(Continuation of Exercise 6.4-2.) In sampling from a

Chapter 6, Problem 3E

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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

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 .

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