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Fix $T<\infty$. By uniform continuity of each path of the continuous local martingale on $[0,T+1]$, its modulustends to zero almost surely. For every $s\le T$, comparison of powers of each increment givesThe squared-increment sum converges in probability to $[X{]}_T$, by the dyadic definition of quadratic variation. Including the last full grid interval instead of the last partial interval changes that sum by a quantity bounded by $2{\eta }_n^2$, so the ceiling convention has the same limit. In particular these squared-increment sums are bounded in probability.

Since $p-2>0$, a quantity tending to zero almost surely multiplied by a family bounded in probability tends to zero in probability. HenceThis is uniform convergence on compacts in probability, and proves the superquadratic case of dyadic power variation of a continuous local martingale.

Fix the specified time $t$. With the path modulus ${\eta }_n$ on $[0,t+1]$ as above, comparison of increment powers now givesThe hypothesis makes the second factor eventually bounded almost surely, while the first tends to zero almost surely because $2-p>0$. Thus the squared-increment sums tend to zero almost surely. They also converge in probability to the quadratic variation $[X{]}_t$, so uniqueness of the limit in probability implies $[X{]}_t=0$ almost surely.

The quadratic variation is nondecreasing, hence it vanishes throughout $[0,t]$. To see explicitly that the continuous local martingale must then be constant, localize it to square-integrable stopped martingales. The Itô isometry givesRemove the localization, apply the conclusion at every rational $s\le t$, and use continuity. Since $X_0=0$, this provesThus $X$ is indistinguishable from the zero process on that interval. This is the subquadratic case of dyadic power variation of a continuous local martingale.

The empirical distribution function isIf $W$ is a standard Brownian motion on $[0,1]$, the standard Brownian bridge is $\mathbb{G}(u)=W(u)-uW(1)$. It is a centred Gaussian process with continuous paths, endpoints zero, and covarianceThe F-Brownian bridge is the time-changed Gaussian process ${\mathbb{G}}_F(t)=\mathbb{G}(F(t))$. Its covariance is $F(\min (s,t))-F(s)F(t)$, exactly the covariance of the empirical distribution process ${\alpha }_n=\sqrt{n}(F_n-F)$.

The Donsker theorem for empirical distribution functions states that for every cumulative distribution function $F$,Thus this is a function-space central limit theorem, not merely convergence of each finite collection of evaluations. The limit is a tight centred Gaussian process. No continuity of $F$ is required for this statement; a jump of $F$ can give a jump of ${\mathbb{G}}_F$.

For precision, ${\ell }^{\infty }(\mathbb{R})$ is generally nonseparable, and the finite-sample process need not be a Borel-measurable random element for its norm topology. The displayed convergence in distribution has the usual empirical-process interpretation: for every bounded continuous real functional $H$, the outer expectation and inner expectation of $H({\alpha }_n)$ converge to $\mathbb{E}H({\mathbb{G}}_F)$. The outer expectation is the infimum of expectations of measurable majorants, and the inner expectation is its negative applied to $-H({\alpha }_n)$. For measurable real statistics this gives ordinary convergence in distribution. The limit is tight and separably supported: $g\mapsto g\circ F$ maps the separable space of continuous bridge paths continuously into ${\ell }^{\infty }(\mathbb{R})$.

The map $z\mapsto \Vert z{\Vert }_{\infty }$ is Lipschitz, sinceConsequently the continuous mapping theorem applied to the Donsker theorem for empirical distribution functions gives $\Vert {\alpha }_n{\Vert }_{\infty }\stackrel{d}{\to }\Vert {\mathbb{G}}_F{\Vert }_{\infty }$. The statistic on the left is measurable: both $F_n$ and $F$ are right-continuous, so its supremum equals the supremum over rational $t$.

If $F$ is continuous, its limits at $-\infty$ and $+\infty$ are zero and one. The intermediate value theorem therefore shows that $F(\mathbb{R})$ contains $(0,1)$, even when $F$ is not strictly increasing. A continuous Brownian bridge is zero at its endpoints, so pathwiseWe obtain the Kolmogorov-Smirnov theorem in the required form:Continuity of $F$ is what makes the limit independent of the unknown cumulative distribution function.

Choose $c_{\alpha }$ with $\mathbb{P}(Z\le c_{\alpha })=1-\alpha$. The law of $Z$ has no atoms. For example, the Brownian reflection principle and conditioning a Brownian motion on its endpoint give $\mathbb{P}(\max \mathbb{G}>a)=e^{-2a^2}$ for $a>0$: the reflected endpoint density at zero is the normal density at $2a$, divided by its value at zero. Hence both one-sided maxima have continuous laws, and an atom of $Z$ would require an atom of one of them. Also $\mathbb{P}(Z=0)=0$, since $\mathbb{G}(1/2)$ is nondegenerate. Thus such quantiles give the desired limiting coverage.

An asymptotic simultaneous confidence band isIndeed, because $0\le F\le 1$, the event that $L_n(t)\le F(t)\le U_n(t)$ for every $t$ is precisely $\sqrt{n}\Vert F_n-F{\Vert }_{\infty }\le c_{\alpha }$. Its probability tends to $1-\alpha$. This is simultaneous coverage over all real $t$, rather than separate pointwise confidence intervals.

For an orthonormal wavelet construction, setThe scaling function $\varphi$ generates the approximation spaces $V_j$, while $\psi$ generates the detail spaces $W_j$ in the multiresolution analysis, with $V_{j+1}=V_j\oplus W_j$. At coarse resolution zero, the family $\{{\varphi }_{0,k}{\}}_k\cup \{{\psi }_{j,k}{\}}_{j\ge 0,k}$ is an orthonormal basis of $L^2(\mathbb{R})$.

For the given square-integrable function the coefficients and wavelet series areFor real wavelets these are the real L2 inner products; use complex conjugates for a complex basis. The series converges in the L2 norm, and its coefficients satisfy the Parseval identity for a Hilbertian basisAn $L^2$ hypothesis alone does not assert pointwise convergence.

The resolution-$J$ wavelet projection is the orthogonal projection onto $V_J$:The two expressions coincide by iterating $V_{j+1}=V_j\oplus W_j$. Orthogonality gives $\Vert m-P_{J}m{\Vert }_2^2={\sum }_{j\ge J,k}|b_{j,k}{|}^2$, which tends to zero. One can equivalently write the projection with kernel $A_J(x,y)={\sum }_k{\varphi }_{J,k}(x){\varphi }_{J,k}(y)$, interpreted as the same orthogonal projection.

In fixed-design nonparametric regression, the observations satisfywith independent errors. In the usual Gaussian version ${\epsilon }_i\sim N(0,{\sigma }^2)$. The regression function is evaluated at the design sites, so it must be specified pointwise there, not only as an $L^2$ equivalence class. For $x_i=i/n$, the observations concern $m$ on $[0,1]$. They cannot identify its values outside this interval. Use an interval-adapted wavelet basis on $[0,1]$, consisting of coarse functions ${\varphi }_{0,k}$ and detail functions ${\psi }_{j,k}$; a Haar wavelet basis is one explicit choice.

Here is a wavelet regression estimator valid for every $n$. Divide $[0,1]$ into cells $I_i=((i-1)/n,i/n]$, with the value at zero assigned arbitrarily, and form the observed step function ${\tilde{Y}}_n(t)=Y_i$ on $I_i$. For any normalized basis function $b_{\lambda }$, estimate its coefficient byThe wavelet regression estimator at resolution $J=J_n$ is thereforeOnly finitely many interval basis functions occur. Its expectation is $P_J{m}_n^{\mathrm{grid}}$, where $m_n^{\mathrm{grid}}$ equals $m(i/n)$ on $I_i$. Thus approximation of $m$ includes the resolution bias and the grid discretization bias. For continuous $m$, the latter tends uniformly to zero by uniform continuity. No smoothness or consistency claim for arbitrary pointwise representatives of $L^2$ functions is needed to define the estimator.

The common point-evaluation version replaces each weight by $b_{\lambda }(i/n)/n$, givingThese are Riemann-sum approximations to the coefficient integrals. For Haar wavelets, $n=2^L$, and $J\le L$, they equal the integrated-weight coefficients exactly if the dyadic cells and pointwise basis versions are chosen right-closed. Each retained basis function is then constant on every sampling cell. This gives a concrete implementation by the discrete wavelet transform. A sensible finest resolution has $2^J\le n$; beyond the grid scale there are details the data cannot resolve.

For wavelet coefficient thresholding, retain the coarse coefficients but discard detail coefficients dominated by noise. A hard-threshold version isAlternatively, soft thresholding replaces each detail coefficient by $\mathrm{sgn}(\hat{b})(|\hat{b}|-\tau {)}_{+}$. The unthresholded wavelet regression estimator retains all details at the chosen resolution; thresholding retains evidence for local features while suppressing small fluctuations.

The thresholds can be derived from a simultaneous wavelet coefficient noise bound. Write $q_{\lambda }^2={\sum }_i{w}_{\lambda i}^2$. Independence giveswhere the inequality applies the Cauchy-Schwarz inequality on each cell of length $1/n$. In the dyadic Haar implementation, $q_{\lambda }^2=1/n$ exactly. For Gaussian errors, the normalized coefficient noise is standard normal. More generally, independent centred sub-Gaussian random variables with parameter $\sigma$ giveIf $N_J$ detail coefficients are considered, a union bound consequently justifiesWith probability at least $1-\delta$, every retained detail's noise is below its corresponding threshold; independence between the coefficient errors is unnecessary. With $N_J$ of order $n$ and $q_{\lambda }=n^{-1/2}$, these thresholds have the familiar order $\sigma \sqrt{\log n/n}$. Coefficients larger than their noise level represent potential signal, and a coefficient zero in the discretized mean is not selected on this simultaneous event. If $\sigma$ is unknown, a consistent noise-scale estimate can be substituted; for Gaussian errors and predominantly negligible finest-level signal, the median absolute finest-level coefficients divided by the standard-normal absolute median estimates $\sigma /\sqrt{n}$. Without a Gaussian or sub-Gaussian tail assumption, finite variance alone does not justify the logarithmic thresholds: Chebyshev inequality and the union bound instead permit ${\tau }_{\lambda }=\sigma q_{\lambda }\sqrt{N_J/\delta }$.

A real functional $\Phi$ is Hadamard differentiable at $s_0$ if there exists a continuous linear functional $L={\dot{\Phi }}_{s_0}:S\to \mathbb{R}$ such that, whenever $t_j\ne 0$, $t_j\to 0$ and $h_j\to h$ in the normed vector space,If $\Phi$ is defined only on a subset, require $s_0+t_j{h}_j$ to lie in its domain. The variation of the directions $h_j$ is essential: convergence along each fixed direction alone is a weaker differentiability condition. Tangential Hadamard differentiability restricts the limit directions to a specified subspace; here the derivative is defined on all of $S$.

We prove the functional delta method using compact sets, so do not need an unjustified application of the Skorokhod representation theorem on an arbitrary nonseparable normed vector space. As usual for function-space convergence in distribution, the limiting random element is tight, and the real statistics are measurable, or convergence is understood through outer expectations. These are automatic under the usual separable complete random-element formulation; the empirical-process formulation used below has a tight separably supported limit.

For $t\ne 0$ putHadamard differentiability says $R_{t_j}(h_j)\to 0$ for every convergent sequence of directions, since continuity gives $L(h_j)\to L(h)$. It implies a Hadamard remainder near a compact set: for every compact $K\subset S$ and $ϵ>0$ there are $a>0$ and $\delta >0$ such thatTo prove it, suppose it failed. Choose $t_j\to 0$ and $h_j$ at distance less than $1/j$ from $K$ with $|R_{t_j}(h_j)|\ge ϵ$. Choose $k_j\in K$ with $\Vert h_j-k_j\Vert <2/j$. A subsequence of $k_j$ converges by compactness, so the corresponding $h_j$ converge too. The defining derivative limit contradicts the lower bound on $R_{t_j}(h_j)$.

Now let $Z_n=r_n(X_n-s_0)\stackrel{d}{\to }X$ and $t_n=1/r_n$. Given $\eta >0$, tightness of a probability measure provides a compact $K$ with $\mathbb{P}(X\in K)>1-\eta$. For the open $\delta$-neighbourhood $K^{\delta }$, the Portmanteau theorem givesOn $K^{\delta }$ the preceding uniform remainder is less than $ϵ$ for all sufficiently large $n$. Consequently ${\mathrm{limsup}}_n\mathbb{P}(|R_{t_n}(Z_n)|\ge ϵ)\le \eta$. Since $\eta$ is arbitrary, this remainder tends to zero in probability. We have proved the actual expansionContinuity of $L$ and the continuous mapping theorem give $L(Z_n)\stackrel{d}{\to }L(X)$. The vanishing remainder and Slutsky theorem therefore yieldThis derivation uses the derivative definition, rather than assuming a uniform Fréchet remainder over every bounded set. If outer expectations are needed, the same compact-neighbourhood argument uses outer probabilities and gives the corresponding empirical-process functional delta method.

For the statistical application, use the sample's empirical distribution function and the plug-in estimatorHere the printed $L^{\infty }$ denotes bounded pointwise functions with the supremum norm, namely ${\ell }^{\infty }(\mathbb{R})$, rather than equivalence classes modulo Lebesgue null sets. Assume Hadamard differentiability at the true $F$ and the usual measurability of the scalar estimator. The Donsker theorem for empirical distribution functions and the proved functional delta method giveNo continuity of $F$ is needed for this application. Since the derivative is continuous and linear and the bridge is a centred Gaussian process, $Z_F$ is a centred Gaussian real random variable, possibly identically zero.

Finally, a finite real weak limit ensures the required boundedness in probability. Choose $M$ so large that $\mathbb{P}(|Z_F|\ge M)<ϵ/2$. Applying the closed-set part of the Portmanteau theorem to $\{z:|z|\ge M\}$ givesHence there is $n_0$ such that, for every $n\ge n_0$,In particular the upper-tail condition in the question holds. The plug-in estimator has estimation error $O_{\mathbb{P}}(n^{-1/2})$, even when its limiting variance is zero.

Choose a proposal probability density function $g$ and a finite envelope constant $M$ such that $f(y)\le Mg(y)$ wherever the target has positive density. A rejection sampling trial draws $Y\sim g$ and, independently, $U\sim \mathrm{Uniform}(0,1)$; accept $Y$ whenotherwise repeat with fresh independent draws. For any measurable set $D$,In particular $\mathbb{P}(\mathrm{accept})=1/M$, so the conditional accepted density is $f$. More explicitly, summing over the possible first accepted trial givesThus the algorithm terminates almost surely and its output has exactly the requested law.

The envelope constant satisfies $M\ge 1$ and is the expected number of proposal trials for one output, by the geometric distribution of the waiting time. This is the quantity often called the rejection rate $M$. The actual probability of rejecting a particular proposal is $1-1/M$, and the expected number of rejections before one acceptance is $M-1$; these conventions should not be conflated.

To generate a standard normal distribution, first generate its absolute value, whose half-normal distribution density isWith the given exponential distribution proposal, the density ratio isIts maximum is at $y=\lambda$, hence the normal rejection sampling with an exponential envelope constant isDraw $Y\sim \mathrm{Exp}(\lambda )$ and a fresh uniform $U$; accept when $U\le e^{-(Y-\lambda {)}^2/2}$. An accepted $Y$ has density $h$. Independently choose a sign, positive or negative with probability $1/2$, and output $Z$ with that sign and magnitude $Y$. On either half-line its density is $h(|z|)/2=(2\pi {)}^{-1/2}{e}^{-z^2/2}$, proving the normal output law.

To optimize the rejection rate, differentiateIts derivative is $\lambda -1/\lambda$, negative before one and positive after one; equivalently its second derivative is $1+1/{\lambda }^2>0$. ThusThe rejection probability is consequently about $0.23983$.

To use only uniforms, apply inverse transform sampling: for an independent $U_1\in (0,1)$,has survival probability $\mathbb{P}(Y>y)=e^{-\lambda y}$ for $y\ge 0$. Use a second independent uniform for acceptance, and, after acceptance, a third independent uniform to choose the sign. Repeat with fresh uniforms after any rejection. No exponential generator is then needed.

An unordered bootstrap sample is determined by the multiplicities $N_i\ge 0$ of the $n$ distinct observations, with ${\sum }_i{N}_i=n$. Conversely each such vector specifies exactly one sample up to rearrangement. Place $n$ identical marks into $n$ boxes using $n-1$ separators. The stars and bars argument therefore givesThese bootstrap count vectors are not equally likely: the probability of a vector is $n!/(n^n{\prod }_i{N}_i!)$, counting its possible ordered sequences. The ordered index sequences themselves number $n^n$.

The first simulation uses independent standard Cauchy distributions and returnsThe tail probability and estimator moments areThe indicator count has a binomial distribution with parameters $n,\theta$; the infinite variance of a Cauchy observation is irrelevant because this estimator averages bounded indicators.

For the next code block, each row is an independent sample of size $n$ from the empirical distribution of the original $x_i$. Its entry in the vector $v$ is the corresponding bootstrap indicator mean ${\hat{\theta }}_b^{\ast }$. The last expression is the variance of these $B=199$ replicate estimates, with denominator $B-1$:Condition on the original data and write a star on bootstrap expectations. Every resampled indicator has a Bernoulli distribution with success probability $\hat{\theta }$, and the $n$ draws within a row are independent. Thus the conditional bootstrap variance of a sample mean givesIt estimates the sampling variance of the tail-probability estimator, not the variance of the raw Cauchy data. Its square root is an estimated standard error. Finite $B$ leaves bootstrap simulation noise. Averaging over the original data, its expectation is $(n-1)\theta (1-\theta )/n^2$, so it is not exactly unbiased for the true variance $\theta (1-\theta )/n$ at finite $n$.

In the importance-sampling block, the transformation $Y=2/(1-U)$ gives, for $y\ge 2$,The variable $Y$ is a proposal draw supported entirely in the integration region. The deterministic value computed from it is the importance sampling weightConsequentlyThe code uses ordinary importance sampling of a Cauchy tail, not self-normalization; $Y$ itself is neither an observation from the target Cauchy law nor the estimator's summand. The weight compensates for sampling from its proposal law.

The final block bootstraps these same proposal observations $Y_i$ and recomputes the weight average in each row. Since $W$ is a deterministic function of $Y$, this is equivalent to resampling the observed weights $W_i$. If ${\tilde{\theta }}_b^{\ast }$ denotes a replicate, the final output is its sample variance over the $B$ replicates. Its conditional expectation isIts expectation over the original proposal sample is $(n-1)\mathrm{Var}(W)/n^2$.

The variance reduction can be calculated, rather than inferred just from the appearance of the weights. Write $A=\arctan (1/2)$. Thenbecause an antiderivative of the integrand without its prefactor is $\frac{1}{2}(\arctan y-y/(1+y^2))$. HenceBy comparison, $\theta (1-\theta )\approx 0.125803$. For $n=100$ the true estimator variances are approximately $9.55253\times 10^{-7}$ and $1.25803\times 10^{-3}$ respectively. The importance estimator has about 1/1,317 of the indicator estimator's variance, because all proposals land in the tail and the weights remain in the narrow interval $[2/(5\pi ),1/(2\pi ))$.

Thus the last bootstrap output should usually be much smaller than the first, and its expected value is smaller by the same variance ratio. This is not a deterministic ordering of two finite-run outputs: if all the original Cauchy draws happen to be at most two, the first bootstrap variance is zero, while unequal importance weights can still give a positive second variance.

The Metropolis–Hastings algorithm proposes $y\sim q(\cdot \mid x)$ and moves from the current state $x$ with acceptance probabilityOn rejection it retains $x$. The target need only be known up to its normalizing constant, but it must define a proper probability distribution. The accepted off-diagonal probability flow iswhich is symmetric in $x,y$ and proves detailed balance. This permits considerable flexibility in proposals, including joint moves when a posterior distribution has strongly correlated parameters, but requires proposal tuning and can waste computation on rejections.

A Gibbs sampler instead updates a coordinate or block from its exact full conditional distribution, holding the other coordinates fixed. This is a special Metropolis–Hastings algorithm update with acceptance probability one: the joint density factors into the unchanged marginal density and the proposed conditional density, which cancel in the acceptance ratio. The coordinate update preserves the target; composing such updates therefore preserves it too, although an entire deterministic sweep need not itself be reversible.

Use Gibbs sampling when the conditional laws are easy to simulate and suitably chosen blocks mix well. Scalar updates can mix poorly under strong posterior distribution correlation, whereas a blocked Gibbs sampler can alleviate that dependence. Prefer a flexible Metropolis proposal when conditional draws are difficult or a well-tuned joint proposal is more effective. Variations include random or systematic coordinate scans, random-walk Metropolis algorithms, independence proposals, and using Metropolis updates for difficult blocks inside a Gibbs scheme. Both methods generate dependent draws; usable long-run estimates require an appropriate irreducible, aperiodic chain and assessment of mixing, not merely a formal update rule.

For the change-point calculation, put $v={\sigma }^2$ andMultiplication of the Gaussian likelihood function by the specified prior distribution gives the formal posterior distribution kernel, with respect to $d\theta d\varphi dv$ and counting measure on $m$,

There is an empty component under an improper prior at $m=n$. The sum $S_n$ is independent of $\varphi$, whose prior distribution is flat on all of $\mathbb{R}$. For every fixed finite $\theta$ and $v>0$, the kernel is a strictly positive constant in $\varphi$, henceThe configuration has positive prior probability $1/n$. Integrating over, for example, a bounded interval of $\theta$ and $1\le v\le 2$ already makes the joint normalization infinite. Thus the printed prior distribution does not produce a posterior probability distribution. This failure is present for every data set, not just for a special arrangement of observations.

The formal full conditionals expose exactly where the requested sampler breaks. For $m<n$, completing the two mean squares givesThe variance kernel is that of $\mathrm{IG}(n/2,S_m/2)$ when $S_m>0$, and the discrete conditional probabilities are proportional to $e^{-S_m/(2v)}$. At $m=n$ the conditional kernel for $\varphi$ is constant and cannot be normalized; it is not a normal distribution with an infinite variance. Therefore no valid Gibbs sampler, and no value of $P(m=n\mid x)$, exists for the model as printed. A chain that simply substitutes an arbitrary distribution for this missing conditional would be sampling a different model.

Here is a fully specified proper-prior repair retaining the no-change configuration. Choose fixed hyperparametersindependently, with $t_{\theta },t_{\varphi },a_0,b_0>0$. Its Gaussian change-point posterior with proper priors isIt is normalizable even for degenerate data: the likelihood function is at most $(2\pi {)}^{-n/2}{v}^{-n/2}$, and its integral against the variance prior distribution is bounded byThe normal mean priors integrate to one, and the normalizing constant is positive. This proves posterior propriety for the repaired model.

Completing the conditional squares gives the Gaussian change-point Gibbs updates. DefineThe four updates areAt $m=n$, $M_{\varphi }={\mu }_{\varphi },V_{\varphi }=t_{\varphi }$, giving its proper prior distribution unchanged. Starting from any finite means, positive variance and allowed $m$, cycle through these four updates using the latest values. The sum-of-squares terms can be evaluated from cumulative data sums; normalize the discrete log weights after subtracting their maximum to avoid numerical underflow. The proper positive conditionals allow movement among all split values and throughout the continuous parameter support.

Under this specified repair, $P(m=n\mid x)$ is the posterior probability of no change within the observed sequence: all $n$ data points belong to the first mean regime. For $N$ retained draws after initialization has ceased to affect the estimates, its Markov chain Monte Carlo estimator isAlternatively average the displayed conditional probability of $m=n$ at the sampled continuous parameters, giving an estimate by Rao-Blackwellization. Autocorrelation must be accounted for in its Monte Carlo uncertainty. These are probabilities and estimators for the explicitly repaired model; the original unnormalizable kernel supplies neither.

The odds ratio compares event odds, not event probabilities. Let group 1 denote the elective group and group 0 the reference group. Their fitted event probabilities are ${\hat{p}}_1=9/1515\simeq 0.00594$ and ${\hat{p}}_0=1215/75057\simeq 0.01619$. ThusThe risk ratio is similarly about $0.367$. The unadjusted observations show a lower adverse-event frequency in the elective group, rather than an increase. This is an association, not evidence that choosing the procedure prevents adverse outcomes. The direction is opposite to the reported adjusted odds ratio above one; the two analyses compare different populations or different conditional risks.

Zero observed events do not mean that the underlying event probability cannot be estimated. For independent observations having a common Bernoulli distribution, the binomial likelihood with no events is $(1-p{)}^{1515}$, maximized at $\hat{p}=0$. A zero-event binomial upper confidence bound shows the remaining uncertainty: the one-sided 95% upper bound solves $(1-p_U{)}^{1515}=0.05$, givingThus the point estimate is zero, but a nonzero mortality risk is compatible with the observations. This bound is an illustration under independent equal-risk sampling; clustered data or unequal risks require their own uncertainty calculation.

The empirical unadjusted odds ratio is also defined:Its log odds ratio is $-\infty$, so the ordinary normal approximation for a log-odds confidence interval is unavailable. Exact likelihood-based limits can still describe uncertainty; a specified continuity correction would instead yield a finite, method-dependent point estimate.

In an unpenalized logistic regression with a separate coefficient for elective exposure, every exposed death outcome is zero. Decreasing that coefficient toward $-\infty$ increases the exposed observations' likelihood contributions and leaves the reference observations unchanged. This is quasi-complete separation: there is no finite ordinary maximum-likelihood estimate of the exposure coefficient, even after adding background predictors. It is reasonable to say that the usual finite adjusted coefficient could not be fitted; it is too strong to say that no risk estimate or statistical information is possible. A specified penalized likelihood or Bayesian inference with proper coefficient priors can provide a finite adjusted estimate, whose dependence on the regularization must be reported.

Using the reference event rate as the background-risk benchmark givesThis is the expected count if the elective group had the reference group's baseline risk and the procedure added no effect. It is substantially above the observed count, illustrating the importance of the background-risk comparison.

Similarity of background factors alone does not fix the elective group's event risk if the procedure itself changes it. If one additionally retained the reported common adjusted odds ratio $R=2.7$ and used $p_0=1215/75057$ as the common baseline risk, then conversion from odds to probability would giveThese are different assumptions: about 25 is the baseline/no-effect benchmark; about 64 includes an odds multiplier of 2.7.

Use a concurrent comparison: test the same specimens for both cocaine and mephedrone, rather than treating the historical cocaine percentage as an exactly known 2010 parameter. Assume representative sampling of privates, effectively independent specimens, stable underlying rates during the sampling period, and sufficiently accurate assays. I use a two-sided 5% test of equal positive-test probabilities, with 80% statistical power at $p_C=0.010$ and $p_M=0.005$. This is a design to distinguish the rates at a twofold alternative; a significance test does not prove that their ratio is exactly two.

The paired design needs the joint positive-test distribution. For specimen $i$, write $C_i,M_i\in \{0,1\}$ and $D_i=C_i-M_i$. Put $q_{11}=P(C_i=M_i=1)$ andThen $\mathbb{E}{D}_i=\delta$ and $\mathrm{Var}(D_i)=q-{\delta }^2$. Under equal marginal rates, the two discordant outcomes have equal probabilities; conditional on the total number of discordant specimens, their split is a fair binomial distribution. This gives the McNemar test. Its large-sample version rejects for a sufficiently large absolute difference between the discordant counts, divided by their estimated standard deviation.

For planning, suppose simultaneous positives are negligible, so $q\simeq 0.015$. Under the alternative the difference in positive counts has mean $n\delta$ and standard deviation $\sqrt{n(q-{\delta }^2)}$; the approximate upper null rejection boundary is $z_{0.975}\sqrt{nq}$. Requiring 80% upper-tail rejection probability gives the paired binary sample size calculationAdd the mephedrone assay to about 5,000 existing tests, roughly ten weeks at 500 per week. An exact conditional-power calculation under the negligible-overlap model gives about $77.6\%$ at 4,707 specimens and $80.2\%$ at 5,000, so the rounding also compensates for the conservative discrete test. If positivity for the two drugs were independent within a specimen, $q_{11}=0.00005$, giving a very similar requirement of about 4,676. For the stated marginal rates, $q\le 0.015$, so negligible overlap is a conservative variance choice within this approximation. Substantial co-use reduces the discordant proportion and changes the necessary size. Repeated tests on the same person, unit-level clustered data or inaccurate assays can instead reduce effective information and require more specimens.

A pre-specified one-sided 5% comparison would use $1.645$ in place of $1.96$ and require about 3,708 specimens before allowing for discreteness under the negligible-overlap assumption. These are different testing conventions. Treating the historical 1% as fixed would give a one-sample calculation, but would ignore uncertainty or changes in the actual 2010 cocaine rate.

Set $\alpha =\log \{{\beta }^{(1)}/{\beta }^{(0)}\}$ and absorb the common multiplier into $h_0(t)=m(t){\beta }^{(0)}$. The model becomesIt is a Cox proportional-hazards model; the unknown baseline does not need a parametric form. The equality hypothesis is precisely $\alpha =0$. The absolute multipliers and $m$ have a scale ambiguity, but their hazard ratio $e^{\alpha }$ is identifiable when both groups supply information.

For an untied event at $a_k$, let $R_k$ be the risk set just before the event, and let $i_k$ be the subject who fails. The conditional event probability is proportional to that subject's hazard, so the common baseline cancels:Multiply these factors to form the Cox partial likelihood. One can maximize it and compare its maximum with its value at $\alpha =0$ using a partial likelihood-ratio test, or use the corresponding score test.

Explicitly, with group risk-set sizes $r_{k0},r_{k1}$ and observed event indicator $d_{k1}$ for group 1, the score at zero and its information areUnder the null the event label has exactly this Bernoulli distribution conditional on its current risk set. Hence $U^2/V$ is the usual one-degree-of-freedom log-rank test statistic, with an asymptotic chi-squared distribution when $V>0$ and enough informative events are observed. Censoring can differ between groups, provided it is independent of event time within the conditioning model. If there are recorded ties, use the appropriate tied-risk-set version: for $d_k$ events the null variance contribution is $d_k(r_k-d_k)r_{k0}{r}_{k1}/[r_k^2(r_k-1)]$.