Posterior Probability Distribution from Geometric DistributionSuppose an experimenter believes that p ∼ U n i...

agantisbz

agantisbz

Answered

2022-07-17

Posterior Probability Distribution from Geometric Distribution
Suppose 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?

Answer & Explanation

sviudes7w

sviudes7w

Expert

2022-07-18Added 12 answers

Step 1
Your 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 2
Considering 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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