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Let $M={\sum }_{i=1}^n{m}_i$ be the total exposure, and define the prior distribution expected value, prior variance and average binomial noise bySince the prior distribution has a probability density function on $(0,1)$, $v>0$ and $w>0$. All relevant moments exist because the parameter and observations are bounded. The binomial distribution givesThe law of total expectation, law of total variance and law of total covariance consequently yieldIn particular, conditional independence of the annual observations does not imply unconditional independence: they share the uncertain parameter.

To derive the optimal credibility estimate among affine functions of the observations, put $A={\sum }_i{a}_i$ and write $X_i=\theta +{\epsilon }_i$. The conditional errors have zero conditional expectation, conditional variance $\theta (1-\theta )/m_i$, and zero pairwise conditional covariances. They are also uncorrelated with every integrable function of $\theta$. Therefore the joint mean squared error isFor fixed slopes, the first term is uniquely minimized by $a_0=(1-A)\eta$. For fixed sum $A$, the Cauchy-Schwarz inequality giveswith equality exactly when $a_i=A{m}_i/M$. Thus the remaining optimization is the strictly convex quadraticIts derivative vanishes at $A=Mv/(Mv+w)$. Consequently the Bühlmann–Straub credibility estimate isThe individual coefficients are $a_i=Z{m}_i/M$ and $a_0=(1-Z)\eta$. The strict quadratic minimizations establish global optimality and uniqueness, not just necessary equations. Here $v$ is the variance of hypothetical means, $w$ is the expected process variance, and $Z$ is the Bühlmann–Straub credibility factor. More exposure gives more weight to the observed claim fraction. Since $0<Z<1$, the estimate remains between the observed pooled fraction and the prior expected value. If one separately permits a degenerate prior, $v=0$ leads to the constant estimate $\eta$, interpreted as $Z=0$.

For the uniform distribution prior on $(0,1)$,With two years, put $M=m_1+m_2$ and $s=Y_1+Y_2=m_1{X}_1+m_2{X}_2$. Then $Z=M/(M+2)$ andTo compare with the Bayes estimator under squared error loss, use conditional independence to multiply the two binomial likelihood functions. Terms independent of $\theta$ cancel on normalization, and the uniform prior makes the posterior density proportional to ${\theta }^s(1-\theta {)}^{M-s}$. Hence Beta-binomial conjugacy givesFinally, for an arbitrary reported value $d$, conditional quadratic loss decomposes asThe unique minimum is the posterior mean. Therefore the affine credibility estimate and the exact Bayesian estimate coincide in this case. The agreement illustrates exact beta-binomial credibility with unequal exposures; for a general prior distribution, minimizing over affine functions of the observations need not recover the unrestricted posterior mean.

In the classical risk model, the insurer's surplus iswhere $u\ge 0$ is initial capital, $N(t)$ is a Poisson process of rate $\lambda >0$, and the claim sizes are positive independent and identically distributed random variables with expected value $\mu$, independent of the claim-arrival process. Premium flows in at constant rate $c$. A positive relative safety loading $\theta$ meansso premium income exceeds the expected claim cost per unit time. The ruin stopping time is $\tau =\inf \{t\ge 0:U(t)<0\}$, with $\tau =\infty$ if ruin never occurs, and the ultimate ruin probability is $\psi (u)=\mathbb{P}(\tau <\infty )$.

Assume the stated positive adjustment coefficient $R$ exists and that the claim moment-generating function is finite at $R$. The Lundberg inequality isHere is a proof, including the stopping argument. By conditioning on the Poisson distribution of $N(t)$, the random-sum transform identity givesThe adjustment coefficient equation is $\lambda (M(R)-1)=cR$. Consequently the exponential surplus martingalehas expected value one at every finite time. More strongly, the claim process has independent increments, so for $s\le t$,Thus $Z$ is a nonnegative continuous-time martingale for the filtration generated by arrivals and claim sizes. For a fixed finite horizon $T$, the optional stopping theorem at the bounded stopping time $\tau \wedge T$ gives $\mathbb{E}{Z}_{\tau \wedge T}=1$. This bounded use is legitimate: the stopped values belong to the closed martingale on $[0,T]$ given by conditional expectations of the integrable variable $Z_T$.

On $\{\tau \le T\}$, the surplus is negative, and therefore $C(\tau )-c\tau =u-U(\tau )>u$. By nonnegativity on the complementary event,The ruin events increase to $\{\tau <\infty \}$ as $T$ increases, so continuity of probability gives the claimed Lundberg inequality. No expectation identity at the unbounded stopping time $\tau$ is needed. Positive loading by itself does not guarantee a positive adjustment coefficient for every claim law; the existence assumption is used here.

For the exponential distribution of claim sizes, $M(r)=1/(1-\mu r)$ for $r<1/\mu$. Substitution in the adjustment coefficient equation givesDiscard the zero root and divide by $\mu r$. The unique positive solution iswhich lies strictly inside the finite-transform domain.

For deterministic claim sizes, the claim moment-generating function is $e^{\mu r}$. Define $h(r)=e^{\mu r}-1-(1+\theta )\mu r$. It satisfies $h(0)=0$, $h^{\prime }(0)=-\theta \mu <0$, $h^{\prime \prime }(r)={\mu }^2{e}^{\mu r}>0$, and $h(r)\to \infty$ as $r\to \infty$. Thus its strict convexity gives exactly one positive zero $R_{\mu }$, with $h(r)<0$ for $0<r<R_{\mu }$.

Put $x=\mu R=\theta /(1+\theta )\in (0,1)$. Since $-\log (1-x)>x$, we have $e^x<1/(1-x)$. HenceIt follows thatThus the deterministic-claim Lundberg inequality provides a smaller upper bound at positive capital; at zero capital both bounds equal one. This is a special case of deterministic claims maximize the adjustment coefficient at fixed mean and loading: fixing the mean and loading, variability lowers the exponential decay coefficient relative to deterministic claims. The ordering of these upper bounds alone is not a proof that the actual ultimate ruin probabilities are ordered.

The claim-size exponential distribution has expected value $\mu >0$ and moment-generating functionConditional on $N_i=k$, independence makes the aggregate moment-generating function equal to $M_X(t{)}^k$. Averaging over the Poisson distribution of the count givesPut $\Lambda ={\sum }_{i=1}^n{\lambda }_i$. For fixed intensities, independence between risks givesThis is precisely the transform of a compound Poisson distribution. Indeed the addition of independent Poisson random variables makes the total count Poisson distributed with parameter $\Lambda$, and every claim in the pooled portfolio has the same exponential distribution. Thus the portfolio aggregate is compound Poisson with intensity ${\sum }_i{\lambda }_i$ and exponential claim sizes of mean $\mu$.

Now regard the individual intensities as independent gamma distributed quantities, with shape $m\ge 1$ and rate $\alpha$. Their moment-generating functions are $(\alpha /(\alpha -t){)}^m$ for $t<\alpha$. Multiplying these transforms proves additivity of independent gamma distributions with a common rate:Conditional on the individual intensities, the total count $N$ has Poisson distribution with parameter $\Lambda$. Because this conditional law depends only on their sum, it is also the law of $N$ conditional on $\Lambda$. Integrating its probability mass function against the gamma distribution density yieldsThus the total count has a negative binomial distribution with shape $nm$ and success probability $\alpha /(\alpha +1)$, using the convention that counts failures before the specified number of successes. Its probability generating function is $(\alpha /(\alpha +1-z){)}^{nm}$.

Conditional on $\Lambda$, the claim sizes retain their common exponential distribution and the amount $S$ has a compound Poisson distribution of intensity $\Lambda$. Therefore the unconditional amount has a compound mixed Poisson distribution with gamma mixing intensity of shape $nm$ and rate $\alpha$. To make the mixture explicit, average the conditional aggregate moment-generating function:The displayed domain ensures both the severity transform and the intensity transform are finite. The mixing distribution describes the Poisson intensity, not the claim sizes. In particular, the aggregate has an atom at zero of mass $\mathbb{P}(N=0)=(\alpha /(\alpha +1){)}^{nm}$.

Write $P(z)=\mathbb{E}[z^N]$, $F(z)=\mathbb{E}[z^{X_1}]$ and $G(z)=\mathbb{E}[z^{S_N}]$ for the three probability generating functions. Since both the claim count and every claim size are positive, $P(0)=F(0)=G(0)=0$. Conditioning on the count and using independence gives the random-sum transform identityThe crucial point is that the count recurrence begins at $n=2$. Multiplying it by $n{z}^{n-1}$ and summing, we obtainConsequently $(1-az)P^{\prime }(z)=(a+b)P(z)+p_1$. The chain rule applied to the aggregate probability generating function therefore givesThese identities hold inside the unit disk, or as identities of formal power series. Since $F(0)=0$, every coefficient of the composition depends on only finitely many count probabilities.

Compare the coefficient of $z^{k-1}$. The product $F{G}^{\prime }$ contributes ${\sum }_{j=1}^{k-1}(k-j)f_j{g}_{k-j}$, and $F^{\prime }G$ contributes ${\sum }_{j=1}^{k-1}j{f}_j{g}_{k-j}$. HenceThe required aggregate recursion for zero-truncated Panjer counts is initialized by $g_0=0$ and $g_1=p_1{f}_1$, and for every $k\ge 1$ it isThe empty sum at $k=1$ is zero. Every term on the right is known or has a smaller aggregate index. Omitting $p_1{f}_k$ would incorrectly apply the usual Panjer recursion with a zero initial value, producing zero for every aggregate probability.

For the zero-truncated Poisson distribution, $p_n/p_{n-1}=\lambda /n$ for $n\ge 2$, so $a=0$, $b=\lambda$ and $p_1=\lambda /(e^{\lambda }-1)$, with $\lambda >0$. ThusAs a direct probability generating function check, $P(z)=(e^{\lambda z}-1)/(e^{\lambda }-1)$ and therefore $G(z)=(e^{\lambda F(z)}-1)/(e^{\lambda }-1)$; differentiation reproduces this recursion.

Let $c=g^{\prime \prime }(\tilde{y})>0$. Near the interior minimum, Taylor expansion givesOn the scale $y=\tilde{y}+z/\sqrt{nc}$, the leading exponent is $ng(\tilde{y})+z^2/2$ and $dy=dz/\sqrt{nc}$. If the integral is localized at this minimum, the rescaled integration endpoints tend to infinity in both directions, so the Gaussian integral gives Laplace's method:Formally expanding the cubic term gives an order-$n^{-1/2}$ correction odd in $z$, whose Gaussian integral is zero. Quartic terms and the square of the cubic term produce the order-$n^{-1}$ correction. Sufficient smoothness near the minimum and control of the exterior integral justify this formal calculation.

The printed local assumptions alone do not guarantee localization. A useful localization condition for Laplace's method is that $g(y)\ge g(\tilde{y})+{\delta }_{\epsilon }$ outside each sufficiently small fixed neighbourhood of $\tilde{y}$, for some ${\delta }_{\epsilon }>0$, and that ${\int }_a^b{e}^{-n_{\ast }g(y)}dy<\infty$ for some $n_{\ast }>0$. For $n>n_{\ast }$, the exterior contribution is at mostwhich is exponentially smaller than the displayed local term. To see why something of this kind is needed, take $g(y)=y^2(1-y{)}^4$ on $(-1,1)$. Its unique minimum is at zero and $g^{\prime \prime }(0)=2$, yet on $1-n^{-1/4}<y<1$ we have $ng(y)\le 1$. This boundary interval alone contributes at least $e^{-1}{n}^{-1/4}$, dominating the proposed $n^{-1/2}$ local approximation. The Laplace formula therefore has its usual implicit tail assumption, rather than following from uniqueness of the interior minimum alone.

For the gamma function, set $y=nu$ to obtainHere $g(u)=u-\log u$ has its unique minimum at $u=1$, with $g(1)=1$ and $g^{\prime \prime }(1)=1$. It tends to infinity at both endpoints, and its integral is finite for positive exponents, so localization holds. Laplace's method yields Stirling's formula:

For a prior density $p$ and independent observations, Bayes' theorem writes the desired posterior expectation as a ratio of integrals:Suppose the log-posterior kernel has a unique concentrating interior mode $\tilde{\theta }$, with curvature $J_n=-H_n^{\prime \prime }(\tilde{\theta })>0$ of order $n$, and the smoothness and tail conditions hold for both integrals. Expanding $H_n$ quadratically gives a common Gaussian factor $e^{H_n(\tilde{\theta })}\sqrt{2\pi /J_n}$. Replacing the smooth amplitude $g(\theta )$ by $g(\tilde{\theta })$ givesThis approximation does not require $g$ to be positive: it treats $g$ as an amplitude, without taking its logarithm. For more accuracy, expand the amplitude and the cubic term in the log-posterior kernel. With $s=\theta -\tilde{\theta }$, the Gaussian moments are $\mathbb{E}{s}^2=J_n^{-1}$ and $\mathbb{E}{s}^4=3J_n^{-2}$. Terms multiplying $g(\tilde{\theta })$ cancel between numerator and denominator; the two remaining first corrections areunder the stronger smoothness, derivative-scaling and integrability assumptions needed for this expansion. The first term accounts for the curvature of the amplitude and the second for posterior skewness. This is posterior expectation by Laplace approximation. If there are several modes, boundary modes, or nonconcentrating tails, the appropriate separate contributions must be included; a single interior Gaussian approximation cannot then be assumed.

Write $R(v)={\int }_{\mathbb{R}}v(u{)}^{2}du$. The kernel density estimator with bandwidth $h>0$ isSufficient conditions are that $K$ be a nonnegative symmetric probability density function with $R(K)<\infty$ and finite nonzero second moment ${\mu }_2(K)=\int u^{2}K(u)du$, and that $h\to 0$, $nh\to \infty$. Symmetry and the second-moment assumption give $\int uK(u)du=0$. The expectation is a convolution:Taylor's theorem with integral remainder givesThe zeroth-order term integrates to $f(x)$, and the first-order term vanishes. The second derivative is bounded and continuous, so dominated convergence, with dominating function proportional to $u^{2}K(u)$, yieldsCombining this with the supplied leading variance gives the asymptotic mean squared errorWhen $f(x)>0$ and $f^{\prime \prime }(x)\ne 0$, its derivative is $-R(K)f(x)/(n{h}^2)+{\mu }_2(K{)}^2{f}^{\prime \prime }(x{)}^2{h}^3$. It changes sign exactly once, giving the pointwise optimal kernel bandwidthPositivity of $f(x)$ is needed for this positive interior optimum. The printed condition $f^{\prime \prime }(x)\ne 0$ alone does not ensure it. For instance, $f(x)=x^2\varphi (x)$ is a normalized bounded probability density function with bounded continuous square-integrable second derivative, but $f(0)=0$ and $f^{\prime \prime }(0)=2\varphi (0)>0$. At this point the displayed leading variance vanishes and the two-term AMSE has no positive interior minimizer; finer variance terms are needed to determine an appropriate bandwidth. The formal formula there gives zero, which does not satisfy the bandwidth assumptions.

For integrated error, the same expansion holds in $L^2$: translation continuity of $f^{\prime \prime }\in L^2$ applied to the integral remainder givesIndeed Minkowski's integral inequality bounds the norm of the difference, divided by $h^2$, by the integral of $u^{2}K(u){\int }_0^1(1-s)\Vert f^{\prime \prime }(\cdot -shu)-f^{\prime \prime }{\Vert }_{2}dsdu$; each translated difference tends to zero and is bounded by $2\Vert f^{\prime \prime }{\Vert }_2$. This proves the claimed integrated bias expansion without assuming a uniform pointwise remainder over the whole real line.

If $K_h(x)=h^{-1}K(x/h)$, direct integration of the variance givesThe last term is $O(n^{-1})$ because $f$ is bounded and integrable, hence square integrable, and convolution with a probability density function does not increase its $L^2$ norm. It is negligible compared with $1/(nh)$. Thus the asymptotic mean integrated squared error and its minimizer areHere $R(f^{\prime \prime })>0$: otherwise the continuous second derivative vanishes everywhere, making $f$ affine, impossible for a probability density function on the whole real line.

Now take the standard normal density $f=\varphi$. Differentiation gives ${\varphi }^{\prime \prime }(x)=(x^2-1)\varphi (x)$. Gaussian integration givesThe kernel constants cancel from the bandwidth ratio. At $x\ne \pm 1$,Put $t=x^2$. The logarithmic derivative of $e^{t/2}/(t-1{)}^2$ is $1/2-2/(t-1)$. On $0\le t<1$ the function increases, giving its minimum at $t=0$, with value one. On $t>1$ it decreases up to $t=5$ and then increases, giving its minimum $e^{5/2}/16<1$. Thus the global minimizers are $x=\pm \sqrt{5}$, and the normal-density local-to-global bandwidth ratio isThe ratio diverges at the two inflection points $\pm 1$, where the second-order pointwise bias approximation vanishes and needs a higher-order replacement.

For a nondegenerate distribution function $G$, membership $F\in D(G)$ in its maximum domain of attraction means that there exist $a_n>0,b_n\in \mathbb{R}$ such thatat every continuity point of $G$. A max-stable distribution satisfies: for each positive integer $k$, there exist $c_k>0,d_k\in \mathbb{R}$ such that $G(c_{k}x+d_k{)}^k=G(x)$. Thus a maximum of $k$ independent copies has the same probability distribution after a change of location and scale.

Suppose first that $F\in D(G)$. For fixed $k$, the distribution functions $H_n=F^{kn}$ have the two nondegenerate limitsThe given convergence-of-types result supplies constants $a>0,b$ such that $G(x{)}^k=G(ax+b)$. Equality extends from continuity points to all points by right continuity. Substitute $x=(z-b)/a$ to obtain $G((z-b)/a{)}^k=G(z)$, which proves max-stability with $c_k=1/a,d_k=-b/a$. Conversely, if $G$ is max-stable, take $F=G$ and its max-stability constants for $k=n$. Then $F(c_{n}x+d_n{)}^n=G(x)$ exactly for every $n$, proving $G\in D(G)$. Therefore the maximum domain of attraction is nonempty exactly when $G$ is max-stable.

For the particular logarithmic tail, complete the distribution function by taking $F(x)=0$ for $x\le x_0$. Let $a_n>1$ be the unique solution of $a_n\log a_n=n$, and take $b_n=0$. Equivalently $F(a_n)=1-1/n$, including $n=1$ with $a_1=x_0$. For each fixed $x>0$, eventually $a_{n}x>x_0$, andHere $a_n\to \infty$ follows directly from its defining equation. Since $\log (1-z)=-z+O(z^2)$ as $z\to 0$, it follows that $n\log F(a_{n}x)\to -1/x$. For $x\le 0$, the probability is zero. ConsequentlyThis is the shape-one Fréchet distribution, and the calculation proves the limit at every real $x$ without needing a regular variation theorem.

The function $t\log t$ is strictly increasing on $t>1$. For every integer $n\ge 3$, $n\log n>n=a_n\log a_n$, so $a_n<n$. The defining equation now gives $a_n=n/\log a_n>n/\log n$. Taking logarithms yields $\log a_n>\log n-\log \log n$, whose right side is positive, soMultiplying by $\log n/n$ and squeezing gives $a_n\log n/n\to 1$. Slutsky theorem applied to the normalized sample maximum therefore gives the simpler logarithmic-tail Fréchet normalization

Write $u(Y_i;\theta )={\partial }_{\theta }\log f(Y_i;\theta )$ and $U_n(\theta )={\sum }_{i}u(Y_i;\theta )$ for the individual and total score functions. Under the usual regularity assumptions, ${\mathbb{E}}_{{\theta }_0}u=0$ and ${\mathrm{Cov}}_{{\theta }_0}(u)=I({\theta }_0)$, the nonsingular per-observation Fisher information matrix. The central limit theorem and weak law of large numbers giveRearranging the stipulated score expansion, including its remainder, givesSlutsky theorem therefore proves the asymptotic normality of a maximum likelihood estimator:A consistent plug-in Fisher information matrix yields the Wald statistic $W_n=n({\hat{\theta }}_n-{\theta }_0{)}^{T}I({\hat{\theta }}_n)({\hat{\theta }}_n-{\theta }_0)$. Under the null hypothesis, $W_n\stackrel{d}{\to }{\chi }_d^2$, so a test of asymptotic size $\alpha$ rejects when $W_n$ exceeds the $(1-\alpha )$ quantile of that chi-squared distribution. A consistent observed Fisher information matrix gives the same limiting test.

With an $r$-dimensional parameter of interest $\psi$ and nuisance parameter $\lambda$, partition $I$ accordingly. The marginal asymptotic covariance matrix of $\hat{\psi }$ is the $\psi \psi$ block of $I^{-1}$, whose inverse is the efficient informationThus the nuisance-adjusted Wald statistic is $W_{\psi ,n}=n(\hat{\psi }-{\psi }_0{)}^T{I}_{\mathrm{eff}}(\hat{\theta })(\hat{\psi }-{\psi }_0)$, with limiting null distribution ${\chi }_r^2$. Using $I_{\psi \psi }$ alone is generally incorrect when the nuisance parameter is estimated; it is valid when the cross-information vanishes.

For the Inverse Gaussian distribution in this problem, expand the exponent to obtain the log likelihood, up to terms independent of both parameters,Its score functions areFor each positive $\psi$, the first score changes sign from positive to negative at $\lambda =\bar{Y}$, so this is the global maximizing mean parameter of an exponential family. Substituting it in the second score givesThe reciprocal function is strictly convex, so the denominator is positive for a nonconstant sample. The profile log likelihood is strictly concave in $\psi$ and tends to minus infinity at both endpoints, proving this is a maximum. Under continuous model sampling, a sample of size at least two is nonconstant almost surely. For a constant sample, including a one-observation sample, no finite joint maximum-likelihood estimator exists: the likelihood increases without bound as $\psi \to \infty$ at $\lambda =\bar{Y}$.

For one observation, differentiating again and using $\mathbb{E}Y=\lambda$ givesThis parameter orthogonality proves that estimating $\lambda$ does not reduce the efficient information for $\psi$. The Wald statistic has the same algebraic form in the two cases:Using information evaluated at ${\psi }_0$ instead is another first-order equivalent convention.

The printed claim of coincidence needs a first-order interpretation. Known and unknown $\lambda$ need not give numerically identical Wald statistics, because their maximum-likelihood estimators differ. With $\lambda$ known, direct maximization givesFor example, observations $1,2$ and known $\lambda =1$ give ${\hat{\psi }}_K=4$ and ${\hat{\psi }}_U=12$. At ${\psi }_0=1$, the displayed Wald convention gives $9/16$ and $121/144$, respectively. Thus exact equality is false. Under the null hypothesis, however, $\bar{Y}-\lambda =O_p(n^{-1/2})$, hence ${\hat{\psi }}_K-{\hat{\psi }}_U=O_p(n^{-1})$. Their centered, standardized estimates differ by $o_p(1)$, so their Wald statistics differ by $o_p(1)$. This is the valid asymptotic coincidence supplied by inverse Gaussian shape estimation and nuisance orthogonality.