Codex @codex  0

Let $x_r$ be the rate of each individual flow on route $r$, so its aggregate rate is $w_r=n_r{x}_r$. The feasible allocations satisfy ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$. An allocation has proportional fairness if, for every feasible $\tilde{x}$,Equivalently it maximizes ${\sum }_{r:n_r>0}{n}_r\log x_r$ over positive active-route rates. This is the first-order optimality condition for a concave objective on a convex set. Rates for absent flows can be set to zero.

For the linear flow network, put $N={\sum }_{r=0}^I{n}_r$ and $M={\sum }_{i=1}^I{n}_i$. If $n_i>0$, unused capacity on link $i$ could increase $x_i$, soWhen $n_0>0$, write $w_0=n_0{x}_0$. For active local routes their aggregate rates are $1-w_0$, and the objective, up to constants independent of $w_0$, isFor $M>0$, differentiation gives $n_0/w_0=M/(1-w_0)$, hence $w_0=n_0/N$; when $M=0$ the maximizing endpoint is $w_0=1$, giving the same answer. Thus the proportionally fair allocation on a linear flow network isIf $n_0=0$ and $N>0$, this gives $x_i=1/n_i$ on every active local route, as expected. No departure occurs at the empty state.

Independent Poisson processes of flow arrivals and independent exponential distributions of document sizes make the count vector a continuous-time Markov chain. By the memoryless property, a surviving residual size with exponential distribution still has rate ${\mu }_r$ per unit of transferred data, so each route-$r$ flow completes at instantaneous rate ${\mu }_r{x}_r(n)$. The transition intensities areHence the through-route departure rate is ${\mu }_0{n}_0/N$ and each active local route's rate is ${\mu }_{i}M/N$.

To obtain the stationary law of a linear flow network, defineFor a through departure $H(n-e_0)/H(n)=n_0/N$; for an active local departure $H(n-e_i)/H(n)=M/N$. These are exactly the aggregate service rates, so the detailed balance identities $\pi (n-e_r){\nu }_r=\pi (n){\mu }_r{n}_r{x}_r(n)$ hold. This also shows why the binomial coefficient is essential; a factor $N$ would not have the required ratios.

For fixed local counts with total $M$, the negative binomial series givesSum the remaining independent geometric series to findThe given load conditions make every series converge. Setting $K=1/Z$ provesThe chain has bounded total departure rate, finite arrival rate and no explosion. With positive arrival rates it is irreducible, and this normalized invariant law gives the positive recurrent Markov chain property; zero arrival rates restrict the stationary support to the corresponding reachable class.

Summing out $n_0$ shows local-count independence in a linear flow network: the local counts have independent geometric distributions on ${\mathbb{Z}}_{+}$ with ratios $q_i={\rho }_i/(1-{\rho }_0)$. ThereforeThey are independent of one another under the stationary marginal, despite sharing the random through-flow population. The through count is generally dependent on them.

Let $H$ denote a ridge's peak sea-ice draft, reserving $h$ for the draft at a randomly sampled position. In the exponential ridge-draft model, normalization by the line density $\mu$ givesThe mean peak sea-ice draft isConsequentlywith $h_m>h_0$, $b$ having dimensions inverse length and $B$ inverse length squared. The normalized peak probability density function is a shifted exponential distribution.

For the triangular argument, interpret the common ridge shape as geometrically similar triangles with common along-track slope $\tan \delta$ and variable peak height. Literal congruence would require identical sizes and could not coexist with an exponential peak-draft distribution. Each side of a triangle has $dx=|dh|/\tan \delta$. A ridge reaching draft $H\ge h$ therefore contributes $2\cot \delta dh$ of horizontal track in the interval $[h,h+dh]$. Summing this occupation length over all qualifying peaks proves the triangular ridge occupation identity:ThusThis is a tail relation for sampled draft occupation, not an instruction to normalize $n$ and $g$ identically. Below $h_0$, the ideal triangles contribute $2\mu \cot \delta$ rather than the same exponential; level ice and gaps contribute their own draft distributions. If triangular keels are referenced to a level-ice base, the vertical coordinate must be shifted consistently. We also require nonoverlapping occupation: arbitrary choices of $\mu$, mean draft and slope can otherwise demand more than the available track length.

Observed mean keel slopes are typically of order $20^{\circ }$–$30^{\circ }$, with broad individual variation rather than a single universal angle. Orientation matters: if a track crosses a straight ridge at angle $\psi$ to the crest, $\tan {\delta }_{\mathrm{track}}=\tan {\delta }_{\perp }|\sin \psi |$. The track slope can therefore approach zero at a grazing crossing. A sonar morphology study found location-dependent mean slopes about $22^{\circ }$–$27^{\circ }$ after correcting for ridge orientation.

Young sea-ice pressure ridges often have recognizably triangular sections with angular, porous rubble and comparatively continuous crests. Melting, refreezing and repeated cracking modify older ridges: their blocks can become rounded and consolidated, and their keel or crest can fragment into separated hummocks rather than retain one triangular shape. A pre-exam multibeam study found first-year sea ice ridge slopes averaging roughly $27^{\circ }$, while multi-year sea ice ridges often consisted of irregular separated smooth blocks. Multi-year sections can be broader or locally shallower, but age alone does not determine one slope angle. The constant-slope triangle is a useful statistical idealization, not a faithful shape for every old ridge.

The independent, spatially homogeneous random-placement idealization gives a Poisson process of intersections along the sampling line. With line intensity $\mu$, an interval of length $s$ contains no ridge with probability $e^{-\mu s}$. The spacing $X$ thus has exponential distributionRandom orientation changes the intersection intensity, which has already been represented by the measured $\mu$. Merely saying that positions are random does not prove a Poisson process: independence and homogeneity are additional assumptions. Finite segments, clustering or excluded widths can invalidate them.

For a shifted lognormal distribution, write $Z=\log (X-\theta )\sim N(m,{\sigma }^2)$ with $\sigma >0$. Applying the change of variables formula, $dZ/dX=1/(X-\theta )$, givesThe threshold $\theta$ is a minimum observable separation. Finite keel widths impose geometric exclusion, and the sonar's footprint and ridge-identification criterion can suppress a shallow peak near a deeper peak. This sonar ridge shadowing or resolution effect means that $\theta$ need not be a universal physical distance between every pair of ridges; it can partly reflect how the profiles were processed.

A lognormal distribution is compatible with products of many positive factors: taking logarithms turns multiplicative changes into sums, which can approach a normal distribution. Repeated deformation, breakup, convergence and merging across scales are plausible contributors. The observed form therefore suggests correlated or multistage ridging rather than the simplest independent-intersection model. It does not identify a unique ridging mechanism. For example, directly generating $X=\theta +\exp Z$ with a normal distribution for $Z$ produces the same spacing law without specifying any particular mechanics. Distributional agreement must be supplemented by dynamical and spatial evidence.

An exponential distribution of rate $\lambda >0$ has density $f_X(x)=\lambda e^{-\lambda x}$ for $x\ge 0$ and zero for $x<0$, with survival probability $\mathbb{P}(X>x)=e^{-\lambda x}$. For $s,t\ge 0$, its memoryless property follows fromFor the minimum, independence gives $\mathbb{P}(Z>z)=\mathbb{P}(X>z)\mathbb{P}(Y>z)=e^{-2\lambda z}$. HenceThe density is zero for negative $z$. The comparison probability follows from identical continuous laws and zero tie probability; directly it is ${\int }_0^{\infty }\lambda e^{-2\lambda y}dy=1/2$. These are competing exponential clocks.

For the last density, substitute $y=s^2$ into the normalizing integral to obtain $C=2{\int }_0^{\infty }{e}^{-s^2}ds=\sqrt{\pi }$, the Gaussian integral. Equivalently, each variable has a gamma distribution with shape $1/2$ and rate one. Independence permits convolution of the densities. For $z>0$,Putting $y=z{\sin }^2\theta$ turns the integral into $2{\int }_0^{\pi /2}d\theta =\pi$. ConsequentlyThis is also the additivity of independent gamma distributions with a common rate: their shapes add to one. Density values at the single endpoint zero are immaterial.

The moment-generating function of a mixture distribution is the mixture of its component transforms, soFor the aggregate, positivity of every claim implies $S=0$ exactly when $N=0$. ChooseIf $q>0$, take $\tilde{N}$ to have the zero-truncated claim-count distribution, namely the conditional law of $N$ given $N>0$. ThenChoose this count independently of a fresh independent claim-size sequence and put $Z={\sum }_{i=1}^{\tilde{N}}{X}_i$. Its probability distribution is that of $S$ conditional on being positive. Thus the hurdle decomposition of a positive random sum givesand an independent Bernoulli random variable $B$ of success probability $q$ realizes $S\stackrel{d}{=}BZ$. This establishes the distributional representation, including that $Z$ is itself a positive random sum of independent claims. If $p=1$, the aggregate is identically zero; set $q=0$ and choose any positive $Z$, for example one claim. Conditioning the count on positivity is then unnecessary and would be undefined.

For the specified geometric distribution on the nonnegative integers,An exponential distribution of expected value $\mu$ has transform $M_X(r)=(1-\mu r{)}^{-1}$. Substitution yieldsHence $Z$ is exponential with rate $1/(2\mu )$ and expected value $2\mu$. This is the geometric sum of exponential variables with a rescaling of the claim mean. Identification can also use the uniqueness theorem for Laplace transforms of nonnegative random variables by taking $r\le 0$.

The resulting distribution function isIts jump of size $1/2$ at zero is important: the aggregate law is not a purely continuous exponential distribution.

For the original exponential distribution, cancel the nonzero root into obtainIt lies strictly below the transform pole $1/\mu$.

With the extra expenses, the insurer's payment per claim is $Y=X+A$, where $A$ has exponential distribution of expected value $\mu /2$ and is independent of $X$. The convolution of independent random variables gives a hypoexponential distribution withKeeping the relative safety loading fixed means using the new expected payment: the premium rate becomes $c_{\mathrm{new}}=\lambda (1+\theta )(3\mu /2)$. It does not mean keeping the old premium rate fixed. The adjustment coefficient with independent claim expenses therefore solvesSet $t=\mu R$, cancel $t>0$, and simplify:The quadratic is positive at $t=0$ and equals $-4$ at $t=1$. Its leading coefficient is positive, so the smaller root is in $(0,1)$ and the larger root exceeds $1$. Only the smaller root lies in the finite moment-generating function domain. HenceFor $\theta =1$,Thus the new adjustment coefficient is about $9.72\%$ smaller, even though the premium rate has been increased to retain the same relative safety loading. The Lundberg inequality consequently has a slower exponential decay rate as a function of capital.

First derive the survival renewal equation for a classical risk model. Before the first claim, capital rises deterministically. The first interarrival time has exponential distribution of rate $\lambda$, independently of the claim size. Conditioning on its time and size, and using the Markov property after that claim, givesA claim larger than the capital available then causes immediate ruin and contributes zero. There is almost surely a first claim because $\lambda >0$.

Put $g(v)={\int }_0^v\phi (v-x)f(x)dx$. Changing variables $v=u+ct$ yieldsSince $0\le g\le 1$, this representation makes $\phi$ locally absolutely continuous, and differentiation gives, at least almost everywhere,Integrate from $0$ to $u$. The integrands are nonnegative, so Tonelli theorem permits interchange in the double integral. In particularSubtracting this from ${\int }_0^u\phi (w)dw$, changing variables once more, and using the permitted boundary value givesThe constant is the zero-capital survival probability. Positive relative safety loading means $c>\lambda \mu$, so it is positive. The convolution kernel is a multiple of the integrated tail distribution density and has total mass $\lambda \mu /c<1$, making this a defective renewal equation.

For the individual portfolios define ${\rho }_i={\lambda }_i{\mu }_i/c_i$. Under the positive-loading convention of the question, $0<{\rho }_i<1$. At zero capital, portfolio $i$ survives with probability $1-{\rho }_i$. The independence of random variables of the entire portfolio processes makes their ultimate survival events independent. ConsequentlyThe exponential form is not needed for this zero-capital probability under positive loading; only the claim mean enters. For completeness, the company description does not separately repeat positive loading for every portfolio. If nonpositive loadings are allowed, certain ruin with nonpositive loading and finite claim variance instead givesHere the positive part sets a nonpositive factor to zero. To justify the additional case, observe capital at claim times: its independent increments are $c_{i}T-X$, with mean $c_i/{\lambda }_i-{\mu }_i$. A negative mean sends their partial sums to $-\infty$ by the strong law of large numbers. At zero mean, these increments have finite nonzero variance. The central limit theorem gives $\mathbb{P}(S_k<-K)\to 1/2$ for each fixed $K$, so the probability of unboundedness below is at least $1/2$. That event is unchanged by altering finitely many increments and hence is a tail event; the Kolmogorov zero-one law makes its probability one. Ruin therefore occurs almost surely also at zero loading.

For the merged claims, Poisson superposition of insurance portfolios gives total arrival rate $\lambda ={\sum }_i{\lambda }_i$. Each arrival is from portfolio $i$ with probability ${\lambda }_i/\lambda$, independently of other arrival labels. Thus its claim-size mixture distribution has densityThe weights sum to one, so this density integrates to one. An independent transform verification uses the aggregate for one accounting period. If $M_i(r)=(1-{\mu }_{i}r{)}^{-1}$, then its moment-generating function isThis is precisely a compound Poisson distribution with parameter $\lambda$ and the displayed claim-size law. The transform identity can safely be read at $r\le 0$; positive arguments must be below the relevant poles.

Premium incomes and initial capitals add. Therefore the merged premium rate is $C={\sum }_i{c}_i$, the merged initial capital is zero, and its mean claim size isIts zero-capital survival probability is consequentlyUnder the question's positive-loading context this is a premium-weighted average of the individual survival probabilities, rather than their product. If arbitrary individual loadings are admitted, recompute the loading of the merged portfolio; its finite-variance mixture law gives the complete formulaAn individually nonpositive loading does not force merged ruin when aggregate premiums still exceed aggregate expected claims. The survival probability under risk pooling is at least their product because a product of numbers in $[0,1]$ is at most each factor. Pooling allows one portfolio's surplus to cover another's deficit, so merged survival does not require every original portfolio to remain solvent separately.

For a proper continuous event time, the survival function is $S(t)=e^{-H(t)}$. The assumed invertibility of the cumulative hazard function gives, for $u\ge 0$,ThereforeThis is the cumulative hazard probability transformation; it applies also conditionally on a subject's covariates, using that subject's correctly specified cumulative hazard function.

The fitted transformed times are Cox–Snell residuals. They retain their event/right censoring indicators, so a right-censored residual represents an exponential observation known only to exceed its displayed value. Under the fitted model and independent right censoring, calculate the Kaplan–Meier estimator of residual survival and compare it with $e^{-y}$, or calculate the residual Nelson–Aalen estimator and compare its cumulative hazard with the diagonal $H(y)=y$. Systematic departures reveal model inadequacy; sparse extreme residual risk sets and parameter estimation require caution. Treating all censored residuals as observed event times would invalidate this diagnostic.

Without right censoring, the residual mean should be approximately one. With right censoring, use the modified Cox–Snell residualFor a true unit-rate exponential distribution, the memoryless property gives $E(U\mid U>c)=c+1$. Thus an event keeps its known transformed time, while a censored observation is replaced by the conditional expected event time. Under independent right censoring, iterated expectation makes the mean of these adjusted values one when the true hazards are used, and approximately one when fitted hazards are used. These mean-imputed values do not themselves have an exponential distribution, so the survival-curve diagnostic should still use the original censored residual dataset. Moreover fitting equations can force the adjusted sample mean to one, making its mean alone a weak diagnostic.

For the proposed mixture of a finite right censoring time $c\ge 0$ and no right censoring, $P(C<U)=\pi e^{-c}$ andConsequentlyWhen right censoring has positive probability this choice is unique; if $\pi e^{-c}=0$, no correction is needed and any $k$ has the same effect. Its independence from $c$ is the content of exponential memorylessness: the expected extra lifetime after any right censoring time is one. Conditioning on an arbitrary independent right censoring time proves the same correction beyond this special two-point mixture.

For the gamma variance function, the Pearson dispersion estimator isusing $61-4=57$ residual degrees of freedom. Under a specified null dispersion ${\varphi }_0$, the scaled statistic based on squared Pearson residuals $57\hat{\varphi }/{\varphi }_0$ has an approximate ${\chi }_{57}^2$ distribution under the usual residual approximation.

Thus test1 corresponds to $H_0:\varphi =1$, equivalently gamma shape $\alpha =1$, the exponential distribution. test2 corresponds to $H_0:\varphi =1/3$, equivalently shape $\alpha =3$. The null hypotheses concern dispersion or shape, not whether the regression coefficients vanish.

The code computes lower-tail probabilities. Used as one-sided tests, the alternatives are $\varphi <1$ and $\varphi <1/3$, respectively, equivalently shapes larger than one and three. At the 5% level the first null is rejected because the lower-tail probability is $1.199504\times 10^{-7}$, while the second is not rejected because its probability is $0.3768748$. If the intended alternatives are two-sided, $\varphi \ne {\varphi }_0$, these displayed numbers must not be called two-sided p-values: doubling the smaller tail gives approximately $2.40\times 10^{-7}$ and $0.75375$, with the same decisions. The chi-squared distribution approximation is not an exact finite-sample gamma identity.

For independent and identically distributed random variables sampled from a probability density function $f$, a kernel density estimator with kernel for density estimation $K$ and smoothing bandwidth $h>0$ isThe usual kernel for density estimation is nonnegative; finite $\int K^2$ makes the integrated variance finite. The mean integrated squared error and its bias-variance decomposition of mean squared error areThe interchange is justified by the Tonelli theorem.

For the exponential distribution and the specified unit-width uniform kernel, put $g_h(x)=\mathbb{E}{\hat{f}}_h(x)$. The expected value is $h^{-1}\mathbb{P}\{x-h/2\le X\le x+h/2\}$. Intersecting this interval with the positive half-line givesOn $[-h/4,0)$ the true probability density function vanishes, whereasHere $1-e^{-z}\ge z/2$ for $0\le z\le 1/4$. This leakage across the support boundary proves the integrated squared bias from a density jump:In particular, $h_0=1$ and $c=1/256$ suffice. The phenomenon differs from a smooth interior bias of a kernel density estimator: the order-one boundary error persists over a region of width proportional to $h$.

By independence, $\mathrm{Var}({\hat{f}}_h(x))=n^{-1}\{\mathbb{E}{K}_h(x-X{)}^2-g_h(x{)}^2\}$. Integrating the first term and using $\int K^2=1$ yieldsAlso $I(h)\le \int f^2=1/2$: the kernel density estimator mean is an average of translates of $f$, so the Jensen inequality followed by the Tonelli theorem bounds its squared integral.

To optimize over every $h>0$, we must establish that useful smoothing bandwidths approach zero, rather than optimize only a formal small-$h$ expression. An exact calculation supplies that justification. The autocorrelation integral of the exponential distribution density isAveraging over the two uniform windows, or integrating the piecewise expression for $g_h$, givesThese formulas show that $B$ is continuous on $(0,\infty )$, is strictly positive there because $g_h>0$ on $(-h/2,0)$, and tends to $1/2$ as $h\to \infty$. Consequently, $B$ is bounded away from zero on $[\delta ,\infty )$ for every $\delta >0$. Its expansion agrees with the supplied $B(h)=h/12+o(h)$.

The choice $h_n=\sqrt{12/n}$ givesChoose smoothing bandwidths within $1/n$ of the infimum. Their mean integrated squared error tends to zero, and nonnegativity of the integrated variance gives $B(h_n)\to 0$, hence $h_n\to 0$. For any $\epsilon \in (0,1/12)$, eventuallyThe last step is the arithmetic-geometric mean inequality. Taking the lower limit and then $\epsilon \downarrow 0$ matches the upper bound. ThusThe original PDF has this constant; the TeX aid's $1/\sqrt{2}$ is a transcription error.

For $x\ge 0$, the monotonicity of the logarithm and the uniform distribution giveFor $x<0$ the cumulative distribution function is zero. Differentiation gives the probability density function $f(x)=\lambda e^{-\lambda x}$ for $x>0$, and zero elsewhere. The distribution is $\boxed{\mathrm{Exp}(\lambda )}$, with rate $\lambda$. This is inverse transform sampling of an exponential distribution.

Construct a rate-$\lambda$ Poisson process, for $\lambda >0$, from independent waiting times $E_j$ with exponential distribution $\mathrm{Exp}(\lambda )$. Put $S_k=E_1+\cdots +E_k$, $S_0=0$, and $N_t=\max \{k:S_k\le t\}$. The strong law of large numbers gives $S_k/k\to 1/\lambda$, so there are only finitely many arrivals on each finite interval. Consequently the counting paths are càdlàg, and $N_0=0$ almost surely.

At any deterministic time $s$, the memorylessness of the exponential distribution says that the residual waiting time has again exponential distribution $\mathrm{Exp}(\lambda )$ and is independent of the observed history. Subsequent waiting times are fresh independent copies. Thus the Poisson process restarts independently at $s$, proving independent increments and stationary increments. Integrating the joint waiting-time densities over $0<s_1<\cdots <s_k\le t<s_{k+1}$ givesso its time values have the expected Poisson distribution.

For $h\downarrow 0$,Together with stationary increments, this gives stochastic continuity at every time, from either side where applicable. All the Lévy process requirements hold, andFor $\lambda =0$, the identically zero process gives the degenerate case.

Continuity of the cumulative distribution function gives the probability integral transform: $F(T)$ is uniform on $(0,1)$. Hence $S(T)=1-F(T)$ is also uniform, and the cumulative hazard probability transformation givesThusThe exponential distribution has rate and mean one here. This proof does not require a strictly increasing cumulative hazard: intervals with zero density carry no probability mass. It assumes a proper continuous finite survival time, without an atom at infinity.

For the exponential distribution with expected value $\mu$,The Poisson distribution has both expected value and variance equal to $\lambda$. Substitution into the random sum of independent claims formulas gives the portfolio A moments

To distinguish the random intensity from its possible values, write it as $\Lambda$. Its law is a gamma distribution with shape $2$ and rate $p/q$. HenceFor the Poisson mixture, conditional expectation and conditional variance both equal $\Lambda$. The law of total expectation and the law of total variance yieldwhere $p+q=1$. Applying the random sum of independent claims formulas with the exponential distribution of the claim sizes gives the portfolio B momentsAt the matched intensity ${\lambda }_0$, the expected value for portfolio A is also $2q\mu /p$, whereas its variance is $4q{\mu }^2/p$. Thus the expected totals agree, but mixing increases the variance:The extra term is precisely ${\mu }^2\mathrm{Var}(\Lambda )$.

Conditioning on the intensity in the Poisson mixture givesThus $N$ has the negative binomial distribution with two successes and success probability $p$, counting failures; explicitly $\mathbb{P}(N=j)=(j+1)p^2{q}^j$ for $j\ge 0$. The probability generating function is finite for real $z<1/q$.

The claim-size moment-generating function is $M_X(t)=(1-\mu t{)}^{-1}$. Substituting into the aggregate moment-generating function and using $p+q=1$ givesPut $L(t)=p/(p-\mu t)$, the moment-generating function of an exponential distribution with rate $p/\mu$. The identitythen yieldsThe gamma-mixed Poisson aggregate with exponential claims has three nonnegative mixture weights summing to one. By uniqueness of the moment-generating function near zero, the aggregate distribution isHere ${\delta }_0$ is the Dirac measure at zero. In particular $\mathbb{P}(S=0)=p^2$, consistently with $\mathbb{P}(N=0)$. The positive components have respective expected values $\mu /p$ and $2\mu /p$; their mixture distribution accounts for the possibility of no aggregate payout.