Put $\mathcal{R}=k_B/m_u$, with $m_u$ the atomic mass constant and $a_r$ the radiation energy-density constant. The ideal gas and radiation pressure contributions obey $P_g=\rho \mathcal{R}T/\mu$ and $P_r=a_r{T}^4/3$. With the constant stellar gas-pressure fraction, eliminate $T=\mu \beta P/(\mathcal{R}\rho )$ to getHenceSince $\mu$ and $\beta$ are spatially constant, so is $K$. Comparing with $P=K{\rho }^{1+1/n}$ gives polytropic index $n=3$. This structural stellar polytrope relation does not assert that every perturbed fluid parcel has stellar adiabatic exponent $4/3$.

Both gas and radiation pressures are nonnegative, so $0\le \beta \le 1$. Rearranging the constant-fraction result givesThis is the Eddington luminosity for the effective opacity constant of this model. Positive gas support gives a strict inequality, with equality only in the radiation-only limiting case. Exceeding the limit would require a negative gas-pressure fraction, which is incompatible with the assumed equation of state.

Divide the radiation-pressure gradient by the total-pressure gradient. Radiative diffusion in a star and hydrostatic equilibrium giveThis last quantity is independent of radius under the specified opacity assumption. Integration gives $P_r=fP+C$. With the standard idealized zero-pressure outer boundary, where both components vanish, $C=0$. The stellar gas-pressure fraction is thereforeconstant throughout the model. This is the Eddington standard model idealization. A finite photospheric pressure offset would need its boundary treatment; the differential relation alone does not set the integration constant to zero.

For the geometric distribution supported on positive integers, set $q=1-p$. The geometric series and its derivative giveEquivalently, the sum of the tail probabilities is ${\sum }_{k\ge 0}P(T>k)={\sum }_{k\ge 0}{q}^k=1/p$. For $p=0.02$, the expected wait is 50 years, so this is the 50-year return level. The exceedance probability determines the return period, but does not by itself determine a numerical flow rate.

In the original PDF's probability plot, most points lie close to the diagonal. The quantile plot likewise agrees well over most of the range, but the largest empirical flow, around $68.5$, is well above its fitted quantile, around $66$. The density plot captures the strongly decreasing right-skewed bulk of the excesses. The return level plot shows a broadly reasonable fit to most observations, with one conspicuously high extreme and increasingly wide uncertainty at long return periods. Consequently the generalized Pareto distribution is a reasonable first approximation, but its most extreme tail is less convincingly represented.

Reading the return-level curve at the 100-year tick givesThese are approximate graphical readings, not parameter-based calculations; rounding to whole units would give about $67$ with an interval roughly $65$ to $69$. The two blue curves are uncertainty bounds for the fitted return level, not prediction bounds for individual observations. The daily-data independence assumption and the stability of the fit as the threshold changes should be checked, for example by examining clusters of high flows and using declustering of extremes where necessary. A 100-year level also extrapolates beyond the 25-year record.

This is the generalized Pareto distribution for the excess $X-u$, with scale $\sigma >0$ and shape $\xi$. Its support requires $1+\xi (x-u)/\sigma >0$. The Pickands-Balkema-de Haan theorem says that, for distributions in an appropriate extreme-value domain of attraction, the conditional distribution of excesses above a sufficiently high threshold approaches a generalized Pareto form. This makes it the natural peaks-over-threshold method model; it is an asymptotic justification, not an assertion that every threshold is sufficiently high.

For $\xi >0$, the tail decays as a power and has no finite upper endpoint, giving a heavy tail. For $\xi =0$, the limiting distribution is exponential, with survival probability $\exp \{-(x-u)/\sigma \}$. For $\xi <0$, there is a finite upper endpoint $u-\sigma /\xi$, giving a bounded tail. Thus the sign of the shape parameter distinguishes heavy, exponential-type and bounded tails.

A 100-year return level is a flow threshold whose probability of being exceeded in a given year is one percent, under a stationary model for annual extremes. Equivalently, its long-run average waiting time between years with an exceedance is one hundred years. It does not imply that exceedances occur regularly one hundred years apart, or that one is guaranteed in any particular hundred-year interval. This definition uses yearly exceedance events even when the underlying observations are daily maxima.

The residual deviance is $290.93$ on $98$ residual degrees of freedom. Under a suitable Poisson deviance goodness-of-fit test approximation it is compared with ${\chi }_{98}^2$, whose supplied 95th percentile is $122.1$. Since $290.93$ is much larger, the chosen Poisson model does not provide a satisfactory absolute fit, despite being preferred among the three candidates.

The ratio $290.93/98\approx 2.97$ suggests substantial overdispersion or mean-model misspecification. It is a deviance ratio, not the exact Pearson dispersion estimator, which cannot be computed from the excerpt. Check residual patterns, nonlinear effects, unusual rivers and dependence. A Quasi-Poisson regression could adjust uncertainty under an estimated dispersion, and a negative binomial regression could model extra count variation; a more adequate mean model may also be needed. The nominal Poisson tests in part (a) should be interpreted in light of this failure, rather than treated as a validated final analysis.

The chosen Poisson regression has fitted meanFor zero building cover, its fitted expected number of floods over the 25-year observation window is $e^{0.96657}\approx 2.629$. Increasing the covered proportion by $\Delta$ multiplies the fitted expected count by $e^{3.02055\Delta }$. Thus a ten-percentage-point increase, $\Delta =0.1$, multiplies the expected count by about $1.353$, an increase of $35.3\%$. A one-percentage-point increase multiplies it by about $1.0307$. The full-unit multiplier $e^{3.02055}\approx 20.50$ compares proportions differing by one, not percentages differing by one.

These are associations in the conditional expectation with the recorded final-year building cover. The regression coefficient is not automatically a causal effect of changing development, especially since the count covers 25 years while cover was measured at the end.

The three Poisson regression models have independent counts and a logarithmic link function: their mean predictors include both pc and urban in m1, only urban in m2, and only pc in m3. Nested comparisons through the full model are preferable to a direct comparison of the two nonnested one-predictor models.

To remove urban from m1, test $H_0:{\beta }_{\mathrm{urban}}=0$ against $H_1:{\beta }_{\mathrm{urban}}\ne 0$. The likelihood-ratio test statistic is the deviance differencewith approximate null distribution ${\chi }_1^2$. It is below $3.84$, so do not reject at 5%. The corresponding Wald p-value $0.588$ is consistent with this decision. To remove pc from m1, test $H_0:{\beta }_{\mathrm{pc}}=0$ against a nonzero coefficient, giving $559.17-290.64=268.53$ with approximate null distribution ${\chi }_1^2$; reject decisively. The Akaike information criterion also prefers m3, whose value $710.94$ is smaller than $712.65$ and $979.19$.

Among these three models choose m3, retaining building cover and omitting the urban indicator. This is a comparison within the stated Poisson family; its absolute fit must still be checked, as the next part does.

Use the Aalen–Johansen estimator of the cumulative incidence function, updating the overall Kaplan–Meier estimator for either event:Right censoring changes subsequent risk sets, not the survival function by itself. Starting with the given estimates at $a_k$, the complete calculation isThe middle event is of the competing type: it decreases overall survival while leaving the disease cumulative incidence function unchanged at that instant. The event-free intervals leave every estimate and risk set unchanged.

At the final time, both the event subject and the subject censored at that time are in the just-before risk set, so its denominator is eight. This is the usual event-before-censoring convention for recorded ties. After the event and right censoring, six subjects remain at risk. ConsequentlyThe nine numbered source items are the inputs to this single calculation, not nine further questions.

The bounded optional stopping theorem gives $\mathbb{E}{X}_{T\wedge n}=\mathbb{E}{X}_0$. Decompose the difference from the terminal value:ConsequentlyThe first term tends to zero by hypothesis. The second tends to zero by the dominated convergence theorem, because $X_T$ is integrable and $T$ is finite with probability one. Thus the stopped martingale converges to $X_T$ with convergence in L1, which permits passage of expectations to the limit:The explicit tail condition supplies exactly the missing control for an unbounded stopping time.

For the real-valued variables here, take a $\mathcal{G}$-measurable version of $X=\mathbb{E}[Y\mid \mathcal{G}]$. Since the variables are bounded, the conditional expectation defining identity extends to the bounded $\mathcal{G}$-measurable multiplier $X$, givingThe equality case for conditional second moments now givesA nonnegative random variable with zero expectation vanishes with probability one. Thus $X=Y$ as an almost sure equality. The same proof works for square-integrable variables; boundedness is more than is needed.

The random walk $S_n$ is a martingale, since its integrable increments are independent of the past and have mean zero. For the bounded stopping time $n\wedge \eta$, the bounded optional stopping theorem, proved in the next question, gives $\mathbb{E}{S}_{n\wedge \eta }=x$.

If $0<x<r$, the value immediately before exit lies in $(0,r)$, soBefore exit the stopped value has absolute value less than $r$, and after exit it equals $S_{\eta }$. Consequently $|S_{n\wedge \eta }|\le r+|X_{\eta }|$ for every $n$. This is an integrable dominating random variable by the preceding part. Since $\eta <\infty$ with probability one, the dominated convergence theorem givesFor $r\le x$, $S_{\eta }=S_0=x$ directly, without any convention about $X_0$. Thus the expected stopped position is well-defined and has the stated value for every $r>0$.

For $0<x<r$, the exit time is at least one. Since $\{\eta \ge n\}=\{\eta >n-1\}\in {\mathcal{F}}_{n-1}$, this event is independent of $X_n$. The Tonelli theorem givesTherefore the integrability of a stopped random-walk increment bound isIt is the survival event, not the exit-at-$n$ event, that is independent of the next increment. The selected exit increment need not have the same distribution or mean as $X_1$.

For $r\le x$, the printed variable $X_{\eta }=X_0$ is undefined because the increment sequence starts at one. Either restrict this part to $0<x<r$, or make the harmless additional convention $X_0=0$. With that convention the conclusion also holds in the immediate-exit case. The integrability assertion needs this indexing qualification.

This is a lacunary power series: the coefficient of $z^{n!}$ is $n^n$, while the other coefficients are zero. Along its nonzero coefficients,The factorial dominates $n\log n$; for example $n!\ge n(n-1)(n-2)(n-3)$ for large $n$ already suffices for this limit. Hence the coefficient limsup in the Cauchy-Hadamard theorem is one. The radius of convergence is $\boxed{R=1}$.

Directly, for $|z|<1$ the consecutive term ratio is $(n+1{)}^{n+1}{n}^{-n}|z{|}^{nn!}$, which tends to zero; for $|z|\ge 1$ the term magnitudes are at least $n^n$ and fail to tend to zero. The factorial is the exponent of $z$, not part of its coefficient.

Let $P$ be a monic polynomial of positive degree $m\ge 1$ with distinct roots of a polynomial $r_j$. The partial fraction decompositionfollows by matching the simple pole residues; the difference is entire and vanishes at infinity. Comparing its expansion there gives ${\sum }_j{r}_j^{\ell }/P^{\prime }(r_j)=0$ for $0\le \ell <m-1$ and ${\sum }_j{r}_j^{m-1}/P^{\prime }(r_j)=1$. These identities give the continuity and derivative jumps needed for constant-coefficient Green functions.

The difference of two torsion-free connections is symmetric, because subtracting their zero-torsion identities gives $\Delta (X,Y)=\Delta (Y,X)$. Agreement of parametrized geodesics makes this difference skew-symmetric as well, by parametrized geodesics determine the symmetric part of an affine connection. Over the real numbers it must therefore vanish. Without the torsion-free condition, an arbitrary skew-symmetric difference changes torsion without changing the parametrized geodesics.

If the essential oscillation on concentric balls obeys $\omega (R/2)\le \theta \omega (R)+A{R}^{\alpha }$ with $0<\theta <1$ and $\alpha >0$, dyadic iteration gives $\omega (r)\le C{r}^{\mu }$ for every $0<\mu <\min \{\alpha ,-{\log }_2\theta \}$. The resulting decay gives a Hölder continuous function. Taking the exponent strictly below both decay rates avoids a logarithmic factor when the rates coincide.