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$Y_{j,k}$ is the number in group $k$ still observed and event-free immediately before event time $a_j$: its group-specific risk set size. Set $Y_j=Y_{j,0}+Y_{j,1}$. Under the common hazard function and conditional on exactly one event at that time, each at-risk individual has the same infinitesimal event probability. ThereforeThe indicator $g_{\pi (j)}$ records whether the event came from group 1, while $p(j,1)$ is its conditional expectation under equal hazards. Thus $z_j$ is the observed-minus-expected group-1 event count at that time. Under the null these are martingale differences, with conditional varianceThe log-rank statistic $z={\sum }_j{z}_j$ has null mean zero. Its predictable variance estimate is $V={\sum }_j{p}_j(1-p_j)$, so the equal-hazards test usesUse a two-sided large-sample log-rank test if no direction was prespecified; $V>0$ and adequate information are needed. The variance is the no-ties form. A positive $z$ indicates more group-1 events than expected. Independent censoring conditional on the model is required for these at-risk event intensities to describe the data.

Under the proportional hazards model, the total instantaneous group-$k$ rate is $Y_{j,k}{e}^{\beta k}{h}_0(a_j)$. Conditioning on the next event cancels the baseline hazard, givingKeep $z_j$ defined using its null probability $p(j,1)$, as in the question. Its conditional expectation under the alternative isIt is zero at $\beta =0$, as required, and has the sign of $\beta$ when both groups remain at risk. Replacing $p(j,1)$ inside the definition of $z_j$ with the alternative probability would give a different residual, whose mean would instead be zero under that alternative.

Put $q_t=\mathrm{Pr}(d{N}_{+}(t)=1\mid {\mathcal{H}}_{t-})$. A binary event increment has conditional variance $q_t(1-q_t)$. Since $q_t=Y_{+}(t)h(t)dt+o(dt)$, its squared mean is of second order, givingTreating the predictable risk set size as known given the history, the Nelson–Aalen estimator increment therefore has first-order varianceSubstitute $d\hat{H}=d{N}_{+}/Y_{+}$ for $dH$ to obtain the estimated increment variance $d{N}_{+}/Y_{+}^2$. Summing these estimated predictable variances gives the usual Nelson–Aalen variance estimator:The martingale $d{N}_{+}-Y_{+}dH$ has orthogonal increments, which justifies accumulating the predictable variances of the estimation error. This is an estimated sampling variance on the observed at-risk range, not a claim of exact finite-sample unbiasedness after the risk set becomes empty. For multiple events at an event time the usual extension replaces the numerator 1 by the event count.

In the Gaussian white noise model, observe the entire pathwhere $W$ is standard Brownian motion and the deterministic drift belongs to L2 space on $[0,1]$. Equivalently, $dY(t)=f(t)dt+n^{-1/2}dW(t)$. For every deterministic $h\in L^2[0,1]$, the observed stochastic integral satisfiesThe noise $W(h)=\int hdW$ is an isonormal Gaussian process. In particular its variance is $\Vert h{\Vert }_2^2$. Gaussian white noise is interpreted through these integrals, rather than as an ordinary random function with a pointwise value at every time. No smoothness of $f$ is needed.

Here is the Gaussian maximum bound without independence. Put $M={\max }_{1\le i\le N}|g_i|$. It is integrable since it is bounded by ${\sum }_i|g_i|$. For every $\lambda >0$,The second inequality uses only the moment-generating function of the standard normal distribution, so independence is unnecessary. Jensen inequality gives $e^{\lambda EM}\le E{e}^{\lambda M}$, henceMinimizing at $\lambda =\sqrt{2\log (2N)}$ proves

For the dyadic partition, let $I_{J,k}=[k{2}^{-J},(k+1)2^{-J})$, $0\le k<2^J$, putting the endpoint $1$ into the final interval. The Haar scaling functionsform an orthonormal basis of the space $V_J$ of functions constant on each interval. The Haar wavelets can be written asFor $J\ge 1$, the constant function ${\varphi }_{0,0}=1$ together with ${\psi }_{j,k}$, $0\le j<J$, is another orthonormal basis of $V_J$. Indeed, each successive level splits a cell's two constants into their sum and difference; the number of basis elements is $1+{\sum }_{j=0}^{J-1}{2}^j=2^J$. Thus the Haar projection is the orthogonal projectionOn cell $I_{J,k}$ its value is $2^J{\int }_{I_{J,k}}f(t)dt$. The first formula also applies to $J=0$.

Estimate each coefficient by its observed stochastic integral:This is the Haar projection estimator in Gaussian white noise. The noise integrals over disjoint intervals form a Gaussian vector with zero off-diagonal covariance. The principle that uncorrelated jointly Gaussian variables are independent then makes these coefficients independent. ThereforeTaking expectations shows that ${\hat{\Pi }}_{V_J}f$ is an unbiased estimator of ${\Pi }_{V_J}f$, both coefficientwise and pointwise for the stated step-function representatives.

Only one scaling function is nonzero in each cell, so the supremum norm of the error has the exact formThe endpoint convention ensures the same identity at $1$. Apply the Gaussian maximum bound without independence with $N=2^J$ to obtain the supremum norm risk of a Haar projection estimator:The factor $2^{J/2}$ comes from cellwise scaling, while the extra square root of $J+1$ comes from taking the maximum over $2^J$ Gaussian errors.

A noncausal stationary autoregression exists. On the two-sided time axis defineThis series converges in the sense of mean-square convergence, and direct subtraction gives $X_t-2X_{t-1}={\epsilon }_t$. The expected value and autocovariance areConsequently this is a weakly stationary process and, by Gaussianity, a strictly stationary process. It is an example of noncausal stationary autoregression, since $X_t$ uses future noise. In particular, $X_{t-1}$ is correlated with ${\epsilon }_t$, so the usual causal variance recursion is inapplicable. If an additional assumption required the driving noise to be independent of past observations, this solution would be excluded; that assumption is not stated here.

No weakly stationary process solves the model with nondegenerate noise. Iteration would giveFor a weakly stationary process with finite variance $V$, the Cauchy-Schwarz inequality instead gives $\mathrm{Var}(X_t-X_{t-m})=2V-2\gamma (m)\le 4V$. These two bounds contradict each other for large $m$. This argument does not assume that $X_{t-m}$ is independent of the intervening noise, so it excludes noncausal solutions too. The corresponding unit-root autoregressive process has persistent random walk behavior rather than stationary fluctuations.

Two useful features of autoregressive conditional heteroscedasticity are persistent changes in conditional scale and excess unconditional kurtosis. Its time-varying conditional variance can explain volatility clustering, where large absolute returns occur in groups even when signed returns have little autocorrelation. A homoscedastic autoregressive moving-average model has a fixed innovation variance.

Also, a conditional normal distribution with a random scale is a Gaussian scale mixture. Its unconditional kurtosis can exceed $3$, or its fourth moment can be infinite. A Gaussian autoregressive moving-average model remains jointly Gaussian and cannot reproduce this effect. In the particular lag-two ARCH process, persistence of the squared scale occurs within each parity subsequence; the two parity subsequences are independent.

The useful autoregressive model is for the squares, not for signed observations. Put $Y_t=X_t^2$ andThenThe errors form a martingale difference sequence relative to the noise history, because the current standardized noise is independent of the past. When the fourth moment is finite they have finite variance and are uncorrelated across distinct times, although their conditional variance depends on the regressor. In centered form, $Y_t-m_2={\alpha }_2(Y_{t-2}-m_2)+u_t$.

For ordinary least squares, use the $n-2$ response-regressor pairs $y_t=X_t^2$, $r_t=X_{t-2}^2$, $t=3,\dots ,n$. With their separate means $\bar{y}$ and $\bar{r}$, minimize ${\sum }_{t=3}^n(y_t-a-b{r}_t{)}^2$. If the regressor sum of squares is positive, the estimators areThe two means use the matched pairs; replacing them indiscriminately by a single full-sample mean is not the exact least-squares formula. The noise-series representation supplies an ergodic stationary process. If $3{\alpha }_2^2<1$, finite regressor second moments and the error's zero conditional expectation justify the usual population regression and statistical consistency argument. The observations still define a finite-sample least-squares fit outside that moment range, but the ordinary finite-variance justification must not be claimed there. If parameter constraints are required, minimize the same criterion subject to $a>0$ and $0<b<1$, rather than assert that unconstrained estimates automatically satisfy them.

For a centered weakly stationary process with $\gamma (0)>0$, the autocorrelation function isFor $h\ge 2$, let $P$ be orthogonal projection in L2 space onto the linear span of $X_{t-1},\dots ,X_{t-h+1}$. The lag-$h$ partial autocorrelation function is the correlation coefficient of $X_t-P{X}_t$ and $X_{t-h}-P{X}_{t-h}$. It removes the linear contribution of the intervening observations. At lag one it is just $\rho (1)$. For nonsingular prediction covariance matrices, it is equivalently the last coefficient $a_{hh}$ in the order-$h$ linear predictor of $X_t$ from $X_{t-1},\dots ,X_{t-h}$. Equal residual variances, by stationarity, identify that coefficient with the residual correlation. If a residual has zero variance, this correlation is undefined; the nondegeneracy condition matters.

For the moving-average process of order one, $\gamma (0)={\sigma }^2(1+{\theta }_1^2)$, $\gamma (1)={\sigma }^2{\theta }_1$, and $\gamma (2)=0$. Regressing each endpoint on the single intervening observation leaves residual covariance $\gamma (2)-\gamma (1{)}^2/\gamma (0)$ and residual variance $\gamma (0)-\gamma (1{)}^2/\gamma (0)$. Therefore its lag-two partial autocorrelation function isThe denominator is positive. In particular, the moving-average autocorrelation cutoff at lag one does not make the lag-two partial correlation zero unless ${\theta }_1=0$.

Use cycles-per-time spectral density, with frequency $\omega \in [-1/2,1/2]$. The spectral representation theorem for a stationary time series gives the centered white noise representationwhere disjoint increments are orthogonal. Put ${\theta }_0=1$. Since the filter is finite, substitute each noise representation and interchange the finite sum with the integral:The new orthogonal increment measure is $d{Z}_X=\Theta (e^{-2\pi i\omega })d{Z}_{\epsilon }$. Its variance measure is therefore ${\sigma }^2|\Theta (e^{-2\pi i\omega }){|}^{2}d\omega$. The coefficients are real, so conjugation changes the sign of the exponent without changing the modulus. HenceWith angular frequency $\lambda =2\pi \omega$, the spectral density of a stationary process instead contains the factor $1/(2\pi )$. The convention explains its absence here. Invertibility is not needed for this finite-filter spectral calculation.

The autocovariance vanishes beyond lag $q$. For an ordinary consecutive sum, variance of a sum givesThe last sum is finite. For $h\ge 0$, $\gamma (h)={\sigma }^2{\sum }_{j=0}^{q-h}{\theta }_j{\theta }_{j+h}$. Counting all coefficient pairs givesFor the odd-indexed sample the lag-$h$ autocovariance is $\gamma (2h)$, so the same finite-sum argument gives ${\sum }_h\gamma (2h)$. Only coefficient pairs of the same parity contribute. If $E={\sum }_{j\mathrm{even}}{\theta }_j$ and $O={\sum }_{j\mathrm{odd}}{\theta }_j$, the odd-subsample limit isThis is the odd-subsample long-run variance of a moving average. Equivalently it is $[f_X(0)+f_X(1/2)]/2$ in the cycles convention. Under the printed invertibility assumption and positive noise variance, the polynomial is nonzero at both $1$ and $-1$, so both limits are positive. The same formulas also hold without invertibility, when cancellations can make a limit zero.

Choose a proposal probability density function $q$ positive wherever $|h|f$ is nonzero, up to null sets, and draw independent and identically distributed random variables $Y_1,\dots ,Y_n$ from $q$. The ordinary importance sampling estimator isThe support condition for importance sampling and absolute integrability give ${\mathbb{E}}_q[h(Y)f(Y)/q(Y)]=\int hf=\mu$, so this is an unbiased estimator. Its variance, possibly infinite, isLet $A=\int |h|f$. If $A>0$, the Cauchy-Schwarz inequality yieldsEquality holds for $q$ proportional to $|h|f$. Thus the minimum-variance importance distribution and the minimum variance areFor a constant-sign integrand this is zero variance. If $A=0$, the integrand vanishes almost everywhere and the zero estimator already has zero variance. Requiring the proposal to cover all of the target's support, including where $h=0$, can exclude $q_{\ast }$; in that stricter class, $(1-ϵ)q_{\ast }+ϵf$ approaches the same infimum as $ϵ\downarrow 0$. This distinguishes coverage of the integral from coverage of every target event.

A convenient general-state-space ergodic theorem is the following. For a positive Harris recurrent Markov chain with invariant probability measure $\pi$, and a measurable function $h$ with $\pi |h|<\infty$,Here $\pi h=\int hd\pi$ is the stationary expected value. Harris recurrence means that every set of positive irreducibility measure is visited almost surely from every state; positive recurrence supplies an invariant probability rather than only an infinite invariant measure. The ergodic theorem for a positive Harris recurrent Markov chain holds from any starting state under these Harris hypotheses. aperiodicity is not necessary just for averages.

A useful sufficient form of the central limit theorem for a geometrically ergodic Markov chain adds aperiodicity, geometric ergodicity, and $\pi (|h{|}^{2+\delta })<\infty$ for some $\delta >0$. It givesThese are sufficient hypotheses, not a claim that irreducibility alone ensures a central limit theorem. The Markov chain Monte Carlo asymptotic variance uses stationary covariances ${\gamma }_h(k)={\mathrm{Cov}}_{\pi }(h(X_0),h(X_k))$, with ${\gamma }_h(0)={\mathrm{Var}}_{\pi }h$. The series is absolutely convergent under the stated sufficient assumptions. If ${\gamma }_h(0)>0$, write ${\rho }_h(k)={\gamma }_h(k)/{\gamma }_h(0)$ and $v_h={\gamma }_h(0){\tau }_{\mathrm{int}}$, where the integrated autocorrelation time is ${\tau }_{\mathrm{int}}=1+2{\sum }_{k\ge 1}{\rho }_h(k)$. When $v_h>0$, the effective sample size of a Markov chain is approximately $n/{\tau }_{\mathrm{int}}$. Negative correlations can reduce the asymptotic variance; a zero asymptotic variance gives a degenerate normal limit. For constant $h$, variance is zero and the autocorrelation normalization is undefined.

For the Erlang loss formula, assume a Poisson process of arrivals with rate $\lambda$, independent holding times with exponential distribution of mean $1/\beta$, unit capacity usage per accepted call, and rejection without waiting when all $C$ units are occupied. Here $C$ is a positive integer and the offered load is $a=\lambda /\beta$. The call population on $\{0,\dots ,C\}$ is a birth-death process; detailed balance givesBy Poisson arrivals see time averages, the fraction of arrivals rejected is the time probability of full occupancy. ThereforeThe paper's $\nu$ is the offered load, rather than necessarily an arrival rate. Exponential holding times suffice for this derivation; no unproved insensitivity of loss networks assumption is needed.

A loss network with fixed routing has capacities $C_j$, call types $r$, and a link-route incidence matrix $A$ with $A_{jr}=1$ exactly when a type-$r$ call requires link $j$. An accepted call holds one unit on every link of its route for its holding time; if any required link is full, the call is rejected everywhere. Take independent Poisson processes of calls, independent holding times, and route offered loads ${\nu }_r$. For the finite-variable formulation below, take $C_j\ge 1$; a zero-capacity link and all routes using it may be removed first.

The Erlang fixed point approximation treats blocking at different links as independent and thins the traffic offered to link $j$ by acceptance at the other links. Its reduced offered load isThe equality uses $A_{jr}\in \{0,1\}$: only routes with $A_{jr}=1$ contribute. Treating this reduced stream as Poisson leads to $B_j=E(a_j(B),C_j)$. Independence and Poisson thinning here are the approximation, not exact properties of a general loss network.

To construct the convex potential for the Erlang fixed point, write $e_C(a)=E(a,C)$ and $m_C(a)=a[1-e_C(a)]$, the carried load of an Erlang loss resource, equivalently the mean occupancy in the single-link stationary distribution. Differentiating its finite sums gives, for $a>0$,The variance is positive because the finite occupancy law is nondegenerate for $a>0$. Furthermore $e_C$ maps $[0,\infty )$ increasingly onto $[0,1)$, while $m_C$ increases from zero to $C$. SetIn particular $U(0,C)=0$, $U(\cdot ,C)$ is continuous and strictly increasing, and $U(z,C)\to C$ as $z\to \infty$.

Forthe first sum is convex, and each integrated strictly increasing function is strictly convex. Thus $F$ is a strictly convex function on $y\ge 0$. Since $U$ tends to a positive limit on every coordinate, $F$ is a coercive function there; it has a unique minimizer. Its coordinate derivative isThe Erlang fixed point equation is exactly ${\partial }_{j}F=0$. For an unused link, $a_j=0$ and the optimum is $y_j=0$; for a used link with positive offered load, the derivative at zero is negative and the optimum is positive. Hence every fixed point gives the unique minimizer, and the minimizer gives a fixed point. The Erlang fixed point is unique under these fixed-routing assumptions.

A convergent cyclic substitution for the Erlang fixed point updates one coordinate at a time, repeatedly cycling through all links:Always use the most recently available values of the other coordinates. Start, for example, with $B=0$. In $y$ coordinates this is exact coordinate descent for $F$: with other coordinates fixed, the displayed update is the unique coordinate minimizer.

Here is a convergence proof. All iterates stay in the initial compact sublevel set of the coercive function $F$, and their potential values decrease to a limit. Each single-coordinate update is continuous, since its reduced load is continuous and finite and $E(a,C)<1$. Thus a whole-sweep map $T$ is continuous. If a full-sweep subsequence converges to $z$, continuity gives $F(Tz)=F(z)$, because both consecutive potential values converge to the same limit. Strict coordinate convexity means equality can occur only when no coordinate changes during that sweep. Thus $z$ minimizes every coordinate, satisfies the nonnegative-orthant first-order conditions, and is the unique global minimizer of $F$. Every subsequential limit is therefore the same point, proving convergence of the full-sweep iterates and, by continuity of the updates, all intermediate iterates. This argument applies to sequential substitution; it does not assume simultaneous updates converge.