Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 38 / 3

Let $C_t$ be the cumulative claim amount and $U_t=u+ct-C_t$ the surplus in the classical risk model. The supplied adjustment coefficient satisfies $\lambda (M_X(R)-1)=cR$. The Compound Poisson process has independent increments, and the random-sum transform identity givesIt follows that $Z_t=\exp (R(C_t-ct))$ is a nonnegative martingale with $Z_0=1$: conditional on the past, the multiplicative increment has expected value one. This is the exponential surplus martingale.

Let $\tau =\inf \{t\ge 0:U_t<0\}$ be the ruin stopping time. Surplus decreases only at claim arrivals, so on $\{\tau \le t\}$ the deficit at ruin gives $C_{\tau }-c\tau >u$ and $Z_{\tau }>e^{Ru}$. The optional stopping theorem at the bounded stopping time $\tau \wedge t$ yieldsAs $t$ increases without bound, the events $\{\tau \le t\}$ increase to eventual ruin. Hence the ultimate ruin probability satisfies the Lundberg inequalityThe proof uses only bounded-time stopping; it does not presume that the martingale can be stopped with equality of expected values at an unbounded ruin time.

For a unit-rate exponential distribution, $\mu =1$ and $M_X(r)=1/(1-r)$ for $r<1$. The adjustment coefficient equation isIts nonzero root isIt lies in $(0,1)$ for $\theta >0$, so the moment-generating function is finite there and the Lundberg inequality applies.

The severity is an equal-weight mixture distribution of rate-$2$ and rate-$2/3$ exponential distributions. Its expected value isThe moment-generating function, finite exactly for $r<2/3$, isCanceling the nonzero $r$ in the adjustment coefficient equation and multiplying by the positive denominator on $0<r<2/3$ givesThus the required polynomial isSince $P(0)=4\theta >0$, $P(2/3)=-2<0$, and the leading coefficient is positive, one root lies in $(0,2/3)$ and the other exceeds $2/3$. Only the smaller root is in the moment-generating function domain. ConsequentlyThe rationalized expression avoids subtraction of nearly equal numbers when $\theta$ is small.

The erroneously chosen same-mean exponential distribution gives $R_{\exp }=\theta /(1+\theta )$. Direct substitution shows $P(R_{\exp })=-\theta <0$, so $R_{\exp }$ lies strictly between the two polynomial roots, and in particular $R_{\mathrm{mix}}<R_{\exp }$. Thereforewhile both bounds equal one at $u=0$. The exponential misspecification produces a smaller claimed upper bound. The Lundberg inequality certifies $e^{-R_{\mathrm{mix}}u}$ for the actual mixture; it does not certify the smaller expression computed from the wrong severity law.