Past exam of the probability, statistics and options I course of the University of Oxford / 2011 / O / 1

For the sample likelihood function $L(\theta )={\prod }_{i}f(X_i;\theta )$ and log-likelihood $\ell =\log L$, the expected Fisher information isThe second equality uses the usual regularity assumptions permitting differentiation under the integral and a parameter-independent support. Independent identically distributed observations give $I_n(\theta )=n{I}_1(\theta )$.

Under the standard consistency and regularity hypotheses for the maximum-likelihood estimator,Equivalently, for large samples $\hat{\theta }$ is approximately $N(\theta ,I_n(\theta {)}^{-1})$. In a standard-error estimate, the unknown parameter in the information may be replaced by the consistent estimate $\hat{\theta }$.

Put $Q={\sum }_i{X}_i^2$. For $\theta >0$, the log-likelihood is $\ell (\theta )=\text{constant}-(n/2)\log \theta -Q/(2\theta )$. Its derivative is $-n/(2\theta )+Q/(2{\theta }^2)$, changing from positive to negative at $Q/n$. ThusThe known mean is zero, so no centring at the sample mean is needed. Taking the expected negative second derivative gives $I_n(\theta )=n/(2{\theta }^2)$. The large-sample variance is consequently $2{\theta }^2/n$. With $z_{1-\alpha /2}$ the indicated standard-normal quantile, a Wald confidence interval isThe interval may be restricted to the parameter space $\theta >0$. The parameter is the variance itself, so the standard error is proportional to $\hat{\theta }$, not to its square root.

Since the true mean is known to be zero, the chi-squared distribution gives $Q/\theta \sim {\chi }_n^2$. Let $q_{\beta }$ be the $\beta$ quantile of this law. The probabilityrearranges, using positivity, to the exact confidence intervalThere are $n$ degrees of freedom, because no unknown mean was fitted.