Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 41 / 1

For the mean parameter $\theta =1/\lambda >0$, the log-likelihood is $\ell (\theta )=-n\log \theta -{\sum }_i{Y}_i/\theta$. Its derivative is $n(\bar{Y}-\theta )/{\theta }^2$, positive below $\bar{Y}$ and negative above it. Therefore $\boxed{\hat{\theta }=\bar{Y}}$ is the unique maximum-likelihood estimator.

The exponential distribution has mean $\theta$ and variance ${\theta }^2$. The central limit theorem gives the pivotLet $z={\Phi }^{-1}(1-\alpha /2)$ for fixed $0<\alpha <1$. For $n>z^2$, the event $-z<T_n<z$ is exactlyBoth denominators are then positive, so the algebra neither reverses the ordering nor changes the parameter domain. Taking probabilities and using continuity of the limiting normal distribution proves the exponential-mean confidence interval by pivot inversion has coverage probability tending to $\boxed{1-\alpha }$. This is inversion of the pivot with the true mean in its denominator; it is not the ordinary symmetric Wald interval centered at $\bar{Y}$. For a small sample with $n\le z^2$, the displayed finite positive upper endpoint is not valid, although the large-sample claim is unaffected.

Use the Probabilists' Hermite polynomials, rather than the differently scaled physicists' convention:Thus $H_2=x^2-1$, $H_3=x^3-3x$, and $H_5=x^5-10x^3+15x$. The cumulant-generating function of an exponential variable of mean $\theta$ is $-\log (1-\theta t)$ near zero, so its $r$th cumulant is $(r-1)!{\theta }^r$. Hence the standardized third and fourth cumulants are ${\rho }_3=2$ and ${\rho }_4=6$; the fourth is a cumulant, not the fourth raw central moment.

Write $F_n$ for the cumulative distribution function of the pivot in part (i). Its Edgeworth expansion, valid here because the exponential law is smooth, nonlattice and has exponential moments, isat the fixed endpoints $t=\pm z$. The coverage is exactly $F_n(z)-F_n(-z)$. Since $H_2$ and $\varphi$ are even, the $n^{-1/2}$ terms cancel; the two odd Hermite terms double. Consequently the symmetric Edgeworth coverage cancellation givesIn particular its coverage probability error is $\boxed{O(n^{-1})}$. This is a fixed-confidence-level expansion; it is not a uniform statement for normal cutoffs increasing with $n$.