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For $C(K)=\mathbb{E}(S-K{)}_{+}$ and $\mathbb{E}S<\infty$, split the static call integral at a fixed positive strike. The first moment bounds the integrand near zero; polynomial call decay controls the integral at infinity for $0<\epsilon <\delta$. The endpoint can fail, as shown by a Pareto distribution with survival exponent $1+\delta$.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with expected value $\mu =\mathbb{E}[X_1]$ and finite, strictly positive variance ${\sigma }^2=\mathbb{E}[(X_1-\mu {)}^2]$. Write $S_n={\sum }_{j=1}^n{X}_j$ and ${\overline{X}}_n=S_n/n$ for their sum and sample mean. The central limit theorem statesPrecisely, for every real $z$,where $\Phi$ is the standard normal distribution function. The conclusion is convergence in distribution to the standard normal distribution. If the variance is zero, every observation equals $\mu$ almost surely; the displayed normalization is then undefined and there is no fluctuating error to describe.

The striking feature is that the observations need not have a normal distribution. They may be binary, bounded, or strongly skewed. Their detailed probability distribution disappears from this limit except through the expected value and variance. The subtraction removes predictable growth: $\mathbb{E}[S_n]=n\mu$. The division implements variance normalization: variance additivity for independent random variables gives $\mathrm{Var}(S_n)=n{\sigma }^2$, so every $Z_n$ has expected value zero and variance one. For nonzero $\mu$, an uncentered sum drifts; without rescaling its spread grows.

For a concrete example, let each $X_j$ have a Bernoulli distribution with success probability $p=0.2$. Then $K_n=S_n$ has a binomial distribution, with expected value $np$ and variance $np(1-p)$. Its exact probability mass function and its normalized count areHere $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is a binomial coefficient. Individual trials are conspicuously skewed, yet their centered, rescaled counts approach the same standard normal distribution as sums of fair trials.

At $n=100$, $K_n$ has expected value $20$ and standard deviation $4$. The exact binomial distribution gives $\mathbb{P}(16\le K_n\le 24)=0.740141\dots$. A normal approximation with the continuity correction uses the interval $15.5$ to $24.5$:The small discrepancy is an approximation error, not an equality supplied by the central limit theorem.

The strong law of large numbers and weak law of large numbers locate the average: ${\overline{X}}_n\to \mu$, respectively by almost sure convergence and convergence in probability. The central limit theorem magnifies that shrinking error and describes its shape. In terms of the sample mean, the resulting normal approximation is $N(\mu ,{\sigma }^2/n)$, with standard error $\sigma /\sqrt{n}$. This explains why four times as many independent observations halve the standard error. When $\sigma$ is known, ${\overline{X}}_n\pm 1.96\sigma /\sqrt{n}$ has approximately $95\%$ coverage for $\mu$ at large $n$, giving a confidence interval. This coverage refers to repeated samples, not a guarantee for an individual interval.

A useful application is the Monte Carlo method: estimate a success probability by the proportion of successful simulated trials. Nicholas Metropolis and S. Ulam's original paper, "The Monte Carlo Method" (1949), illustrates this with games of solitaire and estimation of a region's volume by randomly selected points, and discusses why estimates require probabilistic error control. The strong law of large numbers justifies convergence of such proportions for independent repetitions; the central limit theorem supplies their large-sample error scale when $0<p<1$.^[ref]

There are three useful limits to this picture:

A historical original-paper reference is J. W. Lindeberg's "Eine neue Herleitung des Exponentialgesetzes in der Wahrscheinlichkeitsrechnung" (1922), published in Mathematische Zeitschrift 15, pp. 211–225. The Lindeberg condition and Lindeberg-Feller central limit theorem below describe a version for sums whose summands need not share a probability distribution.^[ref]

A brief proof outline uses the characteristic function of $Y=(X_1-\mu )/\sigma$. Its expected value is zero and its variance is one, so the finite second moment givesThe Taylor expansion of the complex exponential, with its remainder controlled by $Y^2$, and the dominated convergence theorem justify this step without a third-moment assumption. Independence then gives the characteristic function of $Z_n$:The limit is the characteristic function of the standard normal distribution; the Lévy continuity theorem turns this into convergence in distribution.

Let $S_n={\sum }_{i=1}^n{X}_i$, where the summands are independent and identically distributed with Pareto distribution $\mathrm{Pareto}(\alpha ,\beta )$. Their positivity implies $\{X_1\ge na\}\subseteq \{S_n\ge na\}$, so for fixed $a>0$ and $na\ge \beta$,Both outer expressions tend to zero. A single large summand already prevents a strictly negative exponential rate. No positive exponential moment exists, so the exponential-moment form of the Cramér theorem cannot be applied. The bound proves this logarithmic rate and does not assert an exact tail asymptotic. In particular, it remains true when $a$ lies below a finite mean, where the event need not even be rare. The stronger fixed-summand-count asymptotic in a heavy-tailed subexponential distribution is a separate statement.

The Pareto distribution describes a positive random variable whose tail probability falls as a power law. In the shape–minimum convention used here, $X\sim \mathrm{Pareto}(\alpha ,\beta )$ has shape parameter $\alpha >0$ and minimum $\beta >0$, with probability support $[\beta ,\infty )$. Its probability density function isThe minimum $\beta$ has the same units as $X$; the shape parameter $\alpha$ is dimensionless. At fixed $\alpha$, changing $\beta$ gives a scale family, since $X/\beta \sim \mathrm{Pareto}(\alpha ,1)$. This convention is also used in uniform-Pareto conjugacy.^[ref]

The survival function and cumulative distribution function areFor $0<p<1$, solving $F(x)=p$ gives the quantile functionIts right-hand limit at $p=0$ is $\beta$; it tends to infinity as $p\uparrow 1$.^[ref] There is no probability mass at $\beta$: changing a probability density function's value at that single point changes no probabilities.

Choose $\alpha =2$ and $\beta =10$ as an original theoretical example. Then $\mathbb{P}(X>20)=1/4$, $\mathbb{P}(X>100)=1/100$, the median is $10\sqrt{2}\simeq 14.142$, and the 0.99 quantile is $Q(0.99)=100$. These chosen values and all figures below describe mathematical probability distributions, not actual wealth observations.

On this log-log plot, write $r=x/\beta$ to obtain $\log \overline{F}(\beta r)=-\alpha \log r$. The slope is $-\alpha$, whereas the probability density function has power exponent $\alpha +1$. Multiplying a threshold by ten divides its tail probability by $10^{\alpha }$. A larger $\alpha$ therefore concentrates the Pareto distribution nearer $\beta$; a smaller $\alpha$ makes extreme values more plausible. This exact scaling is useful when modelling large positive losses or sizes over several orders of magnitude.

The Pareto distribution is a heavy-tailed distribution: for every $t>0$, the exponential moment $\mathbb{E}[e^{tX}]$ diverges because $e^{tx}{x}^{-\alpha -1}$ eventually grows without bound. In contrast, the Laplace distribution has exponentially decreasing tails and some finite positive exponential moments. The Student's t-distribution also has power tails, but on both sides of the real line, with a smooth central region rather than a positive minimum.

For the Pareto distribution of shape $a>0$ and lower bound $b>0$, the density $a{b}^a{x}^{-a-1}$ is strictly decreasing on $[b,\infty )$. Its highest-density interval of probability $1-\alpha$, $0<\alpha <1$, must begin at $b$. The tail probability $\mathbb{P}(X>u)=(b/u{)}^a$ makes the upper endpoint $u=b{\alpha }^{-1/a}$. If this is a posterior distribution, the result is its highest posterior density interval. In particular, Uniform-Pareto conjugacy supplies posterior shape $a+n$ and lower bound $\max (b,{\max }_i{y}_i)$.

A Pareto distribution is a proper probability distribution for every $\alpha ,\beta >0$, even when its mean is infinite. To see both facts, integrate the probability density function against $x^q$. For real $q\ge 0$,The improper integral converges exactly when $q-\alpha -1<-1$. At $q=\alpha$ it diverges logarithmically; above that it diverges as a positive power of the upper endpoint. Setting $q=0$ gives one, proving normalization regardless of whether higher moments exist.

Taking $q=1,2$ and using $\mathrm{Var}(X)=\mathbb{E}[X^2]-(\mathbb{E}[X]{)}^2$ when both terms are finite givesFor $0<\alpha \le 1$, the nonnegative random variable has expected value $+\infty$; a centered variance about a finite mean is unavailable. For $1<\alpha \le 2$, the mean is finite and the variance is infinite.^[ref] A finite total probability and a finite expected value answer different questions.

At the chosen minimum $\beta =10$, $\alpha =3$ gives $\mathbb{E}[X]=15$, $\mathbb{E}[X^2]=300$ and variance $300-15^2=75$, but an infinite third moment. At $\alpha =3/2$, the mean is 30 and the variance is infinite; at $\alpha =3/4$, even the mean is infinite, while the median remains the finite number $102^{4/3}\simeq 25.198$. At the boundary $\alpha =2$, the earlier example has mean 20 but infinite variance. A finite sample contains finite values almost surely; that fact cannot establish that the underlying mean or variance is finite.

Conditioning a Pareto distribution on exceeding a threshold changes its minimum while preserving its shape parameter. For $u\ge \beta$ and $x\ge u$, the definition of conditional probability givesEquivalently, for $r\ge 1$,The ratio to the threshold has the same Pareto distribution at every admissible threshold.^[ref] For the chosen $\alpha =2$, $\beta =10$, only $1/4$ of outcomes exceed 20, but among those outcomes the probability of exceeding 40 is again $1/4$. When $\alpha >1$, the same identity gives conditional expectation $\mathbb{E}[X\mid X\ge u]=\alpha u/(\alpha -1)$.

This multiplicative scaling differs from the additive memoryless property of an exponential distribution. Indeed $Y=\log (X/\beta )$ has exponential distribution of rate $\alpha$, because $\mathbb{P}(Y>y)=e^{-\alpha y}$ for $y\ge 0$. Thus logarithms turn multiplication into addition and explain the scaling property. Conversely, if $U$ has uniform distribution on $(0,1)$, inverse transform sampling gives $X=\beta (1-U{)}^{-1/\alpha }$.

The excess $Z=X-u$ conditional on $X\ge u$ has survival function $(1+z/u{)}^{-\alpha }$ for $z\ge 0$. In the generalized Pareto distribution convention $\mathbb{P}(Z>z)=(1+\xi z/\sigma {)}^{-1/\xi }$, this means $\xi =1/\alpha$ and $\sigma =u/\alpha$. Here the excess starts at zero; the original Pareto distribution starts at $\beta >0$.

For a nonnegative random variable with finite positive mean, the share of its total amount carried by the largest fraction $q$ of outcomes can be defined by dividing their contribution to the expected value by the whole expected value. For a Pareto distribution, let $0<q\le 1$ and choose $x_q=\beta q^{-1/\alpha }$, so that $\mathbb{P}(X\ge x_q)=q$. If $\alpha >1$, direct integration givesThis is a share of amount, distinct from the fraction of outcomes. It depends on $\alpha$, but not on the unit of measurement $\beta$.^[ref]

For the chosen $\alpha =2$, the top 20% starts at $x_{0.2}=10\sqrt{5}\simeq 22.361$ and carries $T(0.2)=\sqrt{0.2}\simeq 44.72\%$ of the amount. To make the top 20% carry exactly 80%, solveThe resulting Pareto distribution has a finite mean and infinite variance. The 80/20 rule selects one shape parameter; it is not a universal law, or even a property of every Pareto distribution. For $\alpha \le 1$, the population-share definition above is unavailable because its denominator is infinite; substituting such an $\alpha$ into the finite-mean formula is invalid.

The left panel compares three quantiles in units of the minimum. The right panel shows $T(0.2)$ only where the mean is finite, and marks the chosen 80/20 value. The thresholds $\alpha =1$ and $\alpha =2$ separate the moment regimes; they do not separate valid and invalid probability distributions.

A straight-looking empirical log-log plot does not by itself establish a Pareto distribution or a reliable tail exponent; estimating an exponent and assessing goodness of fit are separate tasks.^[ref] If a Pareto distribution is used only above a fitted lower threshold, $T(q)$ describes that conditional tail population, not automatically the whole population. A genuine finite upper cutoff gives a truncated distribution with finite nonnegative moments; its survival function, quantiles and top shares must be recalculated. Truncation is a modelling change, distinct from merely observing a finite sample. None of the chosen numbers here estimates an actual wealth distribution.

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.

Integrating the Pareto distribution proposal yields $G(x)=1-(a/x{)}^b$ for $x\ge a$. For an independent uniform random variable $U$, solve $G(Y)=U$ to obtain the inverse transform sampling ruleEquivalently $Y=a{U}^{-1/b}$ because $1-U$ is also uniformly distributed. In implementation choose $U$ strictly between zero and one so that the inverse is finite.

Write $Z={\int }_a^{\infty }{x}^{\alpha -1}{e}^{-\beta x}dx$. A proper target requires $\beta >0$, and a proper proposal requires $b>0$. Normalization gives $f(x)=x^{\alpha -1}{e}^{-\beta x}/Z$ and the Pareto distribution proposal $g(x)=b{a}^b{x}^{-b-1}$ on $x>a$. Their ratio isThe derivative of its logarithm is $(\alpha +b)/x-\beta$. It is positive before $x_{\ast }=(\alpha +b)/\beta$ and negative after it. Since $a<\alpha /\beta <x_{\ast }$, the maximum occurs within the proposal support. Consequently the tight envelope for truncated gamma rejection sampling with a Pareto envelope isThe accompanying acceptance rule can be evaluated without $Z$:

Let $M_X(t)=\mathbb{E}{e}^{tX}$. Conditioning on $N\sim \mathrm{Pois}(\lambda )$ givesWhere the moment-generating function is finite around zero, differentiating the cumulant-generating function givesIn particular $\mathbb{E}S=\lambda \mathbb{E}X$ and $\mathrm{Var}S=\lambda \mathbb{E}{X}^2$. These compound Poisson cumulants use raw claim moments. The first two formulas also follow by conditioning: $\mathbb{E}[S\mid N]=N\mathbb{E}X$ and $\mathrm{Var}(S\mid N)=N\mathrm{Var}X$, soThis is the law of total variance. They therefore remain valid with finite first two moments even when no positive exponential moment exists.

Under per-claim excess of loss reinsurance, put $Y=\min (X,M)$ and let $W=(X-M{)}_{+}$ be the positive part of the excess. The insurer and reinsurer totals are sums of these payments over the same Poisson distribution count. Write $\overline{F}(x)=\mathbb{P}(X>x)$. Using the tail integral formula for moments on the tail probabilities of $Y$ and $W$ givesFor example, $\mathbb{E}{Y}^2={\int }_0^{\infty }\mathbb{P}(Y^2>t)dt=2{\int }_0^{M}x\overline{F}(x)dx$ after $t=x^2$, and shifting by $M$ gives the reinsurer formula. The moment formulas are finite whenever the corresponding claim moments are finite.

For the specified Pareto distribution, with shape three and minimum $d>0$,If $0\le M\le d$, every claim exceeds the reinsurance retention. ThereforeIf $M\ge d$, the tail integrals instead giveConsequently the requested sum of variances isOn the first branch, $V^{\prime }(M)=\lambda (4M-3d)$; it decreases until $M=3d/4$ and then increases. On the second branch $V^{\prime }(M)=\lambda d^3/M^2>0$. For $\lambda >0$ the minimum is unique; if $\lambda =0$, all variances are zero and every reinsurance retention minimizes them. ThusThe curve starts at $3\lambda d^2$, reaches this minimum, passes through $V(d)=2\lambda d^2$ with matching slope $\lambda d$ on both branches, and then increases concavely toward the horizontal asymptote $3\lambda d^2$. The variance-minimizing retention for shape-three Pareto claims is shown below.

The two aggregate payouts are dependent. Conditioning on the common count gives covariance $\lambda \mathrm{Cov}(Y,W)+\lambda \mathbb{E}Y\mathbb{E}W=\lambda \mathbb{E}[YW]=\lambda M\mathbb{E}W$, so $V(M)+2\mathrm{Cov}(S_I,S_R)=3\lambda d^2$ for every reinsurance retention, as required by $S_I+S_R=S$. The covariance of two components of a compound Poisson sum accounts for this distinction between a sum of variances and the variance of a sum.

There is a moment domain qualification for this heavy-tailed example. For every $t>0$, ${\int }_d^{\infty }{e}^{tx}3d^3{x}^{-4}dx=\infty$, and moments of order $j\ge 3$ also diverge. Thus its positive moment-generating function is not finite and higher finite cumulants cannot be obtained by differentiating around zero. Its mean and variance calculations above are valid by conditioning or by the first two right derivatives at zero of the Laplace transform of a nonnegative random variable.

On a product extension of the probability space, take a uniform distribution variable $U$ on $(0,1)$ independent of the original sigma-algebra. For $x>0$,These are the same conditional tails as in the maximal identity for a continuous nonnegative local martingale tending to zero. Taking expectations identifies the unconditional probability distributions:If $M_0=0$, define $M_0/U=0$; the endpoint $U=0$ is a null event if one uses $[0,1]$ instead. The possible atom at zero has mass $\mathbb{P}(M_0=0)$, since the conditional probability of a positive maximum is zero exactly on that event. For fixed $M_0=m>0$, the conditional law is the Pareto distribution of shape one and minimum $m$, with density $m/x^2$ on $x>m$.

The Brownian reflection principle, applied to the negative increments of the Brownian motion started at $a$, giveswhere $\Phi$ is the standard normal distribution function. Hence $T_0<\infty$ almost surely. The stopped Brownian motion $M_t=B_{t\wedge T_0}$ is a continuous nonnegative martingale, starts at $a$, and eventually vanishes. The maximal identity for a continuous nonnegative local martingale tending to zero therefore gives, for $H={\sup }_{0\le t\le T_0}{B}_t$,Equivalently,There is no atom at $a$. This is the Pareto distribution of shape one, or the translated version of the maximum before a lower Brownian barrier.

For $Y=2/(1-U)$ with $U$ uniform on $(0,1)$,Thus $Y$ has a Pareto distribution and is the importance sampling proposal, concentrated on the required tail. The target tail integrand is $f(y)=1/(\pi (1+y^2))$, so its importance weight isThe function returns $\tilde{\theta }=n^{-1}{\sum }_{i}w(Y_i)$. Since ${\mathbb{E}}_{q}w(Y)={\int }_2^{\infty }f(y)dy=\theta$, it is an unbiased estimator. For its second moment,An antiderivative is $\frac{1}{2}(\arctan y-y/(1+y^2))$, so the integral is $\frac{1}{2}(\pi \theta +2/5)$. Independence of the proposals givesThis is importance sampling of a Cauchy tail, with weights between $2/(5\pi )$ and $1/(2\pi )$. Their narrow range explains the much smaller variance than the indicator estimate, without appealing to nonexistent Cauchy moments.

The transformation of the uniform distribution gives, for $x\ge 1$,Thus $X_1$ has a Pareto distribution with lower endpoint $1$ and shape $2$. It has mean $2$, but no positive exponential moments, so the preceding version of the Cramér theorem is unavailable.

At $a=0$, the event in question is certain. For every fixed $a>0$ and all sufficiently large $n$, positivity of the summands gives the one-big-jump polynomial lower boundTaking logarithms and dividing by $n$ gives an upper bound $0$ and a lower bound $-2\log (na)/n\to 0$. HenceThis covers thresholds below, at and above the mean. Above the mean it means decay is slower than exponential speed $n$, not that the tail probability tends to $1$.

Multiplying the likelihood function by the Pareto distribution prior givesThe integral of this kernel is $m^{-(\alpha +n)}/(\alpha +n)$, so the normalized posterior distribution isThus it is $\mathrm{Pareto}(\alpha +n,m)$. This proves uniform-Pareto conjugacy: applying Bayes' theorem preserves the family of prior distributions.

Here $m_i=\beta$ for every breed, so $S=0$ andThe Bayesian model evidence increases towards its supremum as $\alpha \to \infty$. ThusFor a Pareto distribution, $\mathbb{P}(\theta >\beta (1+\epsilon ))=(1+\epsilon {)}^{-\alpha }\to 0$. The limiting empirical prior, and each corresponding posterior distribution, collapses onto $\beta$.

The boundary result is coherent within the assumed model: the observations favour the smallest allowed upper limits. Nevertheless, reporting exact concentration on a prespecified bound is overconfident for finite data and ignores hyperparameter uncertainty. A proper hyperprior on $\alpha$, sensitivity analysis for $\beta$, or a scientifically justified restriction on concentration avoids treating this limit as certain biological knowledge.

Let $A_{\delta }={\sup }_{K\ge 0}{K}^{\delta }C(K)<\infty$. The call-price decay and moment threshold follows by splitting the preceding integral at one. Since $(S-K{)}_{+}\le S$, for $0<\epsilon <\delta$,For $K\ge 1$, the decay bound givesThe power payoff static call representation therefore yieldsThe $\epsilon =0$ case is the given finite first moment. The strict endpoint matters: a Pareto distribution with $\mathbb{P}(S>x)=x^{-(1+\delta )}$ for $x\ge 1$ has $C(K)=K^{-\delta }/\delta$ for $K\ge 1$, but its moment of order $1+\delta$ is infinite. Thus the stated decay condition does not generally imply the endpoint moment.

For an integer $k\ge 1$, $\lfloor X\rfloor \ge k$ is the same event as $X\ge k$. Integration for the Pareto distribution givesThe second identity subtracts two successive tail probabilities, equivalently integrating over $[k,k+1)$. If $\mathbb{N}$ includes zero, the extra case is $\mathbb{P}(\lfloor X\rfloor \ge 0)=1$ and $\mathbb{P}(\lfloor X\rfloor =0)=0$, since $X\ge 1$ almost surely.