agantisbz

Answered

2022-07-17

Posterior Probability Distribution from Geometric Distribution

Suppose an experimenter believes that $p\sim Unif([0,1])$ 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 $[G\mid p]\sim Geo(p)$.

How can you find the distribution of [p∣G]? Do you use Bayes' Theorem?

Suppose an experimenter believes that $p\sim Unif([0,1])$ 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 $[G\mid p]\sim Geo(p)$.

How can you find the distribution of [p∣G]? Do you use Bayes' Theorem?

Answer & Explanation

sviudes7w

Expert

2022-07-18Added 12 answers

Step 1

Your prior is a Beta(1;1)

$\pi (\theta )=1$

$\theta \in [0;1]$

Your likelihood is $p(x|\theta )=(1-\theta {)}^{x-1}\theta $

$x=1,2,\mathrm{3...}$

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 $\pi (\theta |x)\propto \theta (1-\theta {)}^{x-1}$

$\theta \in [0;1]$

That is $\pi (\theta |x)=Beta(2;x)$

The complete expression of your posterior is $\pi (\theta |x)=(x+1)x\theta (1-\theta {)}^{x-1}$.

To derive this expression not any integral is needed as you immediately recognize it is a Beta distribution

Your prior is a Beta(1;1)

$\pi (\theta )=1$

$\theta \in [0;1]$

Your likelihood is $p(x|\theta )=(1-\theta {)}^{x-1}\theta $

$x=1,2,\mathrm{3...}$

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 $\pi (\theta |x)\propto \theta (1-\theta {)}^{x-1}$

$\theta \in [0;1]$

That is $\pi (\theta |x)=Beta(2;x)$

The complete expression of your posterior is $\pi (\theta |x)=(x+1)x\theta (1-\theta {)}^{x-1}$.

To derive this expression not any integral is needed as you immediately recognize it is a Beta distribution

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