 # In Bayesian probability theory, a class of distribution of prior distribution theta is said to be the conjugate to a class of likelihood function f(x| theta) if the resulting posterior distribution is of the same class as of f(theta). Almintas2l 2022-07-17 Answered
Mathematically prove that a Beta prior distribution is conjugate to a Geometric likelihood function
I have to prove with a simple example and a plot how prior beta distribution is conjugate to the geometric likelihood function. I know the basic definition as
'In Bayesian probability theory, a class of distribution of prior distribution θ is said to be the conjugate to a class of likelihood function $f\left(x|\theta \right)$ if the resulting posterior distribution is of the same class as of $f\left(\theta \right)$.'
But I don't know how to prove it mathematically.
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Step 1
Find $f\left(x|\theta \right)$. Using Bayes theorem we know that $f\left(x|\theta \right)=Cf\left(\theta |x\right)f\left(\theta \right)$. C is just a normalisation constant to make it integrate to 1.
$f\left(\theta \right)$ is the PDF of the prior distribution. I.e. beta distribution (with some parameters $\left(\alpha ,\beta \right)$). Here $f\left(\theta \right)={C}^{\prime }{\theta }^{\alpha -1}\left(1-\theta {\right)}^{\beta -1}$
$f\left(\theta |x\right)$ is the likelihood function for $\theta$ given that the data x is distributed by a geometric distribution with parameter $\theta$. Our geometric likelihood function is $f\left(\theta |x\right)=\prod _{i=0}^{n}\left(1-\theta {\right)}^{{x}_{i}}\theta =\left(1-\theta {\right)}^{\sum _{i=0}^{n}{x}_{i}}{\theta }^{n}$.
Step 2
Now were going to find the product of these and we expect it will have the same form as the beta prior but with new parameters ${\alpha }^{\prime },{\beta }^{\prime }$, and we will find the parameters.
So $f\left(\theta |x\right)f\left(\theta \right)={C}^{\prime }{\theta }^{\alpha +n-1}\left(1-\theta {\right)}^{\sum _{i=0}^{n}{x}_{i}+\beta -1}$. We can see the new parameters are ${\alpha }^{\prime }=\alpha +n$, and ${\beta }^{\prime }=\sum _{i=0}^{n}{x}_{i}+\beta$. Mission accomplished.

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