Defining the Distributions

We are given the following probability distributions:

• The conditional distribution of X given θ is Uniform: $X|\theta \sim \text{Uniform}[0, \theta]$. Its probability density function (PDF) is $f(x|\theta) = \frac{1}{\theta}$ for $0 \le x \le \theta$, and $0$ otherwise.
• The prior distribution of θ is Exponential with mean λ: $\theta \sim \text{Exponential}(\text{rate} = 1/\lambda)$. Its PDF is $f(\theta) = \frac{1}{\lambda} e^{-\theta/\lambda}$ for $\theta \ge 0$.

Posterior Distribution Derivation

We observed the value $X = 2025$. The condition $0 \le x \le \theta$ from the Uniform distribution implies that we must have $\theta \ge x$. Therefore, for our observed value, the constraint is $\theta \ge 2025$.

The posterior distribution $f(\theta|x)$ is derived using Bayes' theorem, and it is proportional to the product of the conditional PDF and the prior PDF: $f(\theta|x) \propto f(x|\theta) f(\theta)$.

Substituting the known PDFs and considering the constraint $\theta \ge 2025$:

• $f(x|\theta) = \frac{1}{\theta}$ (since $\theta \ge 2025 \ge x$)
• $f(\theta) = \frac{1}{\lambda} e^{-\theta/\lambda}$

The joint PDF is proportional to their product:

$f(x, \theta) \propto \frac{1}{\theta} \cdot \frac{1}{\lambda} e^{-\theta/\lambda}$ for $\theta \ge 2025$.

The posterior PDF for θ, given X, is proportional to the terms involving θ:

$f(\theta|x) \propto \frac{1}{\theta} e^{-\theta/\lambda}$ for $\theta \ge 2025$.

Finding the Posterior Mode

The posterior mode is the value of θ that maximizes the posterior PDF $f(\theta|x)$. We need to find the maximum of the function $g(\theta) = \frac{1}{\theta} e^{-\theta/\lambda}$ on the interval $\theta \ge 2025$.

To determine where the maximum occurs, we analyze the derivative of $g(\theta)$ with respect to θ:

$\frac{dg(\theta)}{d\theta} = \frac{d}{d\theta} \left( \theta^{-1} e^{-\theta/\lambda} \right)$

Applying the product rule:

$\frac{dg(\theta)}{d\theta} = (-\theta^{-2}) e^{-\theta/\lambda} + (\theta^{-1}) e^{-\theta/\lambda} (-\frac{1}{\lambda})$

Factor out common terms:

$\frac{dg(\theta)}{d\theta} = - e^{-\theta/\lambda} \left( \frac{1}{\theta^2} + \frac{1}{\lambda \theta} \right)$

Simplify the expression in the parenthesis:

$\frac{dg(\theta)}{d\theta} = - e^{-\theta/\lambda} \left( \frac{\lambda + \theta}{\lambda \theta^2} \right)$

Consider the domain $\theta \ge 2025$ and the given condition $\lambda > 3$. In this domain:

• $e^{-\theta/\lambda} > 0$
• $\lambda > 0$
• $\theta > 0$
• $\lambda \theta^2 > 0$
• $\lambda + \theta > 0$

Therefore, the entire expression for the derivative, $\frac{dg(\theta)}{d\theta}$, is strictly negative for all $\theta \ge 2025$.

A negative derivative indicates that the function $g(\theta)$ is strictly decreasing over its domain $[2025, \infty)$.

For a strictly decreasing function on an interval of the form $[a, \infty)$, the maximum value occurs at the smallest value in the interval, which is $a$.

In this case, the interval is $[2025, \infty)$, so the maximum occurs at $\theta = 2025$.

The posterior mode is therefore $2025$.

Final Answer

The posterior mode equals $2025$.