Proof of geometric probability mass functionShow that P ( X = n + x ∣ X &gt; n ) = P ( X = x ) for x = 1 , 2 , 3 … and n = 1 , 2 , 3 …

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Proof of geometric probability mass functionShow that   P  (  X  =  n  +  x  ∣  X  &amp;gt;  n  )  =  P  (  X  =  x  ) for   x  =  1  ,  2  ,  3  … and   n  =  1  ,  2  ,  3  …

vibrerentb · Accepted Answer

Step 1A geometric distribution is that of the count of trials until the first success in an indefinite sequence of independent Bernoulli trials with constant success rate.Let X be such a count and p be the success rate.Use conditional probability.   P  (  X  =  n  +  x     and     X  &amp;gt;  n  )      /    P  (  X  &amp;gt;  n  )Yes, and further, by Bayes&#039; Rule:      P    (  X  =  n  +  x  ∣  X  &amp;gt;  n  )     =                                 P                (        X        =        n        +        x        )                          P                (        X        &amp;gt;        n        ∣        X        =        n        +        x        )                              P                (        X        &amp;gt;        n        )            Step 2How does the numerator   P  (  X  =  n  +  n     and     X  &amp;gt;  n  ) just simplify to   P  (  X  =  n  +  x  )  ?What is the probability that   X  &amp;gt;  n when given that   X  =  n  +  x, for any positive values of n and x?Step 3Expand using pmf of geometric distribution. BUT how do you expand the denominator in step 1?Well we know       P    (  X  =  k  )     =     (  1  −  p      )          k      −      1          p              1              k      ∈      {      1      ,      2      ,      …      }       because to obtain a count of k we have   k  −  1 consecutive failures and then one success.In a similar manner we can measure       P    (  X  &amp;gt;  k  ) because to have a count greater than k we must have at least k consecutive failures.

Show that P(X=n+x∣X>n)=P(X=x) for x=1,2,3… and n=1,2,3…

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