Your assumption that the maximum likelihood estimator (MLE) of $\Theta$ is $\min(X_i)$ is close, but not entirely correct in this case. Let’s walk through the reasoning and computation.

Problem Setup:

You have $X_1, X_2, \ldots, X_n \sim \text{Uniform}(\Theta, \Theta^2)$, where $\Theta > 1$. The goal is to compute the MLE of $\Theta$, the unknown parameter.

1. Likelihood Function: The probability density function (PDF) of the uniform distribution is:

   \begin{cases} \frac{1}{\Theta^2 - \Theta}, & \text{if } \Theta \leq x \leq \Theta^2, \\ 0, & \text{otherwise}. \end{cases}$$ For the $$n$$ i.i.d. observations, the likelihood function is: $$L(\Theta) = \prod_{i=1}^n f(X_i; \Theta).$$ Since the $$f(x; \Theta)$$ is zero outside the range $$[\Theta, \Theta^2]$$, the likelihood is nonzero only if all observations satisfy: $$\Theta \leq X_{\min} \quad \text{and} \quad X_{\max} \leq \Theta^2,$$ where $$X_{\min} = \min(X_1, \dots, X_n)$$ and $$X_{\max} = \max(X_1, \dots, X_n)$$. Therefore, within this range, the likelihood function becomes: $$L(\Theta) = \left( \frac{1}{\Theta^2 - \Theta} \right)^n, \quad \text{if } \Theta \leq X_{\min} \text{ and } X_{\max} \leq \Theta^2.$$

2. Log-Likelihood: To find the MLE, maximize the likelihood (or equivalently, the log-likelihood): $\ell(\Theta) = -n \ln(\Theta^2 - \Theta).$ However, the constraints $\Theta \leq X_{\min}$ and $\Theta^2 \geq X_{\max}$ are critical for determining the feasible range of $\Theta$.

3. Key Observations:

   • The condition $\Theta \leq X_{\min}$ ensures that $\Theta$ is at least as small as the smallest observed value.
   • The condition $\Theta^2 \geq X_{\max}$ ensures that $\Theta$ is large enough so that the largest observed value is within the distribution’s range.

   Combining these constraints: $\Theta \geq \max(X_{\min}, \sqrt{X_{\max}}).$

4. Maximizing the Likelihood: Within the feasible range, the MLE maximizes the likelihood. Since $\Theta^2 - \Theta$ decreases as $\Theta$ increases, the MLE occurs at the smallest value of $\Theta$ that satisfies the constraints: $\hat{\Theta} = \max(X_{\min}, \sqrt{X_{\max}}).$

Final Answer:

The MLE of $\Theta$ is: $\hat{\Theta} = \max(X_{\min}, \sqrt{X_{\max}}).$

Intuition:

While $\min(X_i)$ alone may seem intuitive, the constraint $\Theta^2 \geq X_{\max}$ means $\Theta$ must be large enough to accommodate the largest observation as well. Thus, the MLE balances both constraints, choosing the smallest $\Theta$ that satisfies $\Theta \leq X_{\min}$ and $\Theta^2 \geq X_{\max}$.

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Do you have further questions or need clarification on this derivation?

Related Questions:

1. How does the constraint $\Theta^2 \geq X_{\max}$ impact the likelihood?
2. Why does maximizing the likelihood result in choosing the boundary of the feasible range for $\Theta$?
3. What happens to the MLE if $\Theta \leq 1$ were allowed (hypothetically)?
4. How does the sample size $n$ affect the MLE of $\Theta$?
5. Could the MLE fail if the sample does not include $X_{\min}$ close to $\Theta$?

Tip:

When working with MLEs for distributions with boundaries, always carefully consider how the observations relate to the parameter constraints.