Say I have X1 to Xn, as n i.i.d. observations of a uniform distribution that goes from an unknown Theta to Theta-squared, with Theta being strictly greater than 1. How do I compute the MLE of Theta? I was under the assumption that the maximum likelihood estimator of Theta would in this case be the smallest possible X_i, so min(X_i), because Theta is the lowest possible value on the uniform.

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:
   \[
   f(x; \Theta) =
   \begin{cases}
   \frac{1}{\Theta^2 - \Theta}, & \text{if } \Theta \leq x \leq \Theta^2, \\
   0, & \text{otherwise}.
   \end{cases}
   \]
   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 \(\Theta^2 \geq X_{\max}\).

---

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.

Explore the detailed derivation of the Maximum Likelihood Estimator (MLE) for Theta in a uniform distribution defined over [Theta, Theta^2]. Learn how constraints on the parameter lead to the estimator Θ = max(Xmin, sqrt(Xmax)). Suitable for undergraduate students in statistics.

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