Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 9

The independent uniform choices from the student list describe sampling with replacement. Under the stated independent normal distribution model for the heights, the population variance is unknown and the given $8.5$ cm is the sample standard deviation $s$. The exact pivotal quantity isThus the exact 95% Student t confidence interval isusing $t_{99,0.975}\approx 1.9842$. The supplied approximation $\Phi (2)=0.975$ instead gives a normal approximation for large sample sizeThis latter confidence interval is approximate even under normality because it substitutes the sample standard deviation for the unknown population value. There is no finite population correction under the specified independent selections. Interpreting the finite student list literally as a discrete population makes normality an idealized model; it does not justify changing the stated sampling design.

At sample size 100, keep essentially the same numerical confidence interval, but treat its calibration as approximate. For independent heights with finite, nonzero variance, the central limit theorem makes the sample mean approximately normally distributed, and the sample variance consistently estimates the population variance. Hence the studentized sample mean is approximately standard normal even without a normal distribution for individual heights.

Using the permitted cutoff 2 givesA Student t-distribution cutoff gives a similar approximation for large sample size, but its justification at fixed sample size is no longer exact. Extreme skewness or influential outliers could make 100 observations insufficient for accurate coverage probability; abandoning normality alone supplies no specific numerical correction. The sampling with replacement assumption still precludes a finite population correction.

For this rate parametrization of the Laplace distribution, $\mu$ is its mean and median. By symmetry, $\mathbb{E}(X-\mu )=0$, while $|X-\mu |$ has exponential distribution with rate $\lambda$, soThe integral follows by two applications of integration by parts. The central limit theorem therefore givesWith known $\lambda$ and the supplied cutoff 2, an approximate 95% confidence interval isThis known-rate Laplace location confidence interval uses the sample mean; the later maximum-likelihood estimator of location need not be used to construct this valid asymptotic confidence interval.

By independence, the likelihood function is the product of the individual probability density functions:Writing $A(\mu )={\sum }_i|x_i-\mu |$, the log-likelihood function is $\ell =n\ln \lambda -n\ln 2-\lambda A(\mu )$. Here $\lambda$ is a rate, the reciprocal of the scale in the usual Laplace distribution notation.

The joint maximum-likelihood estimators are functions of the observations selecting a maximizing parameter pair:The estimates after observation maximize the same likelihood function with $X_i=x_i$. Maximization may be nonunique, as for an even sample size, or may fail to yield a finite admissible pair, as in the one-observation case below. The definition does not turn a boundary supremum into an attained maximum.

For each fixed $\lambda >0$, maximizing the likelihood function over $\mu$ is equivalent to minimizing $A(\mu )={\sum }_i|x_i-\mu |$. Order the distinct observations as $x_{(1)}<\cdots <x_{(n)}$ and write $n=2m+1$. Between successive order statistics, with exactly $k$ observations to the left,Thus $A$ decreases up to $x_{(m+1)}$ and increases after it. Its unique minimum is at the sample median:At a fixed $\mu$ with $A(\mu )>0$, the log-likelihood function has derivative $n/\lambda -A(\mu )$ and second derivative $-n/{\lambda }^2<0$, so its unique maximizing rate is $n/A(\mu )$. The maximized likelihood function is $(n/[2eA(\mu )]{)}^n$, which increases when $A(\mu )$ decreases. Hence the joint maximum-likelihood estimates areThis is Laplace location and scale maximum likelihood, expressed with the reciprocal scale.

A finite rate maximum-likelihood estimate requires a positive total absolute deviation. For distinct observations and odd $n\ge 3$, that condition holds. For $n=1$, however, $\mu =x_1$ gives $A(\mu )=0$ and $L=\lambda /2\to \infty$ as $\lambda \to \infty$. There is no finite joint maximum-likelihood estimate in the stated parameter space; the formula with zero denominator is not an admissible estimate.

Let $n=2m\ge 2$. Between the two middle order statistics, $k=m$ and the slope $A^{\prime }(\mu )=2m-n$ is zero. To the left $A$ decreases and to the right it increases. Consequently all joint maximum-likelihood estimates arewhere the constant minimum isThe strict positivity follows from distinct observations. Thus the location maximum-likelihood estimate is not unique, while the rate maximum-likelihood estimate is unique and the same for every location in the displayed interval. Choosing the arithmetic midpoint of the two middle values is the usual sample median convention and gives one member of this entire maximizing family.