Question

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

The Fisher information \(I(\theta)\) is \(\frac{1}{2\theta^2}\), the MLE estimator of \(\theta\) is efficient, and the asymptotic distribution of \(\sqrt{n}(\widehat{\theta}-\theta)\) is \(N\left(0,2\theta^2\right)\).

Step-by-step solution

Fisher Information

First, let's find out the Fisher information. Note that the probability density function of \(X\), a normal distribution is given by \(f(x; \theta) = \frac{1}{\sqrt{2 \pi \theta}} e^{-\frac{x^2}{2\theta}}\). The derivative of the log-likelihood with respect to \( \theta \) is \(\frac{d}{d\theta} \log f(x;\theta) = -\frac{1}{2\theta} + \frac{x^2}{2\theta^2}\). The Fisher information \(I(\theta)\) is the variance of the derivative of the log-likelihood, which is \(I(\theta) = E\left[\left(\frac{d}{d\theta} \log f(X; \theta)\right)^2\right] = \frac{1}{2\theta^2}\).

Efficiency of the MLE

Let's denote the sample mean of \(n\) observations by \(\overline{X}\). We know from the theory, the MLE of \( \theta \) is \( \widehat{\theta} = \overline{X}^2 \). The variance of \(\overline{X}\) is \( \theta/n \). Therefore, the variance of \( \widehat{\theta} \) is \(4 \cdot \theta^2/n \), which is exactly the Cramér-Rao lower bound. Thus, the MLE \( \widehat{\theta} \) is an efficient estimator of \( \theta \).

Asymptotic Distribution

Because the MLE \( \widehat{\theta} \) is efficient, we can use the Fisher Information to find the asymptotic distribution of \( \sqrt{n}(\widehat{\theta} - \theta) \). By the Central Limit Theorem, we know that \(\sqrt{n}(\widehat{\theta} - \theta)\) converges in distribution to \(N(0, I(\theta)^{-1})\). Substituting \(I(\theta) = \frac{1}{2\theta^2}\), the asymptotic distribution of \(\sqrt{n}(\widehat{\theta}-\theta)\) is \(N\left(0,2\theta^2\right)\).