 agantisbz

2022-07-17

Posterior Probability Distribution from Geometric Distribution
Suppose an experimenter believes that $p\sim Unif\left(\left[0,1\right]\right)$ 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 $\left[G\mid p\right]\sim Geo\left(p\right)$.
How can you find the distribution of [p∣G]? Do you use Bayes' Theorem? sviudes7w

Expert

Step 1
$\pi \left(\theta \right)=1$
$\theta \in \left[0;1\right]$
Your likelihood is $p\left(x|\theta \right)=\left(1-\theta {\right)}^{x-1}\theta$
$x=1,2,3...$
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 $\pi \left(\theta |x\right)\propto \theta \left(1-\theta {\right)}^{x-1}$
$\theta \in \left[0;1\right]$
That is $\pi \left(\theta |x\right)=Beta\left(2;x\right)$
The complete expression of your posterior is $\pi \left(\theta |x\right)=\left(x+1\right)x\theta \left(1-\theta {\right)}^{x-1}$.
To derive this expression not any integral is needed as you immediately recognize it is a Beta distribution

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