Question

Let \( X_1, X_2, \ldots, X_{10} \) be a random sample from a \( U(-\theta, \theta) \) distribution, where \( \theta \in (0, \infty) \). Let \( X_{(10)} = \max \{ X_1, X_2, \ldots, X_{10} \} \) and \( X_{(1)} = \min \{ X_1, X_2, \ldots, X_{10} \} \). If the observed values of \( X_{(10)} \) and \( X_{(1)} \) are 8 and -10, respectively, then the maximum likelihood estimate of \( \theta \) is equal to __________ (answer in integer).

Answer 1

Correct Answer: 10

Step 1: Understand the problem and the uniform distribution.

The given distribution is \( U(-\theta, \theta) \), which implies: \[ f(x; \theta) = \begin{cases} \frac{1}{2\theta}, & \text{if } -\theta \leq x \leq \theta, \\ 0, & \text{otherwise}. \end{cases} \]

The range of the sample, defined by \( X_{(1)} = \min(X_1, X_2, \ldots, X_{10}) \) and \( X_{(10)} = \max(X_1, X_2, \ldots, X_{10}) \), determines the parameter \( \theta \).

Step 2: Derive the maximum likelihood estimate (MLE) of \( \theta \).

For the uniform distribution \( U(-\theta, \theta) \), the likelihood function based on the observed data is: \[ L(\theta) = \prod_{i=1}^{10} f(x_i; \theta) = \begin{cases} \left(\frac{1}{2\theta}\right)^{10}, & \text{if } -\theta \leq X_{(1)} \text{ and } X_{(10)} \leq \theta, \\ 0, & \text{otherwise}. \end{cases} \]

To maximize the likelihood function, \( \theta \) must satisfy: \[ \theta \geq \max(|X_{(1)}|, |X_{(10)}|). \]

Step 3: Apply the observed values.

From the problem, the observed values are: \[ X_{(1)} = -10 \quad \text{and} \quad X_{(10)} = 8. \] Thus, the maximum likelihood estimate of \( \theta \) is: \[ \hat{\theta} = \max(|-10|, |8|) = 10. \]

Conclusion:

The maximum likelihood estimate of \( \theta \) is: \[ \boxed{10}. \]