Chapter 6: Problem 5
Let \(X\) take on the values 0 and 1 with probabilities \(p\) and \(q\), respectively. When it is known that \(1 / 3 \leq p \leq 2 / 3\), (a) find the MLE and (b) show that the expected squared error of the MLE is uniformly larger than that of \(\delta(x)=1 / 2\). [A similar estimation problem arises in randomized response surveys. See Example 5.2.2.]
Short Answer
Step by step solution
Define the likelihood function
Find the log-likelihood function
Derive the MLE from the log-likelihood
Analyze the expected squared error of the MLE
Compare with \( \delta(x) = \frac{1}{2} \)
Conclusion: MLE expected squared error
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Likelihood Function
- \( L(p; x) = p^x (1-p)^{1-x} \)
- \( L(p; x_1, x_2, \ldots, x_n) = \prod_{i=1}^n p^{x_i} (1-p)^{1-x_i} \)
Log-Likelihood
- \( \ell(p; x_1, x_2, \ldots, x_n) = \sum_{i=1}^n \left( x_i \log(p) + (1-x_i) \log(1-p) \right) \)
- \( \hat{p} = \frac{\sum_{i=1}^n x_i}{n} \)
Unbiased Estimator
- \( E(\hat{p}) = p \)
Expected Squared Error
- \( \text{Var}(\hat{p}) = \frac{p(1-p)}{n} \)
- \( \text{Err}(\delta) = \left(\frac{1}{2} - p\right)^2 \)