Posterior Probability Distribution from Geometric DistributionSuppose an experimenter believes that p ∼ U n i f ( [ 0 , 1 ] ) is the chance of success on any given experiment. The experimenter then tries the experiment over and over until the experiment is a success. This number of trials will be [ G ∣ p ] ∼ G e o ( p ).How can you find the distribution of [p∣G]? Do you use Bayes' Theorem?

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Posterior Probability Distribution from Geometric DistributionSuppose an experimenter believes that   p  ∼  U  n  i  f  (  [  0  ,  1  ]  ) is the chance of success on any given experiment. The experimenter then tries the experiment over and over until the experiment is a success. This number of trials will be   [  G  ∣  p  ]  ∼  G  e  o  (  p  ).How can you find the distribution of [p∣G]? Do you use Bayes&#039; Theorem?

sviudes7w · Accepted Answer

Step 1Your prior is a Beta(1;1)  π  (  θ  )  =  1  θ  ∈  [  0  ;  1  ]Your likelihood is   p  (  x      |    θ  )  =  (  1  −  θ      )          x      −      1        θ  x  =  1  ,  2  ,  3...Step 2Considering x as a given number (after seeing the observations), the posterior is the same expression as a function of the parameter, say   π  (  θ      |    x  )  ∝  θ  (  1  −  θ      )          x      −      1        θ  ∈  [  0  ;  1  ]That is   π  (  θ      |    x  )  =  B  e  t  a  (  2  ;  x  )The complete expression of your posterior is   π  (  θ      |    x  )  =  (  x  +  1  )  x  θ  (  1  −  θ      )          x      −      1      .To derive this expression not any integral is needed as you immediately recognize it is a Beta distribution

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