Suppose an experimenter believes that p∼Unif([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]∼Geo(p). How can you find the distribution of [p∣G]? Do you use Bayes' Theorem?

agantisbz

agantisbz

Answered question

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

Beginner2022-07-18Added 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

Do you have a similar question?

Recalculate according to your conditions!

New Questions in High school geometry

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?