Solution Found!
(Continuation of Exercise 6.4-2.) In sampling from a
Chapter 6, Problem 3E(choose chapter or problem)
(Continuation of Exercise 6.4-2.) In sampling from a normal distribution with known mean \(\mu\), the maximum likelihood estimator of \(\theta=\sigma^{2}\) is \(\hat{\theta}=\sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2} / n\)
(a) Determine the Rao-Cramér lower bound.
(b) What is the approximate distribution of \(\hat{\theta}\)?
(c) What is the exact distribution of \(n \hat{\theta} / \theta\), where \(\theta=\sigma^{2}\) ?
Equation Transcription:
Text Transcription:
mu
theta=sigma^2
Hat thea =sum_i=1^n( X_i-mu)^2/n
Hat theta
n hat theta/theta
Questions & Answers
QUESTION:
(Continuation of Exercise 6.4-2.) In sampling from a normal distribution with known mean \(\mu\), the maximum likelihood estimator of \(\theta=\sigma^{2}\) is \(\hat{\theta}=\sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2} / n\)
(a) Determine the Rao-Cramér lower bound.
(b) What is the approximate distribution of \(\hat{\theta}\)?
(c) What is the exact distribution of \(n \hat{\theta} / \theta\), where \(\theta=\sigma^{2}\) ?
Equation Transcription:
Text Transcription:
mu
theta=sigma^2
Hat thea =sum_i=1^n( X_i-mu)^2/n
Hat theta
n hat theta/theta
ANSWER:
Step 1 of 10
Given:- In sampling from a normal distribution with known mean , the maximum likelihood estimator of is .