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Chapter 6: Problem 8

Let \(X\) be \(N(0, \theta), 0<\theta<\infty\). (a) Find the Fisher information \(I(\theta)\). (b) If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from this distribution, show that the mle of \(\theta\) is an efficient estimator of \(\theta\). (c) What is the asymptotic distribution of \(\sqrt{n}(\widehat{\theta}-\theta) ?\)

Short Answer

Expert verified
The Fisher Information, \(I(\theta)\), is obtained from the second derivative of the natural logarithm of the likelihood function of the Normal distribution. To show that the MLE is an efficient estimator, we find the MLE of \( \theta \), and then its variance, which should be equal to the Cramer-Rao Lower Bound (CRLB) to signify efficiency. The asymptotic distribution of the estimator, \( \sqrt{n}(\widehat{\theta}-\theta) \), is Normal according to the Central Limit Theorem.

Step by step solution

01

Find the Fisher Information \(I(\theta)\)

Fisher information, \(I(\theta)\), can be obtained from the likelihood function of Normal distribution \(N(0, \theta)\). The likelihood function, \(L(\theta)\), is:\(L(\theta) = \(1 / (\sqrt{2\pi\theta})\) * \(e^{(-x^2 / 2\theta)}\)Taking the natural logarithm (ln) of the likelihood function, and then finding the second derivative, gives the Fisher Information.
02

Show that MLE is an Efficient Estimator of \( \theta \)

To prove that MLE is an efficient estimator, we start by finding the MLE of \( \theta \). The MLE is obtained by setting the derivative of the natural logarithm of the likelihood function to zero and solving for \( \theta \). Once we have the MLE estimator, we find its variance, i.e. the Cramer-Rao Lower Bound (CRLB). If the variance of the MLE estimator equals the CRLB, it suggests that the MLE is an efficient estimator of \( \theta \).
03

Find Asymptotic Distribution of the Estimator

The asymptotic distribution of an estimator is usually Normal, according to the Central Limit Theorem, with mean equal to the parameter being estimated and variance equal to the inverse of Fisher Information scaled by sample size, i.e., \(n\). Hence, if \( \widehat{\theta} \) is the estimator, its asymptotic distribution is represented as \( \sqrt{n}(\widehat{\theta}-\theta) \). We can confirm this by evaluating the limit as \(n\) approaches infinity of \( \sqrt{n}(\widehat{\theta}-\theta) \).

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Most popular questions from this chapter

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