Suppose the conditional p.d.f. of a random variable X given θ is \(f(x|\theta)=\left\{\begin{matrix}\frac{2x}{\theta ^2} , \ \ 0 <x< \theta\\\ 0,\rm \ \ \ \ otherwise \end{matrix} \right.\) where the prior distribution of θ is Uniform (0, 1). Based on a single observation x from X, which of the following statements are correct?
Bayesian Estimation Problem Setup
The problem asks us to find the Bayes estimates for a parameter $\theta$ based on a single observation $x$ from a random variable $X$. We are given the conditional probability density function (p.d.f.) of $X$ given $\theta$ and the prior distribution of $\theta$.
Given:
- Conditional p.d.f.: \(f(x|\theta)=\left\{\begin{matrix}\frac{2x}{\theta ^2} , \ \ 0 <x< \theta\\\ 0,\rm \ \ \ \ otherwise \end{matrix} \right.\)
- Prior distribution of $\theta$: Uniform (0, 1). This means the prior p.d.f. is \(\pi(\theta) = \left\{\begin{matrix}1 , \ \ 0 <\theta< 1\\\ 0,\rm \ \ \ \ otherwise \end{matrix} \right.\)
We need to find the Bayes estimate for $\theta$ under two different loss functions: Squared Error Loss (SEL) and Absolute Error Loss (AEL).
Recall that:
- The Bayes estimate under SEL is the posterior mean of $\theta$, i.e., $E[\theta|x]$.
- The Bayes estimate under AEL is the posterior median of $\theta$.
Posterior Distribution Calculation
To find the Bayes estimates, we first need to find the posterior distribution of $\theta$ given the observed value $x$. The posterior p.d.f., $\pi(\theta|x)$, is proportional to the product of the likelihood (conditional p.d.f.) and the prior p.d.f.:
\(\pi(\theta|x) \propto f(x|\theta) \pi(\theta)\)Substituting the given functions:
\(\pi(\theta|x) \propto \frac{2x}{\theta^2} \times 1\)This is valid when $0 < x < \theta$ and $0 < \theta < 1$. These conditions together imply $x < \theta < 1$. For this range to be non-empty, we must have $0 < x < 1$.
So, the joint p.d.f. is \(f(x, \theta) = \frac{2x}{\theta^2}\) for $x < \theta < 1$ and $0 < x < 1$. Otherwise, it's 0.
To get the posterior p.d.f., we need to find the normalizing constant, which is the marginal p.d.f. of $X$, $m(x)$.
\(m(x) = \int_{-\infty}^{\infty} f(x, \theta) d\theta\)For $0 < x < 1$, the integral is:
\(m(x) = \int_{x}^{1} \frac{2x}{\theta^2} d\theta\) \(m(x) = 2x \int_{x}^{1} \theta^{-2} d\theta\) \(m(x) = 2x \left[ -\theta^{-1} \right]_{x}^{1}\) \(m(x) = 2x \left[ -\frac{1}{1} - (-\frac{1}{x}) \right]\) \(m(x) = 2x \left[ -1 + \frac{1}{x} \right]\) \(m(x) = 2x \left( \frac{1-x}{x} \right)\) \(m(x) = 2(1-x)\) for $0 < x < 1$.Now, the posterior p.d.f. is:
\(\pi(\theta|x) = \frac{f(x, \theta)}{m(x)} = \frac{2x/\theta^2}{2(1-x)} = \frac{x}{\theta^2(1-x)}\) for $x < \theta < 1$ and $0 < x < 1$.Otherwise, \(\pi(\theta|x) = 0\).
Bayes Estimate under Squared Error Loss
The Bayes estimate under SEL is the posterior mean $E[\theta|x]$.
\(E[\theta|x] = \int_{-\infty}^{\infty} \theta \pi(\theta|x) d\theta\)Using the posterior p.d.f. we found for $0 < x < 1$:
\(E[\theta|x] = \int_{x}^{1} \theta \times \frac{x}{\theta^2(1-x)} d\theta\) \(E[\theta|x] = \frac{x}{1-x} \int_{x}^{1} \frac{1}{\theta} d\theta\) \(E[\theta|x] = \frac{x}{1-x} \left[ \log_e \theta \right]_{x}^{1}\) \(E[\theta|x] = \frac{x}{1-x} (\log_e 1 - \log_e x)\) \(E[\theta|x] = \frac{x}{1-x} (0 - \log_e x)\) \(E[\theta|x] = -\frac{x \log_e x}{1-x}\)This is the Bayes estimate under Squared Error Loss.
Bayes Estimate under Absolute Error Loss
The Bayes estimate under AEL is the posterior median of $\theta$. Let $m$ be the posterior median. It is the value such that:
\(\int_{-\infty}^{m} \pi(\theta|x) d\theta = 0.5\)For $x < \theta < 1$, the integral becomes:
\(\int_{x}^{m} \frac{x}{\theta^2(1-x)} d\theta = 0.5\)where $x < m < 1$ (since the density is positive only between $x$ and 1).
\(\frac{x}{1-x} \int_{x}^{m} \theta^{-2} d\theta = 0.5\) \(\frac{x}{1-x} \left[ -\theta^{-1} \right]_{x}^{m} = 0.5\) \(\frac{x}{1-x} \left[ -\frac{1}{m} - (-\frac{1}{x}) \right] = 0.5\) \(\frac{x}{1-x} \left( \frac{1}{x} - \frac{1}{m} \right) = 0.5\)Simplify the term in the parenthesis:
\(\frac{x}{1-x} \left( \frac{m-x}{xm} \right) = 0.5\) \(\frac{m-x}{m(1-x)} = 0.5\)Multiply both sides by $2m(1-x)$:
\(2(m-x) = m(1-x)\) \(2m - 2x = m - mx\) \(2m - m + mx = 2x\) \(m + mx = 2x\) \(m(1 + x) = 2x\) \(m = \frac{2x}{1+x}\)This is the Bayes estimate under Absolute Error Loss.
Summary of Bayes Estimates
Based on our calculations:
- The Bayes estimate for $\theta$ under Squared Error Loss is \(-\frac{x \log_e x}{1-x}\).
- The Bayes estimate for $\theta$ under Absolute Error Loss is \(\frac{2x}{1+x}\).
Comparing these results with the given options, we find that the expression \(-\frac{x\log _ex}{1-x}\) is the Bayes estimate under squared error loss and \(\frac{2x}{1+x}\) is the Bayes estimate under absolute error loss.
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