Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 39 / 3

In the classical risk model, the surplus with initial capital $u\ge 0$ iswhere $N(t)$ is a Poisson process of rate $\lambda$, and the nonnegative independent claim sizes, of mean $\mu >0$, are independent of that process. Positive relative safety loading $\rho$ makes expected premium income exceed expected claim outflow. With ruin time $T=\inf \{t\ge 0:U(t)<0\}$, the ultimate ruin probability isThe adjustment coefficient is the positive root $R$ in the finite domain of the claim moment-generating function satisfyingThe zero root is excluded by definition.

Under the stated transform-divergence assumption this positive root exists and is unique. Indeed $F(r)=M_X(r)-1-(1+\rho )\mu r$ has $F(0)=0$, derivative $F^{\prime }(0)=-\rho \mu <0$, and strictly positive second derivative on its positive domain. At a finite endpoint the diverging transform makes $F$ positive eventually. At an infinite endpoint, choose $a>0$ with $\mathbb{P}(X\ge a)>0$; the bound $M_X(r)\ge \mathbb{P}(X\ge a)e^{ar}$ makes it dominate the linear term. Strict convexity then gives exactly one positive crossing. This is the secant-slope existence criterion for an adjustment coefficient.

The Lundberg inequality isFor a short justification, the compound-Poisson transform and the defining equation make $e^{-RU(t)}$ a nonnegative martingale starting at $e^{-Ru}$. Stop it at $T\wedge t$, a bounded stopping time. On $\{T\le t\}$ its value is greater than one. The optional stopping theorem therefore gives $\mathbb{P}(T\le t)\le e^{-Ru}$; let $t$ increase to infinity. No expectation identity at an unbounded ruin time is required.

For an exponential distribution with mean $\mu$, the claim moment-generating function is $M_I(r)=(1-\mu r{)}^{-1}$ for $r<1/\mu$. The calculated adjustment coefficient must solveSubtract one, divide by the nonzero factor $\mu R_I$, and obtain $(1-\mu R_I{)}^{-1}=1+\rho$. ThusIt belongs to $(0,1/\mu )$ for every $\rho >0$, as required by the transform domain. The Poisson intensity cancels because the premium rate is specified using the same relative loading and mean claim cost.

Normalizing the true density givesa gamma distribution of shape $1/2$ and rate $\beta$. Its mean is $1/(2\beta )=\mu$, and the Gamma integral givesPut $x=R/\beta$. Since $\mu \beta =1/2$, the positive adjustment coefficient satisfiesBoth sides are positive in this domain. Squaring, multiplying by $1-x$, and dividing out the zero root $x=0$ yieldsThe discriminant is $(1+\rho {)}^2[(\rho -3{)}^2+16\rho ]=(1+\rho {)}^2(\rho +9)(\rho +1)$. The negative constant term makes one root positive and the other negative. The polynomial is negative at zero and has value four at one, so the positive root lies in $(0,1)$ and is not an extraneous solution introduced by squaring. Therefore the adjustment coefficient for shape-one-half gamma claims is

To compare it with the exponential calculation, write $R_I=2\beta \rho /(1+\rho )$. Their difference issince $9(1+\rho {)}^2-(\rho +9)(\rho +1)=8\rho (1+\rho )>0$. Thus the two proposed Lundberg inequality bounds are ordered asOnly ${\psi }_{\mathrm{true}}(u)\le e^{-Ru}$ is justified for the true claim law. The exponential calculation gives ${\psi }_{\mathrm{exponential}}(u)\le e^{-R_{I}u}$ for a different model, not a valid guaranteed bound on the true ruin probability.

This exponential misspecification of shape-one-half gamma claims is optimistic: the true variance is $1/(2{\beta }^2)=2{\mu }^2$, whereas the assumed exponential variance is ${\mu }^2$. Its tail also decays at rate $\beta$, more slowly than the assumed rate $1/\mu =2\beta$. Equivalently, for $0<r<\beta$, the true transform $(1-2\mu r{)}^{-1/2}$ is larger than $(1-\mu r{)}^{-1}$, since $(1-\mu r{)}^2>1-2\mu r$. This stronger transform growth makes the true adjustment root smaller. Comparing upper bounds alone does not by itself prove an ordering of the two actual ruin probabilities. At $u=0$ both displayed bounds equal one.