Almintas2l

2022-07-17

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 $f(x|\theta )$ if the resulting posterior distribution is of the same class as of $f(\theta )$.'

But I don't know how to prove it mathematically.

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 $f(x|\theta )$ if the resulting posterior distribution is of the same class as of $f(\theta )$.'

But I don't know how to prove it mathematically.

Kendrick Jacobs

Beginner2022-07-18Added 16 answers

Step 1

Find $f(x|\theta )$. Using Bayes theorem we know that $f(x|\theta )=Cf(\theta |x)f(\theta )$. C is just a normalisation constant to make it integrate to 1.

$f(\theta )$ is the PDF of the prior distribution. I.e. beta distribution (with some parameters $(\alpha ,\beta )$). Here $f(\theta )={C}^{\prime}{\theta}^{\alpha -1}(1-\theta {)}^{\beta -1}$

$f(\theta |x)$ 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 $f(\theta |x)=\prod _{i=0}^{n}(1-\theta {)}^{{x}_{i}}\theta =(1-\theta {)}^{\sum _{i=0}^{n}{x}_{i}}{\theta}^{n}$.

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 $f(\theta |x)f(\theta )={C}^{\prime}{\theta}^{\alpha +n-1}(1-\theta {)}^{\sum _{i=0}^{n}{x}_{i}+\beta -1}$. We can see the new parameters are ${\alpha}^{\prime}=\alpha +n$, and ${\beta}^{\prime}=\sum _{i=0}^{n}{x}_{i}+\beta $. Mission accomplished.

Find $f(x|\theta )$. Using Bayes theorem we know that $f(x|\theta )=Cf(\theta |x)f(\theta )$. C is just a normalisation constant to make it integrate to 1.

$f(\theta )$ is the PDF of the prior distribution. I.e. beta distribution (with some parameters $(\alpha ,\beta )$). Here $f(\theta )={C}^{\prime}{\theta}^{\alpha -1}(1-\theta {)}^{\beta -1}$

$f(\theta |x)$ 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 $f(\theta |x)=\prod _{i=0}^{n}(1-\theta {)}^{{x}_{i}}\theta =(1-\theta {)}^{\sum _{i=0}^{n}{x}_{i}}{\theta}^{n}$.

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 $f(\theta |x)f(\theta )={C}^{\prime}{\theta}^{\alpha +n-1}(1-\theta {)}^{\sum _{i=0}^{n}{x}_{i}+\beta -1}$. 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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