Actuarial statistics

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.

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.

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.

For positive claims with finite variance and $c\le \lambda \mu$, the classical risk model has ruin probability one from any finite capital. At claim times, surplus increments are independent copies of $cT-X$ with mean $c/\lambda -\mu$. Negative mean sends their partial sums to minus infinity by the strong law of large numbers. At zero mean, the increments have finite nonzero variance. For every fixed $K$, the central limit theorem gives limiting probability $1/2$ of a partial sum below $-K$. The probability of unboundedness below is therefore at least $1/2$; as a tail event it has probability zero or one by the Kolmogorov zero-one law, and hence one.

Independent positively loaded portfolios with zero initial capitals have separate joint survival probability ${\prod }_i(1-{\lambda }_i{\mu }_i/c_i)$. Pool their claims and premium incomes to obtain zero-capital merged survival $1-({\sum }_i{\lambda }_i{\mu }_i)/({\sum }_i{c}_i)$. This is the premium-weighted average of the individual survival probabilities and is at least their product. The event of aggregate solvency permits transfers of surplus between the original portfolios.

The ultimate survival probability satisfies $\phi (u)=\phi (0)+(\lambda /c){\int }_0^u\phi (u-x)(1-F(x))dx$. Condition on the first arrival of the Poisson process, differentiate the resulting exponentially weighted integral, and integrate the convolution derivative equation from zero. Tonelli theorem converts the claim-density convolution to the tail convolution. The kernel mass is $\lambda \mu /c<1$, so this is a defective renewal equation.

In a classical risk model, before the first claim at time $t$, available capital is $u+ct$. A claim of size $x\le u+ct$ leaves future survival probability $\phi (u+ct-x)$ by the Markov property. Integrating over the independent first-arrival exponential distribution and claim density gives $\phi (u)={\int }_0^{\infty }\lambda e^{-\lambda t}{\int }_0^{u+ct}\phi (u+ct-x)f(x)dxdt$.

In a classical risk model with positive relative safety loading, zero initial capital has ultimate survival probability $1-\lambda \mu /c$, where $\lambda$ is the claim-arrival rate, $\mu$ the claim expected value and $c$ the premium rate. It depends on the claim law through its mean alone. It is not the event of never receiving a claim: premium accumulates between arrivals.

The adjustment coefficient is a positive root of $\lambda (M_X(r)-1)=cr$ in the classical risk model. When the moment-generating function is finite at $R$, the process $\exp (R[{\sum }_{j=1}^{N_t}{X}_j-ct])$ is a continuous-time martingale. It yields the Lundberg inequality and, under the relevant tilted integrability, the Cramér–Lundberg ruin asymptotic.

If each claim includes an independent expense $A$, replace its payment law by the convolution of independent random variables $X+A$. Its moment-generating function is $M_X(r)M_A(r)$ and its expected value is $\mathbb{E}X+\mathbb{E}A$. Keeping the relative safety loading $\theta$ fixed therefore changes the premium rate as well. The new coefficient solves $M_X(R)M_A(R)-1=(1+\theta )(\mathbb{E}X+\mathbb{E}A)R$, within the common finite-transform domain.

For positive claims with finite nonzero expected value $\mu$, the function $(M_X(r)-1)/r$ is continuous and strictly increasing on the positive finite-transform domain, starting at $\mu$. If $M_X$ diverges at a finite upper endpoint, or is finite for all positive arguments, the secant slope tends to infinity. In the latter case use $M_X(r)\ge \mathbb{P}(X\ge a)e^{ar}$ for some $a>0$ of positive tail probability. Every target $(1+\theta )\mu$ with positive relative safety loading therefore has exactly one positive root.

For a classical risk model with adjustment coefficient $R>0$, the ultimate ruin probability from capital $u\ge 0$ satisfies $\psi (u)\le e^{-Ru}$. Stop the exponential continuous-time martingale at ruin or a finite horizon, bound its value on the ruin event, and then increase the horizon.

In the classical risk model with positive relative safety loading $\rho$ and adjustment coefficient $R$, tilting the ruin defective renewal equation gives a proper renewal equation. The key renewal theorem yields $e^{Ru}\psi (u)\to \rho /[R{\int }_0^{\infty }x{e}^{Rx}{f}_I(x)dx]$. The constant is positive if the denominator is finite and zero if it is infinite; the claim-size density provides the nonarithmetic hypothesis.

If the expected claim outflow per unit time is $\lambda \mu$, the relative safety loading is $\rho =c/(\lambda \mu )-1$. Positive loading means expected premium income exceeds expected claim outflow. This is the net profit condition in the classical risk model.

Reinsurance transfers part of an insurer’s claim liability to another insurer. If aggregate claims are $S$, a retained payout $g(S)$ with $0\le g(S)\le S$ leaves the reinsurer with $S-g(S)$. Quota share reinsurance retains a fixed fraction, whereas aggregate stop loss reinsurance retains losses only up to a fixed aggregate threshold.

With aggregate retention $M\ge 0$, the direct insurer pays $\min (S,M)$ and the reinsurer pays $(S-M{)}_{+}$, where $+$ denotes the positive part. The threshold applies to the whole annual loss; applying a threshold to individual claims is a different contract.

If $S$ has exponential distribution with expected value $\mu$, the tail integral formula for moments gives $\mathbb{E}\min (S,M)=\mu (1-e^{-m})$ and $\mathrm{Var}(\min (S,M))={\mu }^2(1-2m{e}^{-m}-e^{-2m})$, where $m=M/\mu$. At matching retained expected value, the excess variance under quota share reinsurance is $2{\mu }^2{e}^{-m}(m-1+e^{-m})\ge 0$.

Among retained payouts $g(S)$ with $0\le g(x)\le x$ and the same expected value as $\min (S,M)$, aggregate stop loss reinsurance minimizes the variance. Pointwise $(g(x)-M{)}^2\ge (\min (x,M)-M{)}^2$, and subtracting the identical squared distance of their common expected value from $M$ proves the claim. Equality requires equal payouts almost surely.

Quota share reinsurance retains a fixed proportion $\alpha \in [0,1]$ of each claim. The annual retained aggregate is $\alpha S$, whose expected value and variance are $\alpha \mathbb{E}S$ and ${\alpha }^2\mathrm{Var}(S)$.

An aggregate claims model separates the number $N$ of claims from their sizes $X_j$, with total $S={\sum }_{j=1}^N{X}_j$ and the empty sum equal to zero. Under independent and identically distributed random variables for claim sizes, independent of $N$, conditional expectation gives tractable moments and transforms.

Independent claim Poisson processes of rates ${\lambda }_i$ merge into a Poisson process of rate $\lambda ={\sum }_i{\lambda }_i$, by the Superposition theorem for Poisson point processes. The merged claim law is a mixture distribution of the individual claim laws with weights ${\lambda }_i/\lambda$. Equivalently, multiplication of the individual compound-Poisson transforms produces $\exp (\lambda [{\sum }_i({\lambda }_i/\lambda )M_i(r)-1])$.

For independent claim sizes with common expected value $m$ and variance $v$, independent of the nonnegative integer count $N$, the aggregate satisfies $\mathbb{E}S=m\mathbb{E}N$ and $\mathrm{Var}(S)=v\mathbb{E}N+m^2\mathrm{Var}(N)$. Its moment-generating function is $G_N(M_X(t))$ wherever finite. These identities follow from the law of total expectation and law of total variance.

The law of a sum of independent identically distributed claims with an independent Poisson distribution count of parameter $\lambda$. If the claim moment-generating function is $M_X$, the aggregate transform is $\exp (\lambda (M_X(r)-1))$ wherever finite. It has an atom at zero when the claims are positive. This is the fixed-time law of a Compound Poisson process.

For positive claim sizes, a random sum of independent claims is zero exactly when its count is zero. Its law is a mixture distribution of a zero atom of mass $p=\mathbb{P}(N=0)$ and, with weight $1-p$, a random sum whose count has the zero-truncated claim-count distribution. An independent Bernoulli random variable multiplying that positive component gives the same law. This is distinct from arbitrarily inserting additional zeros into an otherwise unchanged count law.

If $\Lambda$ has gamma distribution with shape $2$ and rate $p/q$, $N\mid \Lambda$ has Poisson distribution with intensity $\Lambda$, and the independent claim sizes have exponential distribution of mean $\mu$, where $p+q=1$, the aggregate law is $p^2{\delta }_0+2pq\mathrm{Exp}(p/\mu )+q^2\mathrm{Gamma}(2,\text{rate }p/\mu )$. This follows by expanding its moment-generating function as $[p+qp/(p-\mu t){]}^2$.