Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / O (Statistics)

For the continuous distribution in the question, $F$ maps the support monotonically from zero to one. For $0<t<1$, strict monotonicity givesThus $F(X)$ is uniform on $(0,1)$, the probability integral transform.

The order statistics are the sample values sorted increasingly, $X_{(1)}\le \cdots \le X_{(n)}$. Monotonicity of $F$ makes $F(X_{(i)})$ the $i$th order statistic of independent uniforms. Its density is $n!t^{i-1}(1-t{)}^{n-i}/[(i-1)!(n-i)!]$, givingThis follows either by beta integrals or by the equal expected lengths of the $n+1$ uniform spacings.

For the stated log-logistic cumulative distribution,Therefore plot $\log [p_i/(1-p_i)]$ vertically against $\log X_{(i)}$ horizontally, with plotting positions $p_i=i/(n+1)$. A plausible model gives an approximately straight line, with slope $\beta -1$ and intercept $-(\beta -1)\log \alpha$. This specifies the axes explicitly, since reversing them would invert the slope.

With immediate replacement and uniform mixing, each capture is a Bernoulli trial with tagged probability $p=k/N$. Counting failures before the $r$th tagged capture gives the stated negative binomial distribution. Independent runs on the successive days give the sample model.

Writing $S={\sum }_i{x}_i$, the log-likelihood is $\ell (p)=nr\log p+S\log (1-p)+\text{constant}$. Its derivative vanishes atThe second derivative is negative. If $S=0$, this formula gives the boundary maximum $p=1$; on the strict physical parameter space $p<1$, it is a supremum rather than an attained interior maximum.

Using the supplied negative-binomial mean, $\mathbb{E}S=nr(1-p)/p$. Thus total Fisher information for the sample is

For an interior true parameter and large $n$, standard asymptotic normality of a maximum likelihood estimator givesEstimate its standard error by $s_e=\hat{p}\sqrt{(1-\hat{p})/(nr)}$. A two-sided approximate Wald confidence interval is $(L,U)=(\hat{p}-z_{1-\alpha /2}{s}_e,\hat{p}+z_{1-\alpha /2}{s}_e)$, intersected with the parameter range $(0,1)$. The normal quantile is chosen so a standard normal lies between its positive and negative values with probability $1-\alpha$.

The relation $N=k/p$ is strictly decreasing. Thus an interval $0<L<U<1$ transforms by reversing endpoints:Keep integer population sizes satisfying $N>k$. If the clipped lower confidence endpoint for $p$ is zero, the transformed upper endpoint for $N$ is infinite. This monotone transformation preserves the approximate coverage probability supplied by the interval for $p$.

An independent sum of normal variables is normal, with the means and variances added. Therefore $\overline{X}\sim N(\mu ,{\sigma }^2/n)$ and $Z\sim N(0,1)$ exactly.

Under the null, $Z_0=\sqrt{n}(\overline{X}-{\mu }_0)/\sigma$ is standard normal. Reject for $Z_0>w$, where $w={\Phi }^{-1}(1-\alpha )$, giving size $\alpha$. For $\alpha =0.05$, the supplied table gives the critical region $\overline{X}>{\mu }_0+1.64\sigma /\sqrt{n}$, using its rounded quantile.

The power function is the probability of rejection when the true parameter is $\mu$. Recenter by that true mean:For $n=100$, $\sigma =2$, power at least $0.70$ means $w-5(\mu -{\mu }_0)\le -0.52$. Hence, with the supplied rounding,The exact unrounded threshold is ${\mu }_0+({\Phi }^{-1}(0.95)-{\Phi }^{-1}(0.30))/5$.