Geometric Distribution Probability ProblemSuppose that X is a geometric random variable with parameter (probability of success) p.Show that Pr ( X &gt; a + b ∣ X &gt; a ) = Pr ( X &gt; b ).

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Geometric Distribution Probability ProblemSuppose that X is a geometric random variable with parameter (probability of success) p.Show that   Pr  (  X  &amp;gt;  a  +  b  ∣  X  &amp;gt;  a  )  =  Pr  (  X  &amp;gt;  b  ).

ruinsraidy4 · Accepted Answer

Step 1There are two ways to get   Pr  (  X  &amp;gt;  n  ), the easy way and the hard way. Since there is something to be learned from both, we do it both ways.The hard way: The probability that   X  =  n  +  1 is the probability of n failures in a row, then success. This is   (  1  −  p      )    n    pSimilarly, the probability that   X  =  n  +  2 is the probability of   n  +  1 failures in a row, then success. This has probability   (  1  −  p      )          n      +      1        p.Continue, and add up. There is a   (  1  −  p      )    n    p &quot;in&quot; each term. Take it out as a common factor. So we get   (  1  −  p      )    n    p      (    1    +    (    1    −    p    )    +    (    1    −    p          )      2        +    ⋯    )    .The geometric series inside the brackets has sum       1          1      −      (      1      −      p      )        =      1    p  Step 2The easy way: We have   X  &amp;gt;  n precisely if we get n failures in a row, probability   (  1  −  p      )    n  .

koraby2bc · Accepted Answer

Step 1Here&#039;s how to handle an infinite geometric series when the index starts at n instead of 0:                              ∑                      k            =            n                    ∞                a                  x          k                                    =        a                  x          n                +        a                  x                      n            +            1                          +        a                  x                      n            +            2                          +        a                  x                      n            +            3                          +        ⋯                                          =        (        a                  x          n                )        +        (        a                  x          n                )        x        +        (        a                  x          n                )                  x          2                +        (        a                  x          n                )                  x          3                +        ⋯                                          =        b        +        b        x        +        b                  x          2                +        b                  x          3                +        ⋯            Now it starts at index 0. And of course b is   a      x    n  .It looks as if you can do the rest.Second method: You mentioned that the probability of &quot;success&quot; is p. That means the probability of success on each trial is p.If X is defined as the number of trials needed to get one success, then the event   X  &amp;gt;  n is the same as the event of failure on all of the first n trials, so that probability of that is   (  1  −  p      )    n  .If X is defined as the number of trials needed to get one failure, then the event   X  &amp;gt;  n is the same as the event of success on all of the first n trials, so the probability of that is       p    n  .

Suppose that X is a geometric random variable with parameter (probability of success) p. Show that Pr(X>a+b∣X>a)=Pr(X>b).

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