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Let $T_i$ be an event time and $C_i$ a potential censoring time. Under independent censoring, observe $Y_i=\min (T_i,C_i)$ and ${\delta }_i=1\{T_i\le C_i\}$. For an exponential distribution of rate $\theta$, an observed failure contributes $\theta e^{-\theta Y_i}$ to the survival likelihood, whereas a right-censored observation contributes $e^{-\theta Y_i}$. The censoring-law factors may be discarded when they are independent of $\theta$. Thus, writing $d={\sum }_i{\delta }_i$ and $V={\sum }_i{Y}_i$ for total observed person-time,For $d,V>0$ the unique maximum-likelihood estimate isCensored times belong in $V$: they record time during which failure was not observed. With $d=0$ and $V>0$, the likelihood decreases with $\theta$, so the supremum is at the boundary $\theta \downarrow 0$; there is no positive interior estimate. This is exponential-rate estimation from censored exposure.

Uninformative censoring means that the potential censoring time supplies no information about the latent failure time after conditioning on the modeled variables. A study ending at a fixed follow-up duration independent of prognosis is an example of administrative censoring. Informative censoring occurs, for example, when a seriously deteriorating patient withdraws because of that deterioration and the prognostic information is omitted from the model. Under informative censoring, those remaining in the risk set are selected by their future failure propensity, so the ordinary survival likelihood and its estimate need not target the true rate.

Using the reported durations and flags as though they were valid right-censored records givesTreating all durations as failures instead givesRelabelling has halved the event count without shortening any observed follow-up. The resulting estimate is therefore exactly half the complete-data estimate.

This is not a simulation of uninformative ordinary right-censoring. For a record declared censored at $Y_i$, ordinary right censoring requires $T_i>Y_i$; the construction instead took $Y_i$ to be the simulated failure time itself. Keeping a failure time and hiding only its flag does not produce the minimum of that time and an independent earlier censoring time. Even choosing labels at random does not fix this: independence of a flag from the recorded duration is different from independent censoring of a latent event time. Indeed the labelled durations still have mean one while the failure fraction is one half, so their ordinary censored estimate converges to $1/2$, not the generating rate one. Thus it is not a valid uninformative censoring scheme for the intended failure distribution; strictly, it has not implemented ordinary right-censoring of those simulated failure times at all. One could reproduce the randomly labelled record distribution with independent latent event and censoring times both of rate $1/2$, but that would be a different failure model, not the claimed rate-one model. Calling the flags random is therefore insufficient to justify the target distribution or its censored likelihood. This is the random relabelling is not independent right censoring issue.

A valid simulation draws independently $T\sim \mathrm{Exp}(1)$ and $C\sim \mathrm{Exp}(1)$, and records $(\min (T,C),1\{T\le C\})$. Sinceit gives independent censoring with the requested probability. Equivalently generate $T=-\log U$ and $C=-\log V$ from independent uniform random numbers. Another valid choice is fixed censoring at $\log 2$, since $\mathrm{Pr}(T>\log 2)=1/2$. These give a random censored fraction with expectation one half; they do not promise exactly 500 censored records. For the independent-exponential scheme, observed durations have rate two and mean one half, so $d/\sum Y_i$ correctly converges to one.

In the Bühlmann model, one policy has a random risk parameter $\Theta$. Conditional on $\Theta$, its annual observations $X_i$ are independent and identically distributed random variables, with mean $m(\Theta )$ and variance $v(\Theta )$. Put $m_0=\mathbb{E}[m(\Theta )]$, let $v=\mathbb{E}[v(\Theta )]$ be the expected process variance, and let $a=\mathrm{Var}(m(\Theta ))$ be the variance of hypothetical means. Assume the required second moments are finite.

The Bühlmann credibility premium is the best affine estimate of $m(\Theta )$ under mean squared error. After choosing the intercept to make the estimate unbiased, write it as $m_0+{\sum }_{i=1}^n{b}_i(X_i-m_0)$. If $B={\sum }_i{b}_i$, the law of total variance and conditional independence give its errorFor fixed $B$, the Cauchy-Schwarz inequality minimizes the last sum at $b_i=B/n$. Differentiating $a(1-B{)}^2+v{B}^2/n$ gives $B=na/(na+v)$. Thus the credibility factor and credibility premium areThe ratio form assumes $a>0$; when $a=0$ and $v>0$, the conditional mean is constant and $Z=0$. Predicting $X_{n+1}$ rather than its conditional mean adds the constant $v$ to the error and leaves the optimal affine estimate unchanged.

Here the latent intensity has a Pareto distribution with shape three and lower bound one. For $r<3$, its moments are $\mathbb{E}[{\Theta }^r]=3{\int }_1^{\infty }{\theta }^{r-4}d\theta =3/(3-r)$. ConsequentlyThe equality for $v$ uses the Poisson distribution conditional variance $v(\Theta )=\Theta$. There are two years of experience, so $Z=1/2$ and the observed sample mean is $k/2$. The Poisson credibility with a shape-three Pareto intensity is thereforeIn particular it equals $2$ when $k=5$.

For the exact Bayesian credibility estimate, the total count conditional on $\Theta =\theta$ has Poisson distribution of mean $2\theta$. Multiplication of its likelihood by the prior distribution density gives a Bayesian posterior density proportional to ${\theta }^{k-4}{e}^{-2\theta }$ on $\theta >1$. At $k=5$, its normalizing integral and first-moment integral areHence the next-year predictive expected value isThe Bühlmann credibility premium optimizes over affine estimates, whereas the Bayesian posterior mean optimizes over all square-integrable estimates. This explains why the two credibility estimates need not agree for this nonconjugate prior.

Write the surplus in the classical risk model as $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, with premium rate $c=(1+\rho )\lambda \mu$. The adjustment coefficient is the positive solution ofFor the exponential distribution of mean $\mu$, $M_X(r)=(1-\mu r{)}^{-1}$ for $r<1/\mu$. Cancelling the nonzero root $R$ in its equation gives $\lambda \mu /(1-\mu R)=c$, and henceIt is strictly between zero and $1/\mu$, as required by the transform domain.

The ultimate ruin probability is $\psi (u)=\mathbb{P}({\inf }_{t\ge 0}{U}_t<0)$. The Lundberg inequality statesRuin can occur only at a claim arrival. Thus the event defining the finite-claim ruin probability ${\psi }_n(u)$ is contained in the ultimate ruin event, giving ${\psi }_n(u)\le \psi (u)\le e^{-Ru}$ for every $n\ge 1$ and $u>0$.

For the explicit calculations, take $\mu =1$, put $d=1+\rho =c/\lambda$ and $h=d+1=2+\rho$, and let $T$ be the first arrival time. It has rate-$\lambda$ exponential distribution, independent of the severity. Ruin on the first claim is the event $X_1>u+cT$, soFor ruin by the second claim, condition on whether the first claim ruins the insurer or leaves capital $u+cT-x$. The independent future arrivals and severities have the same law as a fresh classical risk model. ThereforeThe two terms in the first line refer to disjoint ruin events, so there is no double counting.

For a direct check of the Lundberg inequality, now $R=1-1/d$. The ratio ${\psi }_1(u)/e^{-Ru}=h^{-1}{e}^{-u/d}$ is less than one. For the second finite-claim ruin probability, put $A=h^{-1}+d{h}^{-3}$ and $B=h^{-2}$. Its ratio is $e^{-u/d}(A+Bu)$. Sincethe derivative of this ratio is $e^{-u/d}[B-(A+Bu)/d]<0$ for $u\ge 0$, and its value at zero is $A<1$. This verifies both required bounds directly from the calculated probabilities.

Condition on the claim count. The empty random sum of independent claims is zero, and independence gives $\mathbb{E}[e^{tS}\mid N=n]=M_X(t{)}^n$. Therefore the random-sum transform identity iswherever the composition is finite. This follows directly from the law of total expectation, rather than replacing the random count by its mean.

Put $q=1-p$. For the specified negative binomial distribution, the negative binomial series gives $G_N(z)=[p/(1-qz){]}^k$. A rate-$\lambda$ exponential distribution has moment-generating function $M_X(t)=\lambda /(\lambda -t)$. HenceThe last expression is exactly the moment-generating function of a compound binomial distribution with independent count and severitiesThe moment-generating function exists in a neighborhood of zero, so its uniqueness identifies the distributions. In particular, the new claim sizes have mean $1/(p\lambda )$, not $1/\lambda$, and both aggregate distributions have an atom $p^k$ at zero. This is the negative-binomial sum of exponential claims representation with the severity rate explicitly retained.

Conditional on $N=n\ge 1$, the aggregate has Erlang distribution with density ${\lambda }^n{s}^{n-1}{e}^{-\lambda s}/(n-1)!$ on $s>0$. Repeated integration by parts of its tail givesIndeed, after substituting $v=\lambda s$, the integral ${\int }_{\lambda x}^{\infty }{e}^{-v}{v}^{n-1}dv/(n-1)!$ reduces recursively to the corresponding integral with exponent $n-2$, terminating at $e^{-\lambda x}$. Applying the law of total probability yields the original infinite mixture distribution tailApplying the same argument to the compound binomial distribution gives instead the finite Erlang-mixture tail for a negative-binomial exponential aggregate:The two tails are equal. The binomial representation replaces an infinite sum by a finite exact calculation, eliminating the need to choose a claim-count truncation and control its omitted tail. The inner sums can be evaluated recursively, using $a_{j+1}=a_j(p\lambda x)/(j+1)$ for their exponential-weighted terms.

For a bimatrix game with row and column payoff matrices $A,B$, an equilibrium pair is a pair of mixed strategies $(p,q)$ for which neither player gains by a unilateral change:This is a Nash equilibrium, or mutual best response pair. Every finite game with nonempty finite pure-strategy sets has at least one mixed-strategy equilibrium by Nash's theorem. Equivalently, the strategy domains are compact convex probability simplexes and the expected payoffs are continuous and linear in each player's own strategy. The existence assertion need not hold if only pure strategies are allowed.

Write the probability vectors in the order Bach, Mozart, Schubert. Against $q=(q_B,q_M,q_S)$ the row player's three pure-action payoffs areAgainst $p=(p_B,p_M,p_S)$ the column player's payoffs areA supported action must attain the corresponding maximum, and an omitted action must give no larger payoff.

We first justify that both players have the same support of a mixed strategy, rather than assume it. Each player's maximum payoff is strictly positive for every opponent probability vector. Thus $p_M>0$ implies $q_M>0$, $q_B>0$ implies $p_B>0$, and $p_S>0$ is equivalent to $q_S>0$. If $p_B>0$ but $q_B=0$, then either $q_M>0$, making Mozart's payoff $2q_M$ strictly exceed Bach's $q_M$, or $q_S=1$, making Bach's payoff zero and Schubert's positive. Both contradict best response. Therefore $p_B>0$ also implies $q_B>0$. Similarly, if $q_M>0$ but $p_M=0$, then $p_B>0$ makes the column payoff for Bach $2p_B$ exceed Mozart's $p_B$, or $p_S=1$ makes Mozart's payoff zero and Schubert's positive. Thus $q_M>0$ implies $p_M>0$. The supports coincide, and there are only seven nonempty possibilities.

For singleton supports the three matching pure pairs are equilibria. For each two-element support, solve the two players' indifference equations and normalize their probabilities. For example, on Bach–Mozart, the equations are $4q_B=q_M$ and $p_B=4p_M$, giving $q=(1/5,4/5,0)$ and $p=(4/5,1/5,0)$. On Bach–Schubert they are $4q_B=3q_S$ and $2p_B=3p_S$. On Mozart–Schubert they are $2q_M=3q_S$ and $4p_M=3p_S$. In each case the omitted action gives strictly less than the common supported payoff.

For full support, the row indifference equations give $q_M=4q_B$ and $3q_S=2q_M$; the column equations give $p_B=4p_M$ and $3p_S=2p_B$. Normalization determines the remaining pair. The full list, with row strategy first, isAll displayed supported probabilities are positive. The supported payoff pairs in the four nonpure cases are respectively $(8/5,8/5)$, $(12/7,6/5)$, $(6/5,12/7)$ and $(24/23,24/23)$; these values also verify the omitted-action inequalities directly. The common-support argument excludes every other support pair, and each indifference system has a unique solution. Hence there are exactly seven equilibrium pairs, three pure and four genuinely mixed. This is a complete support enumeration for a bimatrix game, not just a list of selected equilibria.