Posterior Probability Distribution from Geometric Distribution

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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 $[p \mid G]?$ Do you use Bayes' Theorem?

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Your prior is a $Beta(1;1)$

$$\pi(\theta)=1$$

$\theta \in[0;1]$

Your likelihood is

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

$x=1,2,3...$

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