Bayesian Analysis: Lambda Distribution Estimation

This problem involves Bayesian inference for a parameter $\lambda$ from a given probability density function (PDF).

Analyzing the Density Function

The given probability density function for $x > 0$ is $f_\lambda(x) = 2\lambda x e^{-\lambda x^2}$.

Let $Y = \lambda X^2$. The PDF of $Y$ can be derived. The cumulative distribution function (CDF) is $P(Y \le y) = P(\lambda X^2 \le y) = P(X^2 \le y/\lambda)$.

Consider the transformation $y = \lambda x^2$. Then $x = \sqrt{y/\lambda}$. The Jacobian is $dx/dy = \frac{1}{2\sqrt{y\lambda}}$.

The PDF for $Y$ is $f_Y(y) = f_X(x(y)) \left|\frac{dx}{dy}\right| = 2\lambda \sqrt{y/\lambda} e^{-\lambda (y/\lambda)} \cdot \frac{1}{2\sqrt{y\lambda}} = 2\lambda \frac{\sqrt{y}}{\sqrt{\lambda}} e^{-y} \frac{1}{2\sqrt{y}\sqrt{\lambda}} = e^{-y}$ for $y>0$. This is the PDF of a standard Exponential distribution.

Alternatively, let $Z = X^2$. Then $f_Z(z) = \lambda e^{-\lambda z}$ for $z>0$, which is an Exponential distribution with rate $\lambda$. The likelihood function derived from $f_\lambda(x)$ depends on $\sum x_i^2$. The likelihood function for a sample $x_1, \dots, x_n$ is:

$L(\lambda | \mathbf{x}) = \prod_{i=1}^n 2\lambda x_i e^{-\lambda x_i^2} = (2\lambda)^n \left(\prod_{i=1}^n x_i\right) e^{-\lambda \sum_{i=1}^n x_i^2}$

Let $S = \sum_{i=1}^n x_i^2$. The likelihood is proportional to $\lambda^n e^{-\lambda S}$.

Analyzing the Prior Distribution

The prior distribution for $\lambda$ is Exponential with mean 1. The PDF of an Exponential distribution with rate $\alpha$ is $p(\lambda) = \alpha e^{-\alpha \lambda}$. The mean is $1/\alpha$. Given mean = 1, we have $1/\alpha = 1$, which implies $\alpha = 1$. So, the prior PDF is $p(\lambda) = 1 \cdot e^{-1 \cdot \lambda} = e^{-\lambda}$ for $\lambda > 0$.

Deriving the Posterior Distribution

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

$p(\lambda | \mathbf{x}) \propto L(\lambda | \mathbf{x}) p(\lambda)$

$p(\lambda | \mathbf{x}) \propto (\lambda^n e^{-\lambda S}) (e^{-\lambda})$

$p(\lambda | \mathbf{x}) \propto \lambda^n e^{-\lambda(S+1)}$

This is the kernel of a Gamma distribution. Specifically, $p(\lambda | \mathbf{x})$ follows a Gamma distribution with shape parameter $k = n+1$ and rate parameter $\theta = S+1$. Notation: $\lambda | \mathbf{x} \sim \text{Gamma}(n+1, S+1)$.

Evaluating the Options

1. Option A: Posterior distribution of $\lambda$ is exponential.

   The derived posterior distribution is Gamma($n+1$, $S+1$), not Exponential. Thus, statement A is incorrect.

2. Option B: Bayes estimator of $\lambda$ w.r.t. squared error loss exists and is unique.

   The Bayes estimator under squared error loss is the posterior mean. For a Gamma($k$, $\theta$) distribution, the mean is $k/\theta$. Here, the posterior mean is $E[\lambda | \mathbf{x}] = \frac{n+1}{S+1}$. Since $n \ge 1$ and $S = \sum x_i^2 > 0$, the mean is finite and well-defined. The posterior distribution is unique and continuous, ensuring a unique posterior mean. Thus, statement B is correct.

3. Option C: Bayes estimator of $\lambda$ w.r.t. absolute error loss exists and is unique.

   The Bayes estimator under absolute error loss is the posterior median. The Gamma distribution has a unique median. Therefore, the Bayes estimator for absolute error loss exists and is unique. Thus, statement C is correct.

4. Option D: Bayes estimator of $e^{-\lambda}$ does not exist.

   The Bayes estimator for $g(\lambda) = e^{-\lambda}$ is the posterior expectation $E[e^{-\lambda} | \mathbf{x}]$.

   $ E[e^{-\lambda} | \mathbf{x}] = \int_0^\infty e^{-\lambda} p(\lambda | \mathbf{x}) d\lambda $

   Using the posterior $\text{Gamma}(n+1, S+1)$, the integral evaluates to:

   $ \int_0^\infty e^{-\lambda} \frac{(S+1)^{n+1}}{\Gamma(n+1)} \lambda^n e^{-(S+1)\lambda} d\lambda = \frac{(S+1)^{n+1}}{\Gamma(n+1)} \int_0^\infty \lambda^n e^{-(S+2)\lambda} d\lambda $

   The integral is $\frac{\Gamma(n+1)}{(S+2)^{n+1}}$.

   $ E[e^{-\lambda} | \mathbf{x}] = \frac{(S+1)^{n+1}}{\Gamma(n+1)} \cdot \frac{\Gamma(n+1)}{(S+2)^{n+1}} = \left(\frac{S+1}{S+2}\right)^{n+1} $

   This value is finite and exists. Thus, statement D is incorrect.

The correct statements are B and C.