at the first success.Step 2Concerning your question note that   (  1  −  p      )          n      +      k       is the probability that the first   n  +  k trials are failures.However these events for   k  =  1  ,  2  ,  3  ,  … are not mutually exclusive.So summation of the terms leads to multiple counting.

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