91Ó°ÊÓ

A random sample of size n is taken from the Bernoulli distribution with parameter θ, which is unknown, and that the prior distribution of θ is a beta distribution for which the mean is\({\mu _0}\)

Consider an experiment with only two outcomes: Success and Failure.

Let x be the number of success observed out of n conducted trails of the experiment. Let \(\theta \) be the true proportion of successes. If it is assigned a uniform prior distribution over the range (0,1) for \(\theta \)and assume that the likelihood of observing x successes in n trial, given the value of \(\theta \), is a binomial distribution, then the posterior distribution for \(\theta \)is beta with parameters \(\alpha = x + 1\) and \(\beta = n - x + 1\) .

The Bayesian posterior estimate, variance of the estimate of\(\theta \)and posterior distribution of\(\theta \)are given as below,

\(\)\(\begin{array}{l}E\left( {\theta /x} \right) = \frac{\alpha }{{\alpha + \beta }}\\Var\left( {\theta /x} \right) = \frac{{\alpha \beta }}{{{{\left( {\alpha + \beta } \right)}^2}\left( {\alpha + \beta + 1} \right)}}\end{array}\)

Let’s consider a random sample of size n is taken from the Bernoulli distribution with parameter \(\theta \)which is unknown, a prior distribution of \(\theta \), is equal to then beta distribution with mean \({\mu _0}\).

To show that the mean of the posterior distribution of \(\theta \)will be weighted average having the form \({\gamma _n}{\overline X _n} + \left( {1 - {\gamma _n}} \right){\mu _0}\)

Let’s consider the prior distribution of \(\theta \),is equal to the beta distribution with parameter \(\alpha \) and \(\beta \).

Therefore we express mean \({\mu _0}\) as follows\({\mu _0} = \frac{\alpha }{{\alpha + \beta }}\)

Now by theorem, if \({X_1},...,{X_n}\) form a random sample from the Bernoulli distribution with parameter\(\theta \)given that \({X_i} = {x_i}\left( {i = 1,...,n} \right)\) is the beta distribution with parameters \(\alpha + \sum\limits_{i = 1}^n {{x_i}} \) and\(\beta + n - \sum\limits_{i = 1}^n {{x_i}} \).

Then the mean of the posterior distribution is computed as follows

Therefore, the mean of the posterior distribution is computed as follows

\(\begin{array}{c}E\left( {\theta |{X_i} = {x_i}\left( {i = 1,...,n} \right)} \right) = \frac{{\alpha + \sum\limits_{i = 1}^n {{x_i}} }}{{\alpha + \sum\limits_{i = 1}^n {{x_i} + \beta + n - \sum\limits_{i = 1}^n {{x_i}} } }}\\ = \frac{{\alpha + \sum\limits_{i = 1}^n {{x_i}} }}{{\alpha + \beta + n}}\end{array}\)

Rearrange the mean of the posterior distribution as,

\(\begin{array}{c}E\left( {\theta |{X_i} = {x_i}\left( {i = 1,...,n} \right)} \right) = \frac{{\alpha + \sum\limits_{i = 1}^n {{x_i}} }}{{\alpha + \beta + n}}\\ = \frac{{\alpha + \beta }}{{\alpha + \beta + n}} \times \frac{\alpha }{{\alpha + \beta }} + \frac{n}{{\alpha + \beta + n}} \times \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ = \frac{{\alpha + \beta }}{{\alpha + \beta + n}} \times {\mu _0} + \frac{n}{{\alpha + \beta + n}} \times {\overline X _n}\end{array}\)

Now, substitute \({\gamma _n} = \frac{n}{{\left( {\alpha + \beta + n} \right)}}\)

So, the mean of the posterior distribution will be

\(\begin{array}{c}E\left( {\theta |{X_i} = {x_i}\left( {i = 1,...,n} \right)} \right) = \frac{{\alpha + \beta }}{{\alpha + \beta + n}} \times {\mu _0} + \frac{n}{{\alpha + \beta + n}} \times {\overline X _n}\\ = {\gamma _n}{\overline X _n} + \left( {1 - {\gamma _n}} \right){\mu _0}\end{array}\)

Also it is observed that, \({\gamma _n} \to 1\) as \(n \to \infty \)