Let X₁....X₁₀ be a random sample from a distribution with the probability density function $\mathrm{f}(\mathrm{x}|\theta )=\{\begin{array}{cc}\theta x^{\theta -1},& if0<x<1\\ 0,& otherwise,\end{array}$

where θ > 0 Is an unknown parameter. The prior distribution of θ is given by  

$\pi (\theta )=\{\begin{array}{cc}\theta e^{-\theta },& if\theta >0\\ 0,& otherwise,\end{array}$

The Bayes estimator of θ under squared error loss is  

Prior Distribution: In Bayesian estimation, the prior distribution represents our beliefs about the

parameter $\theta$  before seeing any data. In this problem, the prior distribution of $\theta$ is given by 

$\pi (\theta )=\{\begin{array}{ll}\theta e^{-\theta },& \text{if }\theta >0\\ 0,& \text{otherwise}\end{array}$

Likelihood Function:

The likelihood function expresses the probability of observing the data given the parameter $\theta$.

For this problem, the probability density function (pdf) of the observations $X_1,X_2,\dots ,X_n$ given $\theta$ is:

$f(x|\theta )=\{\begin{array}{ll}\theta x^{\theta -1},& \text{if }0<x<1\\ 0,& \text{otherwise}\end{array}$

For a random sample $X_1,X_2,\dots ,X_n$ , the likelihood function is the product of the individual densities

$L(\theta |X_1,X_2,\dots ,X_n)=\prod _{i=1}^n\theta X_i^{\theta -1}$

This simplifies to $L(\theta |X_1,X_2,\dots ,X_n)={\theta }^n\prod _{i=1}^n{X}_i^{\theta -1}$
 

Explanation:

To solve the problem of finding the Bayes estimator of $\theta$ under squared error loss,

let’s break it down step by step. The random sample $X_1,X_2,\dots ,X_{10}$ comes from a

distribution with the probability density function (pdf):

$f(x|\theta )=\{\begin{array}{ll}\theta x^{\theta -1},& \text{if }0<x<1\\ 0,& \text{otherwise}\end{array}$, where $\theta >0$ is an unknown parameter.

The prior distribution of $\theta$ is given by

$\pi (\theta )=\{\begin{array}{ll}\theta e^{-\theta },& \text{if }\theta >0\\ 0,& \text{otherwise}\end{array}$

The likelihood function for a sample $X_1,X_2,\dots ,X_n$ from the given pdf is

$L(\theta |x_1,x_2,\dots ,x_n)=\prod _{i=1}^{n}f(x_i|\theta )$

Substitute the given pdf $f(x|\theta )$:

$L(\theta |x_1,x_2,\dots ,x_n)=\prod _{i=1}^n\theta x_i^{\theta -1}$

$L(\theta |x_1,x_2,\dots ,x_n)={\theta }^n\prod _{i=1}^n{x}_i^{\theta -1}$

This can be rewritten as

$L(\theta |x_1,x_2,\dots ,x_n)={\theta }^n\prod _{i=1}^n{x}_i^{\theta }\prod _{i=1}^n{x}_i^{-1}$

⇒ $L(\theta |x_1,x_2,\dots ,x_n)={\theta }^n\prod _{i=1}^n{x}_i^{\theta }\prod _{i=1}^n\frac{1}{x_i}$

Taking the log of the likelihood function

$\log L(\theta |x_1,x_2,\dots ,x_n)=n\log \theta +\theta \sum _{i=1}^n\log x_i-\sum _{i=1}^n\log x_i$

The posterior distribution is proportional to the product of the likelihood and the prior distribution.

So, we multiply the likelihood function by the prior distribution:

$\pi (\theta |x_1,x_2,\dots ,x_n)\propto L(\theta |x_1,x_2,\dots ,x_n)\times \pi (\theta )$

The prior distribution is $\pi (\theta )=\theta e^{-\theta }$, so

$\pi (\theta |x_1,x_2,\dots ,x_n)\propto {\theta }^n\prod _{i=1}^n{x}_i^{\theta }\theta e^{-\theta }$
 

Taking the logarithm of this $\log \pi (\theta |x_1,x_2,\dots ,x_n)=(n+1)\log \theta +\theta \sum _{i=1}^n\log x_i-\theta$

The Bayes estimator under squared error loss is the mean (expectation) of the posterior distribution. That is

$\hat{\theta }=E[\theta |X_1,X_2,\dots ,X_n]$

From the posterior distribution, which is a Gamma distribution, the expectation of $\theta$ is given by

$\hat{\theta }=\frac{n+1}{\sum _{i=1}^n\log x_i}$

For this problem,  n = 10, so

$\hat{\theta }=\frac{11}{\sum _{i=1}^{10}\log x_i}$
 

Thus, option 1) is correct.