For a group with sample size $n_s$, let $Y_s$ be its overweight count. The natural sampling model is a binomial distribution, $Y_s\sim \mathrm{Bin}(n_s,{\pi }_s)$, with observed proportion $p_s=Y_s/n_s$. A grouped-binomial logistic regression specifiesFitting event/nonevent counts, or fitting the proportions with weights $n_s$, gives the same binomial likelihood. The weights represent trial counts; unweighted regression of the proportions would give small and large groups the same influence. If printed percentages are rounded, exact original event counts are preferable to manufacturing fractional counts from $n_s{p}_s$.

Year should ordinarily be a factor with three levels unless a linear trend over time is specifically intended. Country, sex and age group are also possible factors. The actual subset, reference levels and formula are absent from this PDF. For a year-by-age interaction term, writeWith treatment contrasts, $\alpha$ is the baseline logit, $u_a$ and $v_t$ are effects at the reference level of the interacting factor, and $w_{at}$ is an additional difference of log-odds differences. Exponentiating a main-effect contrast gives an odds ratio; exponentiating an interaction contrast gives a ratio of odds ratios. It is incorrect to interpret a main effect as a constant effect across the other factor when the interaction term is present.

A coefficient table should be read as estimate, standard error, a Wald estimate/error ratio and its reference-distribution probability. A whole multi-level interaction term is better tested by a nested likelihood-ratio test, using the decrease in deviance and the number of added independent coefficients. Null and residual deviances, residual degrees of freedom, convergence and fitted probabilities should all be checked. Large residual dispersion suggests lack of fit or dependence within sampled schools; a quasibinomial regression or an appropriate clustered analysis may then be preferable. Residual deviance is not automatically a valid chi-squared goodness-of-fit statistic for sparse groups or after estimating a free dispersion.

An interaction plot should display the group means or fitted probabilities against year, with separate lines for the interacting factor, and should also inspect the link scale. Nonparallel probability curves alone do not establish a logit interaction: even an additive logit model transforms nonlinearly to probabilities. As an independent descriptive illustration, the original study's Table 2 reports English boys' age-specific percentages. The following original plot shows their widening age differences after 1984 on both scales. It uses published descriptive values, not a reconstruction of the absent examination fit. The data are from pmc.ncbi.nlm.nih.gov/articles/PMC26603/ .

For example, on the probability scale the 1984–1994 increase for the oldest group exceeds that for the youngest by $6.9-0.8=6.1$ percentage points using the rounded table values. On the logit scale the corresponding difference of differences isThis describes the interaction pattern, but its sampling uncertainty requires the group counts. The absent output prevents identifying the exam fit's coefficient estimates and significance tests.

There are two different uses of Poisson regression. A rare-event approximation models $Y_s\sim \mathrm{Pois}(n_s{\pi }_s)$ withThe fixed offset $\log n_s$ accounts for exposure, and exponentiated slopes are risk ratios, not odds ratios. The approximation is most appropriate when probabilities are small; it is not exact for appreciable prevalence because binomial variance is $n_s{\pi }_s(1-{\pi }_s)$ rather than $n_s{\pi }_s$.

An exact alternative stacks event and nonevent counts in each group and fits a log-linear model with a separate group intercept and event-by-predictor effects:Conditioning on their total $n_s$ gives the binomial event probability ${\mu }_{s1}/({\mu }_{s0}+{\mu }_{s1})$, whose logit is exactly $x_s^T\beta$. Profiling the group intercepts enforces the observed totals and reproduces the grouped-binomial logistic regression. This Poisson surrogate for a conditional multinomial model is exact and should not be confused with the rare-event approximation.

Record the discount level at the start of each policy year as the state. A no claims discount system is a Discrete-time Markov chain when, conditional on this state, the next year's claim behavior and update rule do not depend on earlier history. Independent annual accident and loss mechanisms with a fixed state-based reporting rule give this property; one should not equate the probability of an accident with the probability of a submitted claim.

The density identifies $Z=(\log L-\mu )/\sigma$ as a standard normal random variable: substitute $x=e^{\mu +\sigma z}$ and $dx=\sigma xdz$. Therefore, for $a>0$,Here $L$ is the numerical repair cost. The character preceding $L$ in the PDF is a currency symbol, not a separate multiplier.

Order states as $0\%,20\%,30\%$. With the stipulated common annual probability $p$ of no submitted claims, the transition matrix isA claim resets any state to zero, while a no-claim year raises the discount one level or keeps the top level fixed. If claim probabilities depend on the state, the corresponding row instead uses $p_i=\mathbb{P}(\text{no submitted claim}\mid \text{state }i)$; the same transition mechanism applies. The supplied single $p$ is a simplifying model assumption.

For a policyholder currently at zero discount, the current premium $C$ has the same value under either decision and cancels in the comparison. If the repair is self-funded and no later accidents occur, the next two premiums are $0.8C$ and $0.7C$, giving a total cost $L+2.5C$ including the current premium. If a claim is made, the next two premiums are $C$ and $0.8C$, giving total cost $2.8C$, with the repair paid by the insurer. Thus the two-year no-claims threshold with reset to zero isAt equality the holder is indifferent; it has probability zero under the continuous density. This assumes full reimbursement and no excess or deductible, as in the stated model.

Put $a=0.3C$. The conditional expected claim is $\mathbb{E}[L\mid L>a]$. Completing the square in the normal integral gives the truncated lognormal momentDividing by the reporting probability gives the expected size of a submitted claim for the full-premium holder:All monetary amounts must use the same unit despite the PDF's mixed currency notation. The other two current levels both have next-two-year no-claim premiums $0.7C+0.7C$, versus $C+0.8C$ after a claim, so their thresholds are $0.4C$. Consequently incorporating this behavioral decision into claim probabilities generally produces state-specific $p_i$, rather than deriving a common $p$ from the severity density alone; no accident-frequency law is provided.

Write the surplus process in the classical risk model as $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, with positive independent claims. Condition on the first claim time, which has exponential distribution of rate $\lambda$. If that claim occurs at $t$, the available capital is $u+ct$ and survival requires its size not to exceed that capital. The independent future process then givesPut $A(v)={\int }_0^v\varphi (v-x)f(x)dx$ and change variables $v=u+ct$. We obtainThe convolution of bounded $\varphi$ with the integrable density is continuous; differentiation of this expression is therefore justified. It proves the survival version of the ruin integro-differential equation:

The density is a hyperexponential distribution with mixture weights $3/4,1/4$ and exponential rates $4,2$. Its total integral is one and its mean isThe relative safety loading means $c=(1+\rho )\lambda {\mu }_X$. Substitution of $\rho =3/5$ gives $c=\lambda /2$ and $\lambda /c=2$.

Set $A_a(u)={\int }_0^u\varphi (u-x)e^{-ax}dx$. Differentiating in its equivalent form ${\int }_0^u\varphi (v)e^{-a(u-v)}dv$ gives $A_a^{\prime }=\varphi -a{A}_a$. The survival equation becomesApply the constant-coefficient operator $(D+4)(D+2)$, where $D=d/du$. Since $(D+4)A_4=\varphi$ and $(D+2)A_2=\varphi$,Expanding and cancelling yields $\boxed{{\varphi }^{\prime \prime \prime }+4{\varphi }^{\prime \prime }+3{\varphi }^{\prime }=0}$. Thus $\varphi (u)=A+B{e}^{-u}+C{e}^{-3u}$.

Positive loading also fixes the limiting boundary value $\varphi (u)\to 1$. Indeed the aggregate-claims process minus premium income tends to minus infinity almost surely by the strong law of large numbers. Its supremum over all time is consequently finite almost surely, since paths have finitely many jumps on bounded intervals. The probability that this supremum exceeds $u$ tends to zero. Hence $A=1$. At zero, the given value is $\varphi (0)=3/8$, and the integral equation supplies the extra condition ${\varphi }^{\prime }(0)=2\varphi (0)=3/4$. Thereforeso $B=-9/16$ and $C=-1/16$. The nonruin probability isThis is the two-exponential-mixture survival probability. It increases from $3/8$ to $1$ and satisfies the original integral equation, not just the differentiated equation; the zero-capital derivative condition prevents introducing extraneous solutions.

Where the moment-generating function $M_Y(t)=\mathbb{E}{e}^{tY}$ is finite in a neighborhood of zero, define the cumulant-generating function ${\kappa }_Y(t)=\log M_Y(t)$ and the $r$th cumulant by ${\kappa }_r(Y)={\kappa }_Y^{(r)}(0)$. Differentiation givessince $M_Y(0)=1$. Thus the first two cumulants are the mean and variance.

Conditional on $N=n$, independence gives $\mathbb{E}(e^{tT}\mid N=n)=M_V(t{)}^n$. Summing over the Poisson distribution yieldsHence the compound Poisson cumulants giveFor the moment statements, finite second moments suffice even when a moment-generating function does not exist near zero: conditioning gives $\mathrm{Var}T=\mathbb{E}N\mathrm{Var}V+\mathrm{Var}N(\mathbb{E}V{)}^2=\nu \mathbb{E}{V}^2$. This distinguishes the raw second moment from the variance of an individual claim.

For a single flood, the law of total expectation and law of total variance similarly giveCondition on the number of floods and apply the same laws once more. All floods have these moments and are independent of $M$, so the annual total satisfiesThe additional $\nu {\lambda }^2{\mu }^2$ term is the variation in the number of entire flood clusters; it would be lost by treating individual claims as an ordinary Poisson stream.

For the annual count $N={\sum }_{i=1}^M{N}_i$, its probability generating function isThis is the nested Poisson flood count. Differentiating, or conditioning on $M$, gives $\mathbb{E}N=\nu \lambda$ and $\mathrm{Var}N=\nu \lambda (1+\lambda )$. Thus $N$ is not Poisson when $\nu ,\lambda >0$, since its variance exceeds its mean. If either parameter is zero it is identically zero, a degenerate Poisson distribution.

Let $K_i$ be the number of valid claims in flood $i$. Independent validity marking gives $K_i\mid N_i\sim \mathrm{Binomial}(N_i,1-p)$. Its generating function isso Poisson thinning gives $\boxed{K_i\sim \mathrm{Poisson}(\lambda (1-p))}$. With validity independent also of claim amount, each claim contributes expected paid amount $(1-p)\mu$. Therefore the annual expected payment is $\boxed{\nu \lambda (1-p)\mu }$. The independent-mark assumption is needed for this multiplication; a size-dependent invalidity rule would require the joint expected value of amount and validity instead.