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Begin by defining the scientific target: prediction for a new patient, or estimation of adjusted covariate effects. Check event definitions, follow-up origins, right-censoring codes and any delayed entry. The analysis assumes independent patients and censoring that is noninformative conditional on modeled covariates. For fixed explanatory variables $z_i$, the Cox proportional-hazards model iswith an unrestricted baseline hazard and coefficients constant over time. Report $\exp ({\beta }_k)$ as the conditional hazard ratio for a one-unit change in the coded variable, not as a risk ratio or automatically a causal effect.

For untied event times, fit the coefficients by maximizing the Cox partial likelihoodThe baseline cancels within each event risk set. Use an appropriate tied-event method, such as an Efron approximation or an exact method for a genuinely discrete event scale; the Breslow approximation for tied event times is another explicit approximation. Numerical score/information methods fit the model, and inverse information gives conventional covariance estimates for independent subjects. Report coefficient estimates, hazard ratios, uncertainty and the coding that makes their interpretation meaningful. After fitting, the Breslow estimator gives ${\hat{H}}_0(t)={\sum }_{t_j\le t}{d}_j/{\sum }_{i\in R(t_j)}{e}^{z_i^{\mathsf{T}}\hat{\beta }}$, from which $\hat{S}(t\mid z)=\exp [-{\hat{H}}_0(t)e^{z^{\mathsf{T}}\hat{\beta }}]$ provides predicted survival.

Adequacy is broader than significance of coefficients. Inspect influential observations, deviance residuals and data errors. Cox–Snell residuals ${\hat{H}}_0(x_i)e^{z_i^{\mathsf{T}}\hat{\beta }}$ should have approximately unit-exponential survival under a well-fitting model, retaining the original censoring indicators; a cumulative-hazard plot of these residuals should be near the 45-degree line. This is an overall diagnostic, not a substitute for the functional-form and proportionality checks below. For prediction, assess held-out discrimination and survival prediction calibration at prespecified time horizons with methods accounting for censoring. A high concordance index alone does not establish good calibration or correct hazard structure.

Use bootstrap or held-out cross-validation that repeats the complete modeling procedure, including imputation, variable selection and tuning. Validation of only the final fitted coefficients understates overfitting from earlier decisions. Report any substantive lack of fit rather than presenting a single time-independent hazard ratio when the data do not support that representation.

The episodes from one patient share treatment, biological susceptibility and prior history, so they are not independent observations. Treating every row as an unrelated subject can substantially understate uncertainty. Two approaches address this within-patient recurrent-event dependence:

A patient-level bootstrap is another way to retain dependence: resample complete histories, not individual episode rows. It is not necessary to assume that robust marginal/working and frailty-conditional effect estimates target exactly the same parameter.

The log-rank test compares survival experience through the allocation of events within the successive risk sets. Its null is equal hazard functions between the groups over follow-up, with independent individuals and noninformative censoring within groups. At each distinct event time $t_j$, let $n_{Aj},n_{Bj}$ be the numbers at risk immediately before that time, and let $d_{Aj},d_{Bj}$ be event counts. Under the null, conditional on the total $d_j=d_{Aj}+d_{Bj}$, the expected number of A events is $e_{Aj}=d_j{n}_{Aj}/(n_{Aj}+n_{Bj})$.

The signed log-rank statistic is the observed-minus-expected scoreA positive score indicates relatively more A events than expected, hence a higher event hazard for A; B's score is its negative. Censoring times do not produce score terms, but remove people from all later risk sets.

At each event time the conditional null allocation is hypergeometric, as if the $d_j$ events were sampled without replacement from the risk set. Its variance gives the score-increment variance, with the finite-population correction for ties. Summing these conditional variances gives $V$, since distinct-time martingale score increments have zero cross covariance under the null. For sufficiently many informative events, $U_A/\sqrt{V}$ is approximately standard normal and $U_A^2/V$ approximately chi-squared with one degree of freedom. In the no-tie case, each variance contribution is $n_{Aj}{n}_{Bj}/(n_{Aj}+n_{Bj}{)}^2$. A time at which only one group remains at risk contributes no comparative information.

The numerical parts below report signed-score and variance contributions after day 160. They cannot be added as separate chi-squared statistics to the earlier follow-up: add the scores and variances first, and standardize once for the complete dataset. The log-rank test is particularly effective for proportional hazards alternatives; crossing hazards can produce cancelling score contributions.

For an autoregressive moving-average model, the driving-to-output transfer function is $A(z)=\theta (z)/\varphi (z)$, and the inverse transfer function is $A(z{)}^{-1}=\varphi (z)/\theta (z)$. The causality and invertibility root criteria for an ARMA model require these respective rational functions to have power series about zero converging on a disk larger than the unit disk. If the two polynomials have no common factor, the conditions becomeIndeed, outside-disk roots leave a radius of convergence greater than one, so the coefficients decay geometrically and are absolutely summable. Conversely an uncancelled pole inside or on the unit disk prevents the required stable power series. If factors are common, apply the criterion after cancellation, to the noise-driven solution rather than additional homogeneous components. The backshift operator translates these power series into the desired one-sided filters.

The original representation has $\varphi (z)=1-2z$ and $\theta (z)=1+2z$, with roots $1/2$ and $-1/2$. There is no cancellation. Hence it is neither causal nor invertible relative to its specified driving noise. Stationarity is nevertheless possible through a two-sided solution: expanding the autoregressive inverse in negative powers givesThis is an anticausal time series representation with square-summable coefficients.

To establish the alternative representation on the same process, defineThe inverse of $1+\frac{1}{2}B$ is a stable one-sided filter. For $|z|=1$, the identities $|1+2z{|}^2=4|1+z/2{|}^2$ and $|1-2z{|}^2=4|1-z/2{|}^2$ giveThe time-series spectral density of $X$ is ${\sigma }^2|1+2e^{-i\omega }{|}^2/(2\pi |1-2e^{-i\omega }{|}^2)$; therefore the defined $\eta$ has constant time-series spectral density ${\sigma }^2/(2\pi )$. Its mean is zero, its variance is ${\sigma }^2$, and all its nonzero-lag autocovariances vanish. It is thus weak white noise. Its definition directly givesThis root reflection of an ARMA representation has roots $2,-2$, so is causal and invertible. A constant spectrum proves whiteness, not independence of non-Gaussian coordinates; no Gaussian assumption is needed for the required white noise representation.

For best linear prediction from an infinite past, let ${\mathcal{H}}_0$ be the closed linear span of $X_t$ with $t\le 0$. The causal representation expresses every such $X_t$ in present and past $\eta$ values, while invertibility puts every ${\eta }_t$ with $t\le 0$ in ${\mathcal{H}}_0$. In particular ${\eta }_1$ is orthogonal to ${\mathcal{H}}_0$. The representation at time one then gives the orthogonal projectionConsequently the best linear predictor and its error variance areOrthogonality proves optimality among linear predictors in the closed past span, without asserting that the predictor must be the conditional mean for a non-Gaussian process.

For an aggregate claims model, let $X$ denote one claim size, with expected value $\mu$ and variance ${\sigma }^2$. The total $S$ is zero when the claim count is zero. Given $N=n$, independence givesThe law of total expectation and law of total variance, together with $\mathbb{E}N=\mathrm{Var}(N)=\lambda$ for a Poisson distribution, implyThe raw second moment appears because the random count itself contributes $\lambda {\mu }^2$ to the variance.

Conditional on $N$, the moment-generating function of the sum is $M_X(t{)}^N$. Averaging with the Poisson distribution therefore gives the compound Poisson distribution transformThis is valid wherever $M_X(t)$ is finite. Finite first and second moments alone do not ensure positive exponential moments; for positive claims the corresponding Laplace transform always exists.

For the independent portfolios put $\Lambda ={\lambda }_1+{\lambda }_2>0$ and $w_i={\lambda }_i/\Lambda$. Their Poisson distribution counts add to a Poisson distribution with parameter $\Lambda$. By Poisson-multinomial conditioning, conditional on the total count $n$ the first risk count has a binomial distribution with parameters $n,w_1$, and the second is the remaining count. Thus one can generate the same total loss by drawing $n$ independent risk labels with these weights, then drawing each claim from its label's law. The merged severity has mixture distributionIt follows that $T$ has a compound Poisson distribution with count parameter $\Lambda$ and this severity law. This is the fixed-year version of Poisson superposition of insurance portfolios. Its expected value and variance areAlternatively, multiplying the two independent aggregate Laplace transforms yields $\exp \{\Lambda (w_1{M}_{X_1}(t)+w_2{M}_{X_2}(t)-1)\}$ wherever finite, confirming the same compound Poisson distribution.

For the retained compound Poisson aggregate under per-claim reinsurance, replace each claim $X$ by its retained payment $g(X)$. Assume the retention is measurable, with $0\le g(x)\le x$ as usual. The count parameter remains $\Lambda$, and the severity law is the pushforward measure of the mixture severity under $g$. Thus the retained compound Poisson aggregate $T_I$ has a compound Poisson distribution with that transformed severity andIts moment-generating function is $\exp \{{\sum }_i{\lambda }_i(M_{g(X_i)}(t)-1)\}$ on its finite domain. For a general retention $g$, zero retained payments can occur and are allowed as compound-Poisson marks; they may equivalently be removed by Poisson thinning. The two specified contracts retain strictly positive payments for strictly positive claims.

We use the normalized Gowers U3 norm on an interval, for which a quadratic phase has norm one. Let ${\mathcal{C}}_N$ consist of integer tuples $(x,h_1,h_2,h_3)$ whose eight vertices $x+{\omega }_1{h}_1+{\omega }_2{h}_2+{\omega }_3{h}_3$, $\omega \in \{0,1{\}}^3$, all lie in $[1,N]$. If $\mathcal{C}$ denotes complex conjugation, defineEquivalently, choose a prime $8N<M<16N$, extend $f$ by zero to $G=\mathbb{Z}/M\mathbb{Z}$, call that extension $F$, and put $I=1_{[1,N]}$. ThenThe large ambient modulus prevents wraparound in cubes supported on the interval, so the ratio is independent of the chosen such $M$. This also proves nonnegativity and the norm properties by the Gowers uniformity norm on $G$. Some conventions omit the denominator; their interval norm differs by a fixed bounded factor, and the quadratic-phase norm is then $\Vert I{\Vert }_{U^3(G)}$.

For $f_0(x)=e(\alpha x^2)$, the exponent in every conjugated cube product is a third additive difference of a quadratic polynomial and is zero. Thus $\Vert f_0{\Vert }_{U^3(N)}=1$ in the normalized interval convention.

The generalized von Neumann inequality for four-term progressions connects this norm to counting arithmetic progressions. For functions bounded by one on a cyclic group of prime order greater than three,Three applications of Cauchy-Schwarz prove the bound. Zero extension and division by the number of interval progressions give the analogous interval estimate up to an absolute constant. In particular, if $g=1_A-\alpha 1_{[1,N]}$, where $\alpha =|A|/N$, has small Gowers U3 norm on an interval, expansion of the progression count shows that it differs from ${\alpha }^4$ times the count for $[1,N]$ by $O(\Vert g{\Vert }_{U^3(N)}{N}^2)$. A large Gowers U3 norm detects structure capable of changing four-term progression counts; quadratic phases are the basic example.

For the remaining proof use the ambient $G$ just specified, define ${\partial }_{h}F(x)=F(x+h)\overline{F(x)}$, and take averages uniformly on $G$. For frequencies not on the character grid, an exponential is evaluated at the chosen integer representatives; the general proof below selects actual characters $\theta =k/M$. The printed question does not define $G$ or the interval normalization, so these conventions make the assertion precise. If instead ${\partial }_{h}f(x)=f(x)\overline{f(x+h)}$ is used, conjugate the correlations and reverse every frequency sign; the quadratic example then has $\theta (h)=-2\alpha h$.

Lagrange's theorem states that for a subgroup $L$ of a finite group $G$, $|G|=[G:L]|L|$. In particular $|L|$ divides $|G|$; the equal-sized cosets partition $G$.

Apply the theorem to $H\cap K$ inside both subgroups. Its order divides both coprime orders, so $H\cap K=\{1\}$. For $h\in H$ and $k\in K$, the group commutator $hk{h}^{-1}{k}^{-1}$ lies in $K$ by $K$ being a normal subgroup, and in $H$ because $k{h}^{-1}{k}^{-1}\in H$. It is therefore the identity. Thus every element of $H$ commutes with every element of $K$.

Define $\theta :H\times K\to G$ by $\theta (h,k)=hk$. The cross-commutation just proved gives $\theta (h,k)\theta (h^{\prime },k^{\prime })=h{h}^{\prime }k{k}^{\prime }=\theta (h{h}^{\prime },k{k}^{\prime })$, so this is a group homomorphism from the direct product of groups. The product assumption makes it surjective. If $hk=1$, then $h=k^{-1}\in H\cap K$, so both entries are the identity and the kernel of a group homomorphism is trivial. HenceThis proves the internal direct product theorem in the present coprime-order setting, rather than merely matching the numbers of elements.

By the partition and the law of total probability, $\mathbb{P}(A)={\sum }_j\mathbb{P}(A\cap B_j)={\sum }_j\mathbb{P}(A\mid B_j)\mathbb{P}(B_j)$. Also $\mathbb{P}(B_i\mid A)=\mathbb{P}(A\cap B_i)/\mathbb{P}(A)$. Combining these identities proves Bayes theorem:The assumed positive probabilities make the conditional expressions well defined.

Two nonzero vectors are linearly independent when $s\mathbf{x}+t\mathbf{y}=0$ forces $s=t=0$. Their linear span then has dimension two. If they are linearly dependent, one is a nonzero scalar multiple of the other, and their span has dimension one. Thus the two requested dimensions are $\boxed{2\text{ and }1}$.

The Euclidean scalar product and its norm are $\mathbf{x}\cdot \mathbf{y}={\sum }_{j=1}^n{x}_j{y}_j$ and $\Vert \mathbf{x}\Vert =\sqrt{\mathbf{x}\cdot \mathbf{x}}$. To prove the Cauchy-Schwarz inequality, first handle $\mathbf{y}=0$ trivially. Otherwise the squared normgivesEquality holds exactly when the residual vector is zero, meaning the vectors are linearly dependent; this includes the zero-vector cases. Expanding $\Vert \mathbf{x}+\mathbf{y}{\Vert }^2$ and applying Cauchy-Schwarz then givesTaking nonnegative square roots proves the triangle inequality.

For the unit-vector optimization, let $\mathbf{w}=\mathbf{x}+\mathbf{y}$. Independence ensures $\mathbf{w}\ne 0$. The variable part of $S$ is $\mathbf{z}\cdot \mathbf{w}$, which Cauchy-Schwarz bounds below by $-\Vert \mathbf{w}\Vert$. Equality occurs only for the antiparallel unit vector. ThusThis is minimizing a linear functional on a sphere. The two consequences follow below.