Probability and statistics

A probability plot compares order statistics with plotting positions or model quantiles to diagnose whether a proposed probability distribution describes data. If the model can be transformed into a linear identity $T(F(x))=a+bS(x)$, plot $T(i/(n+1))$ against $S(X_{(i)})$. Approximate linearity supports that model's shape; systematic curvature suggests a mismatch.

The plotting position $i/(n+1)$ comes from the exact mean of the $i$th uniform order statistic, by the probability integral transform. It does not assert equality between the expectation of a nonlinear transformation and that transformation of the expectation. Fitted slope and intercept can estimate model parameters, but the visual plot alone is not a calibrated hypothesis test.

For a sample or finite biological population of positive size $n$, the relative frequency of a category represented $n_i$ times is $n_i/n$. It is a proportion and differs from an oscillation frequency.

Random variation represented by a probability distribution or stochastic process, whether from intrinsic events or an uncertain environment.

Descriptive statistics summarizes observed data, for example through location and spread statistics or graphical summaries such as a histogram.

A table of category counts, often accompanied by sample proportions or percentages. Its denominator must be stated, especially when observations contain missing data.

The expected value of the number of observations in a category under a specified statistical model. For $n$ observations with common category probability $p_j$, linearity of expectation gives $E_j=n{p}_j$; the calculation does not itself require independence. In a contingency table, fitting independence gives $E_{ij}=n_{i+}{n}_{+j}/n$. These are model-based counts, which may be fractional, rather than an additional set of observed responses. Pearson's chi-squared test compares observed counts with their corresponding expected frequencies.

The sum of category percentages up to a specified position in an ordering. It has a substantive interpretation for an ordinal categorical variable; the cumulative percentage of an arbitrarily ordered nominal categorical variable changes when the labels are reordered.

A percentage whose denominator excludes observations missing the variable being summarized. It can differ from a percentage calculated over the complete dataset's sample size. Valid percentages for two different variables need not use the same observations.

A mode is a value at which a probability mass function or specified probability density function attains its largest value. There may be more than one. For the normal distribution the unique mode equals the mean; the Maxwell speed distribution has mode $\sqrt{2k_{B}T/m}$ and a larger mean speed. For a continuous probability distribution, a mode maximizes density rather than the probability of an exact value, which is zero. Density modes can change under a nonlinear change of variables, as the distinction between Maxwell-Boltzmann velocity distribution and Maxwell speed distribution illustrates.

A scatter plot represents paired quantitative observations as points in a coordinate plane. Group labels, uncertainty and transformations can reveal nonlinear patterns or heterogeneous associations that a single correlation coefficient conceals. A visual association does not establish causation.

A box plot compresses a statistical sample into a box spanning its lower and upper quartiles $Q_1,Q_3$, a line at its median $Q_2$, and two whiskers. It makes location and central spread easy to compare across groups on a common value axis. The box represents the central half in terms of quantile levels; with ties or interpolated quartiles, it need not contain exactly half the observations. Whisker endpoints require a stated convention: they can be the observed extremes, or they can stop short of separately marked observations. ^[ref]

Here is one explicitly toy dataset, reused under histogram: twelve invented, unitless values, with repeated values counted separately, already arranged as order statistics:For this box plot we choose the interpolated quartiles defined under quartile, using R's type-7 rule. This givesThe interquartile range is $I=Q_3-Q_1=3.5$. Our whisker convention uses the fencesThe whiskers end at the smallest and largest observations in the closed interval $[L,U]$: here $1$ and $8$. Values strictly outside the fences receive separate outlier markers: here only $20$. An observation exactly on a fence is retained within the whisker under this convention. The fences are thresholds, not observed whisker endpoints; the upper whisker is at $8$, not $11.5$. R's documented whisker rule likewise includes observations no more than $1.5$ box lengths away, although its box uses hinges, which can differ from our type-7 quartiles. ^[ref]

A separate outlier marker is a prompt to investigate, not a finding that the value is bad data and not an instruction to delete it. A distant observation can be a recording error, a legitimate unusual outcome, or evidence that the assumed statistical model misses some behavior. Check its provenance and context before deciding how to analyze it. No measurement-error claim is possible for our invented $20$: we deliberately put it there to demonstrate the rule. ^[ref]

The beauty of a box plot is its compact comparison of ordered positions rather than every observation. That compression also loses information: many different statistical samples have the same quartiles and whiskers, so the box alone cannot reveal every cluster or gap that a histogram might show. For example, replacing the two invented $6$ values by $5.5$ and $6$ leaves the type-7 quartiles, whiskers and separate marker unchanged. Displaying the individual values alongside the box plot is especially helpful for such a small statistical sample. Always state both the quartile rule and the whisker rule; changing either can change which values are marked.

Quartiles are the quantiles at probability levels $1/4,1/2,3/4$; $Q_2$ is a median. For a population we can use its quantile function, $Q_r=Q(r/4)$ for $r=1,2,3$. Finite-sample conventions differ. Our box plot uses linear interpolation between order statistics: for $n\ge 2$ and $0<p<1$, setThen $Q_r=\tilde{Q}(r/4)$. This is R's type-7 convention, described in the documentation authored by Ivan Frohne and Rob J Hyndman. It differs from the generalized-inverse sample quantile defined elsewhere in this wiki: for the twelve toy values, that rule selects ranks $3,6,9$ and gives $2,4,6$, rather than $2.75,4.5,6.25$. These are different stated conventions, not arithmetic disagreements. ^[ref]

The interquartile range measures the width between the lower and upper quartiles, in the same units as the observations. It describes central spread rather than the distance between the observed extremes. In our toy box plot, $\mathrm{IQR}=6.25-2.75=3.5$; moving the largest value from $20$ to $200$ would leave the quartiles and interquartile range unchanged under the chosen rule, because the order statistics used in their interpolation would not change.

A histogram groups numerical observations into adjacent intervals called bins and draws one rectangle per bin. Its attraction is that a long list of values becomes a visible pattern of concentration, gaps and skewness. Specify the bin edges, the treatment of values on those edges, and what the vertical axis means: count, proportion, or density. Those choices are part of the graphic. ^[ref]

Let $n>0$ observations be covered by disjoint bins $B_j$ with positive widths ${\Delta }_j$ and counts $n_j$, so ${\sum }_j{n}_j=n$. For example, use left-closed, right-open bins, closing the final right edge too. The three common height choices have different meanings:

For the density convention, a histogram is a piecewise constant probability density function estimate:Its normalization follows by adding rectangular areas:If the observations have units, density has reciprocal units; a density height is not a probability and can exceed $1$. With equal bin widths, count heights and density heights differ only by a common scale factor. With unequal widths, count-height areas $n_j{\Delta }_j$ give wider bins extra visual weight unrelated to their counts. Dividing by width removes that distortion. R's histogram documentation explicitly describes area-one normalization for unequal bins. ^[ref]

Reuse the twelve invented values under box plot, with bins $[0,2),[2,4),[4,8),[8,22]$. The counts are $(1,3,6,2)$ and the widths are $(2,2,4,14)$. Hence the density heights and areas are exactlyThe second and third bars have equal density heights even though the third contains twice as many observations: it is also twice as wide. The last bar contains only two observations; its count-height area would be $2\cdot 14=28$, larger than the six-observation bar's $6\cdot 4=24$. Its correct density area is only $1/6$.

Bin choice trades detail against stability. Broad bins merge distinctions: the toy bin $[8,22]$ hides the empty stretch between $8$ and $20$. Narrow bins expose gaps but can make each individual observation dominate a bar. With one-unit bins starting at $0$, the bins $[2,3),[4,5),[6,7)$ each contain two values and appear equally prominent; that apparent pattern depends on grouping the repeated integers together. Moving the bin origin also changes membership without changing the observations. For density estimation, this is a form of the bias-variance tradeoff: averaging over broad bins loses local detail, while narrow bins give more variable density estimates across repeated statistical samples. ^[ref] A useful practice is to compare a few plausible widths and offsets, keeping common edges when comparing groups. Automatic bin selectors are starting points, rather than guarantees of a unique correct picture. ^[ref]

A histogram retains only bin membership. Its density estimate spreads each bin's fraction uniformly across that bin, so our last bar assigns density throughout $[8,22]$ even though the only toy values there are $8$ and $20$. It does not recover within-bin positions or observation order, and a visually convincing shape does not establish a population probability distribution. A box plot provides a complementary compact summary of quartiles; neither graphic replaces the original observations.

The birthday problem asks for the probability that at least two members of a group share a birthday. Use the explicit model of independent random variables taking $m$ labels with a uniform distribution on a finite set. Let $m\ge 1$ and $n\ge 0$ be integers. For calendar examples below, stipulate $m=365$ equally likely labels and omit leap days. For $0\le n\le m$, the probability of at least one matching pair isAt $n=0$, the empty product is one and $P_{0,m}=0$. At $n>m$, a repeat is certain by the pigeonhole principle. Count the complement to obtain the formula: the first label has $m$ choices, then a distinct second label has $m-1$, and so forth, out of $m^n$ equally likely label sequences.^[ref]

For the stipulated calendar, $P_{22,365}\approx 47.57\%$ and $P_{23,365}\approx 50.73\%$. Only 23 people suffice to put the probability above one half. There are $\left(\genfrac{}{}{0ex}{}{23}{2}\right)=253$ possible pairs, each with matching probability $1/365$. The many comparison opportunities explain why a small group can contain a repeat.

The distinction also matters for hash collisions. If independently generated outputs of a hash function are uniformly distributed over $m$ possible labels, the same counting formula governs repeats. The birthday square-root scale explains why looking for any repeated output requires far fewer labels than trying to match one predetermined output. The uniform-independent model is an assumption about those outputs; the birthday problem does not itself establish that a particular implementation meets it.

In a group of total size $n\ge 1$, designate one member and ask whether any of the other $n-1$ labels equals that member's label. Conditional on the designated label, all the other draws avoid it with probability $1-1/m$. Independence givesFor $n=23$, this is approximately $5.86\%$, while the probability of some matching pair is approximately $50.73\%$. The designated-member question examines $22$ pairs; the any-pair question examines all $253$. A repeated label elsewhere in the group does not require a match with its designated member.

The birthday collision count counts equal-label unordered pairs. Write it as a sum of indicator random variables:This uses linearity of expectation; mutual independence of the pair indicators is unnecessary. For 23 uniform calendar labels, $\mathbb{E}[C]=253/365\approx 0.69315$, whereas $\mathbb{P}(C\ge 1)\approx 0.50730$. A configuration with three equal labels contributes three pairs, so counting pairs and recording whether any pair exists are different random variables.

Treating every pair event as mutually independent would give the wrong exact formula. For three draws, $X_1=X_2$ and $X_1=X_3$ force $X_2=X_3$. The probability that all three equalities hold is $1/m^2$, while the product of their individual probabilities is $1/m^3$ for $m>1$. This distinguishes independence of the original label draws from mutual independence of events built from overlapping draws.

The union bound gives $P_{n,m}\le \min (1,n(n-1)/(2m))$. Using $1-x\le e^{-x}$ on the distinct-label product givesThe lower bound reaches at least one half when $n(n-1)/(2m)\ge \log 2$. The simple approximation $n\approx \sqrt{2m\log 2}$ predicts a group size near 23 for $m=365$. It is the quadratic number of pairs that makes the square root appear.

For a precise limiting statement, let $m\to \infty$ and $n/\sqrt{m}\to c\ge 0$. Expanding the logarithm of the complement yieldsOn this scale the remainder tends to zero. Thus a fixed substantial collision probability appears after order $\sqrt{m}$ draws. For $m=2^b$ uniform output labels, the corresponding scale is order $2^{b/2}$; matching one specified birthday instead involves a number of draws of order $m$.

Suppose two independent random variables have the same categorical distribution, with label probabilities $p_1,\dots ,p_m$. The disjoint equal-label events giveFor the original chosen probabilities $(1/2,1/4,1/8,1/8)$, this is $11/32=0.34375$, rather than the uniform four-label value $1/4$. The Cauchy-Schwarz inequality gives $1=({\sum }_i{p}_i{)}^2\le m{\sum }_i{p}_i^2$, so uniform probabilities minimize this two-draw collision probability. This pair formula alone does not supply the entire group collision probability for a nonuniform model.

Actuarial statistics uses probability theory and statistical inference to quantify insurance claims, premiums, and the uncertainty of future losses. Aggregate claims model describes portfolios, reinsurance redistributes losses, and the Bühlmann model combines experience with population information.

A policyholder's future insurance premiums depend on a finite discount level, updated according to whether a claim is submitted in each year. With independent yearly accidents and repair costs, and a claim decision depending only on the current level and cost, the level is a Discrete-time Markov chain. Accident probability and submitted-claim probability differ when the policyholder pays small repair costs without claiming.

In a no claims discount system with discounts $0,\alpha ,\beta$, a reported claim resets next year's discount to zero. With no later accidents, the next two premiums after reporting are $C+(1-\alpha )C$. At the zero level, not reporting instead gives $(1-\alpha )C+(1-\beta )C$; at either other level it gives $2(1-\beta )C$. Their differences give the displayed repair-cost thresholds. A fully reimbursed loss above its threshold should be reported under this undiscounted two-year rule. Current premiums cancel. Different thresholds generally produce state-dependent submitted-claim probabilities, even with a common accident mechanism.

Let a policyholder have one fixed exponential distribution intensity $\Lambda$ of rate $\nu$, with conditionally independent annual Poisson distribution counts. Its one-year marginal count is geometric, but averaging that count does not make the discount process a Markov chain. In the three-level rule, start at zero discount. At time three, the histories $0,\alpha ,\beta ,\beta$ and $0,0,\alpha ,\beta$ both end at the same top level. The first means three no-claim years, giving next no-claim probability $(\nu +3)/(\nu +4)$. The second means a positive count followed by two zero counts, giving $(\nu +2)/(\nu +4)$: divide the integrals of $(1-e^{-\lambda })e^{-(\nu +3)\lambda }$ and $(1-e^{-\lambda })e^{-(\nu +2)\lambda }$. These are distinct, so the Markov property fails when only the current discount is recorded. Conditioning on the intensity, or recording its Bayesian posterior, restores the required predictive information.

Suppose annual claim counts are independent with geometric distribution mass $p{q}^j$, $j\ge 0$. In a no claims discount system with discounts $0,\alpha ,\beta$, a claim returns either of the first two levels to zero; at the top, one claim sends the level to $\alpha$ and two or more send it to zero. The transition matrix is $\left(\begin{array}{ccc}q& p& 0\\ q& 0& p\\ q^2& pq& p\end{array}\right)$. Solving $\pi P=\pi$ gives the displayed stationary distribution. With full premium $c$, the stationary expected value of the premium is $c[1-(\alpha pq+\beta p^2)/(1-p^{2}q)]$. Independent annual sampling is essential; a persistent random intensity does not justify this marginal transition matrix.

For a three-level birth-death chain whose down-or-stay claim probabilities are $p_1,p_2,p_3$, with each in $(0,1)$, detailed balance is ${\pi }_1(1-p_1)={\pi }_2{p}_2$ and ${\pi }_2(1-p_2)={\pi }_3{p}_3$. Normalizing gives the displayed stationary distribution. The finite chain is irreducible and has endpoint holding probabilities, so it is aperiodic and its level proportions converge to this distribution from any initial level.

Let $b$ be the undiscounted annual premium and $d_i$ the discount fraction at level $i$. Compare the next two levels $i_r^0$ after no current claim with $i_r^1$ after a current claim, assuming no later accidents. Their total additional premium is the displayed threshold. A repair cost above it makes a claim advantageous under this two-year rule. For discounts $0,\alpha ,\beta$ and one-level moves with reflecting endpoints, the thresholds at levels one, two, and three are respectively $b\beta$, $b(2\beta -\alpha )$, and $b(\beta -\alpha )$. The premium scale must be specified or fixed as the monetary unit.

For a no claims discount system with discounts $0,\alpha ,\beta$, assume a policyholder starts at $\beta$, the first claim lowers next year's discount to $\alpha$, and a second claim in the same year lowers it to zero. With no further losses and undiscounted future premiums, claiming for the first loss costs $c(\beta -\alpha )$ in extra premium. Once that claim is submitted, the second claim adds $c\alpha$ next year and $c(\beta -\alpha )$ the year after, a total $c\beta$. Thus the second decision uses the incremental threshold, not the cost of both claims relative to no claim. For independent unit-rate exponential distribution losses the probabilities are $e^{-c(\beta -\alpha )}$ and $e^{-c\beta }$, respectively; the latter remains the conditional probability given submission of the first claim.

The expected value premium principle charges the expected value of the covered loss plus a proportional margin. For a nonnegative random variable $L$ with finite expected value, and a fixed dimensionless loading $\theta \ge 0$, it setsHere $\theta =0$ gives the expected claim cost alone; $\theta >0$ gives a positive margin when $\mathbb{E}[L]>0$. The loading is a fraction of the expected value, not of the largest possible claim size. Specify the covered period and use a consistent monetary valuation date. The rule itself does not determine $\theta$. ^[ref].

Consider an explicitly toy one-year policy: it pays £1,000 with probability $p=0.1$, and zero otherwise. Write $L=1000B$, where $B$ has the Bernoulli distribution with parameter $p$. The expected value is $1000(0.1)=\pounds 100$. A loading $\theta =0.2$ adds £20, making the premium £120. Neither the £100 expected value nor the £120 premium is a possible realized claim: the actual payment is £0 or £1,000. The first diagram separates these quantities. All numerical examples and both diagrams here are exact consequences of invented assumptions, not measured insurance data.

Why pay £120 for a loss whose expected value is only £100? Conditional on the insurer meeting its obligation, the policyholder replaces an uncertain £1,000 bill with a known £120 payment. The insurer can combine many such policies, so a few claims are funded by premiums from the whole pool. Winkel develops this motivation and the importance of independence in §15.2 of ^[ref].

To see the pooling mechanism precisely, suppose there are $n$ toy policies with mutually independent Bernoulli random variables $B_1,\dots ,B_n$, each of parameter $0.1$. Every claim size is £1,000, with at most one claim per policy. The aggregate claims model isUsing linearity of expectation and addition of variances for independent random variables, the expected value and standard deviation of this binomial distribution aggregate areThus 100 policies have expected aggregate claims £10,000 and aggregate standard deviation £3,000; 10,000 policies have expected aggregate claims £1,000,000 and aggregate standard deviation £30,000. Total uncertainty grows, but much more slowly than the expected value. Per policy, $S_n/n$ has expected value £100 and standard deviation $\pounds 300/\sqrt{n}$. Equivalently, the aggregate coefficient of variation, its standard deviation divided by its positive expected value, isIt falls from $300\%$ for one policy to $30\%$ for 100 policies and $3\%$ for 10,000. This is the useful beauty of independent pooling: an erratic individual cost becomes a comparatively stable average cost. The law of large numbers describes the corresponding convergence of the average toward its expected value.

At the same fixed loading $\theta =0.2$, total premium is $\Pi (S_n)=120n$, with expected surplus before other outgoings $120n-100n=20n$. The loading margin divided by the aggregate standard deviation is $20n/(300\sqrt{n})=\sqrt{n}/15$. This ratio rises with $n$ under our independence assumption; it is not a specified loss probability. For $n=100$, premiums total £12,000, but 13 claims cost £13,000. The one-year probability that claims exceed collected premiums is exactlyA positive loading therefore leaves about a $19.8\%$ chance of a one-year claims deficit in this toy pool. A claims deficit need not mean insolvency if capital is available. This calculation assumes premiums are collected before claims and ignores expenses and investment income; it is not an ultimate ruin probability.

The expected value premium principle also has a revealing limitation. With a common fixed $\theta$, linearity of expectation gives $\Pi (L_1+L_2)=\Pi (L_1)+\Pi (L_2)$ even without independence. Pooling therefore does not automatically reduce the premium per policy under this rule, and two losses with the same expected value receive the same premium regardless of their variance or large-loss probabilities. Any decision to change the loading after pooling requires information beyond the formula.

Dependence can remove the stability benefit. For arbitrary square-integrable losses, the aggregate variance includes their covariances:In the extreme toy common-shock model, every policy has $L_i=1000B$ for the same Bernoulli random variable $B$. Each policy still has expected value £100, so total premium is still $120n$, but $S_n=1000nB$ has standard deviation $300n$ and coefficient of variation $3$ for every $n$. With probability $0.1$, all policies claim together and losses exceed premiums by $880n$. The second diagram compares this model with independent pooling. Real shared events can invalidate independence; Winkel uses flood claims as an example in §15.2 of Actuarial Science II. The common-shock curve is an illustrative limiting model, not a fit to flood data.

The earlier $n^{-1/2}$ relative-variability calculation needs finite variance and the stated model of independent and identically distributed random variables. A finite expected value alone suffices to define the premium but does not imply finite variance. The rule makes no promise of solvency: losses, available capital and the timing of payments must be modelled separately. In the classical risk model, claim arrival rate $\lambda$ and mean claim size $\mu$ give premium rate $c=(1+\theta )\lambda \mu$; relative safety loading is then $\theta$. That model tracks capital through time and asks for ultimate ruin probability, a different question from calculating an expected margin. ^[ref]. Under quota share reinsurance, the reinsurer can apply the expected value premium principle to its ceded fraction $(1-\alpha )S$ using its own loading, which need not equal the direct insurer's loading.

A credibility estimate combines a risk's individual experience with population information. In the basic equal-exposure form, the credibility factor $Z$ weights the observed sample mean, while $1-Z$ weights a prior or collective expected value. The Bühlmann credibility premium is the best affine estimate under mean squared error; an exact Bayesian posterior mean can also have this form, as in the natural conjugate credibility identity.

Bayesian credibility estimates a risk's conditional mean by its Bayesian posterior expected value under squared-error loss. When that posterior mean is an affine combination of exposure-weighted individual experience and a prior mean, the coefficient of the experience is its credibility factor. Exact Bayesian credibility is distinct from merely selecting the best affine estimator when the true posterior mean is nonlinear.

Conditionally independent normal distribution observations of mean $\theta$ and known variance ${\sigma }_1^2$, with normal prior mean $\mu$ and variance ${\sigma }_2^2$, give normal-normal conjugacy with known observation variance. Their posterior mean has the displayed exact credibility estimate form. It minimizes posterior squared-error loss and predicts the conditional mean of the next observation.

For conditionally independent annual averages $X_j\mid \theta \sim N(\theta ,v/m_j)$ and prior $\theta \sim N(\mu ,{\sigma }^2)$, the posterior mean is $Z{\overline{X}}_w+(1-Z)\mu$, with ${\overline{X}}_w={\sum }_j{m}_j{X}_j/W$. Completing the quadratic likelihood gives posterior precision ${\sigma }^{-2}+W/v$. This is an exact Bayesian credibility version of the Bühlmann–Straub credibility estimate. Increasing the between-risk prior variance increases the experience weight; increasing process variance decreases it. The next total pure premium is $m_{n+1}$ times the estimated per-life mean.

For a common gamma distribution (shape $\alpha$, rate $\beta$) frequency prior distribution and conditionally independent Poisson distribution yearly counts of means $w_j\Theta$, observed average amounts $y_j$ determine counts $k_j=w_j{y}_j/c_j$. The posterior distribution shape is $\alpha +{\sum }_j{k}_j$ and rate is $\beta +{\sum }_j{w}_j$. The future per-policy mean is the display, equal to credibility factor $W/(W+\beta )$ times ${\sum }_j(w_j/W)(c_{n+1}/c_j)y_j$ plus the complementary weight times $c_{n+1}\alpha /\beta$. The data must lie on their count lattice; impossible noninteger recovered counts have zero likelihood.

// Destination: probability-theory.bigb

For a common success chance with beta prior and conditionally independent binomial observations of sizes $m_i$, put $M={\sum }_i{m}_i$ and $s={\sum }_i{x}_i$. Beta-binomial conjugacy gives posterior mean $(\alpha +s)/(\alpha +\beta +M)$, equal to $Z(s/M)+(1-Z)\alpha /(\alpha +\beta )$. Multiplying by the next exposure predicts its expected count. The factor tends to one as exposure increases and to zero as prior concentration increases at fixed prior mean.

Given a latent risk parameter $\Theta$, observations are independent and identically distributed random variables with conditional expectation $m(\Theta )$ and conditional variance $v(\Theta )$. The Bühlmann credibility premium estimates $m(\Theta )$ by an affine function of past observations under mean squared error. Its population parameters are $m=\mathbb{E}m(\Theta )$, expected process variance $v=\mathbb{E}v(\Theta )$, and variance of hypothetical means $a=\mathrm{Var}(m(\Theta ))$.

Conditional on a common latent variable $\Theta$, annual averages $X_j$ are independent, have common conditional expectation $\mu (\Theta )$, and have conditional variance ${\sigma }^2(\Theta )/m_j$ for known positive exposures $m_j$. Its structural parameters are $m_0=\mathbb{E}\mu (\Theta )$, expected process variance $v=\mathbb{E}{\sigma }^2(\Theta )$ and variance of hypothetical means $a=\mathrm{Var}(\mu (\Theta ))$. It generalizes the equal-exposure Bühlmann model.

For total exposure $W={\sum }_j{m}_j$ and weighted average ${\overline{X}}_w={\sum }_j{m}_j{X}_j/W$, the best affine mean squared error estimate of the conditional mean is $Z{\overline{X}}_w+(1-Z)m_0$, where $Z=Wa/(Wa+v)$. If the coefficients sum to $B$, their error is $a(1-B{)}^2+v{\sum }_j{b}_j^2/m_j$. Cauchy-Schwarz inequality minimizes the latter term at $b_j=B{m}_j/W$, and a one-dimensional quadratic minimization then gives $B=Z$.

With a Poisson-gamma conjugacy with unequal exposures model, $m_0=v=\alpha /\beta$ and $a=\alpha /{\beta }^2$, so the Bühlmann–Straub credibility factor is $W/(W+\beta )$. The credibility estimate simplifies to $(\alpha +{\sum }_j{Y}_j)/(\beta +W)$, exactly the posterior mean and hence the Bayes estimator under squared error loss. The agreement is exact because that posterior mean is already affine in the exposure-weighted data.

The experience weight is $Z=Wa/(Wa+v)$, where $W$ is total observed exposure, $a$ the variance of hypothetical means, and $v$ the expected process variance. With positive $a$, it is $W/(W+v/a)$. More exposure raises the weight; larger process noise lowers it. If $a=0$, the unknown conditional mean is constant and one can set $Z=0$.

The Bühlmann credibility premium is the best affine mean squared error estimate of a risk’s conditional claim expected value. In the Bühlmann model it is $Z\overline{X}+(1-Z)m$, where $Z=na/(na+v)$. The linear least-squares projection equations use $\mathrm{Var}(X_j)=a+v$ and $\mathrm{Cov}(X_i,X_j)=a$ for distinct years.

For independent comparable risks observed over $n\ge 2$ years, estimate the expected process variance $v$ by the average within-risk sample variance. The sample variance $B$ of the risk means estimates $a+v/n$, where $a$ is the variance of hypothetical means. Thus $B-\hat{v}/n$ is unbiased for $a$; truncating at zero produces a practical nonnegative estimate but introduces bias. Use the grand mean for $m$ and plug these estimates into the Bühlmann credibility premium $\hat{Z}{\overline{X}}_s+(1-\hat{Z})\hat{m}$. At least two risks are needed for $B$; a single year cannot estimate within-risk variance this way.

Let a fixed policy intensity have prior density $3{\theta }^{-4}$ on $\theta >1$, and let $n$ annual claim counts be conditionally independent with Poisson distribution mean $\theta$. The expected process variance is $3/2$ and the variance of hypothetical means is $3/4$, so the Bühlmann credibility factor is $n/(n+2)$. A total count $k$ gives the displayed credibility premium. By contrast, the Bayesian posterior density is proportional to ${\theta }^{k-4}{e}^{-n\theta }$ on $\theta >1$. Its mean is the ratio ${\int }_1^{\infty }{\theta }^{k-3}{e}^{-n\theta }d\theta /{\int }_1^{\infty }{\theta }^{k-4}{e}^{-n\theta }d\theta$. For $n=2,k=5$, that Bayesian credibility estimate is $5/3$, whereas the affine estimate is $2$.

In the Gamma prior for an inverse-gamma scale model, assume $k>2$. The variance of hypothetical means is $a=\alpha /[{\lambda }^2(k-1{)}^2]$, and the expected process variance is $v=\alpha (\alpha +1)/[{\lambda }^2(k-1{)}^2(k-2)]$. The Bühlmann credibility factor is $Z=n/(n+v/a)$, giving the displayed expression. The optimal affine credibility estimate is $(1-Z)\alpha /[\lambda (k-1)]+Z\overline{X}$. Its fixed-weight arithmetic-mean form differs from the reciprocal-based posterior mean.

With fixed structural parameters and $K=v/a$, the Bühlmann credibility estimate has form $({\sum }_{i=1}^n{x}_i+K{m}_0)/(n+K)$. The next observation updates it by the displayed rule. Consequently the estimate decreases exactly when the new observation is less than the previous credibility estimate, with strict equality handled separately.

The credibility factor in the Bühlmann model is the weight $Z=na/(na+v)$ placed on the sample mean after $n$ observations. The remaining weight is placed on the population expected value. More experience or larger variance of hypothetical means increases $Z$; larger expected process variance decreases it.

The variance of hypothetical means in the Bühlmann model is the between-risk variance of the conditional claim expected value. It creates the shared covariance $a$ between different years of the same risk and determines how much individual experience should influence the premium.

The expected process variance in the Bühlmann model averages within-risk conditional variance over the population of risks. It measures variation that repeated observations can average out and contributes $v/n$ to the variance of the sample mean.

The classical risk model has surplus $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, where $N_t$ is a Poisson process of rate $\lambda$, independent claim sizes are positive, and premiums flow in at constant rate $c$. The relative safety loading is $\rho =c/(\lambda \mathbb{E}{X}_1)-1$. Ruin is the first time surplus becomes negative.

In a classical risk model, $T_j$ is the time of the $j$th claim. The displayed probability restricts ruin to the first $n$ claims and increases to the ultimate ruin probability. Conditioning on the first arrival and severity gives ${\psi }_{n+1}(u)=\mathbb{E}[{\overline{F}}_X(u+c{T}_1)+{\int }_0^{u+c{T}_1}{\psi }_n(u+c{T}_1-x)f_X(x)dx]$, with ${\psi }_0=0$. For exponential distribution claims of mean $\mu$ and relative safety loading $\rho$, put $d=1+\rho$ and $h=d+1$. Then ${\psi }_1(u)=e^{-u/\mu }/h$ and ${\psi }_2(u)=e^{-u/\mu }(h^{-1}+(u/\mu )h^{-2}+d{h}^{-3})$. The recursion follows because a surviving first claim leaves fresh independent future arrivals and claims. The formulas follow by integrating the rate-$\lambda$ first-arrival density; the claim-arrival rate cancels with $c=d\lambda \mu$.

Starting from capital $u$, ruin occurs exactly when the aggregate loss exceeds $u$. Positive relative safety loading and the strong law of large numbers make $S(t)-ct$ tend to minus infinity, so its supremum is finite almost surely. For exponential distribution claims of rate $\alpha$, $L$ has an atom $1-\lambda /(\alpha c)$ at zero, and, conditional on $L>0$, has exponential distribution of rate $\alpha -\lambda /c$. The ultimate ruin probability determines this mixture directly.

For a nonnegative claim size $X$ of finite positive mean $\mu$, its equilibrium claim-size distribution has this density and survival function ${\overline{F}}_I(x)={\mu }^{-1}{\int }_x^{\infty }{\overline{F}}_X(t)dt$. The tail integral formula for moments normalizes the density. Its moment-generating function is $(M_X(r)-1)/(\mu r)$ wherever finite, so at an adjustment coefficient it equals one plus the safety loading.

A nonnegative lifetime or claim law is new better than used in expectation if its mean residual life at every age with positive survival is at most its original mean. The displayed equivalent inequality says that its equilibrium claim-size distribution is no larger in stochastic order than the original law. For an Erlang distribution of shape two and rate one, the two survival functions are $(1+x/2)e^{-x}$ and $(1+x)e^{-x}$, so the condition holds.