0
$\begingroup$

I've been working on Durrett Exercise 5.7.8 without any avail: Let $S_n$ be the total assets of an insurance company at the end of year $n$. In year $n$, premiums totaling $c>0$ are received and claims $\zeta_n\sim N(\mu,\sigma^2)$ are paid where $\mu<c$. To be precise, if $\xi_n= c - \zeta_n$ then $S_n = S_{n-1} + \xi_n$. The company is ruined if its assets drop to 0 or less. Show that if $S_0 > 0$ is nonrandom, then

$P($ruin$)\leq \exp(-2(c-\mu)S_0/\sigma^2)$.

The way I've thought about this problem is that if $\tau$ is the time of bankruptcy, we are looking for $P(\tau < \infty)$. Obviously, Wald's Identity isn't very useful.

If I define $X_n = S_n - n(c-\mu)$, then $X_n$ is a martingale. However, I can't find any theorems regarding the stopping times of martingales or the probability of them being infinite. Durrett's theorems mostly involve expectations of $\tau$, which in our case are clearly infinite.

$\endgroup$
  • $\begingroup$ To make $e^{-\tau S_n}$ a martingale, find a positive root of $E(e^{-\tau \xi})=1$. Since $S_n\to\infty$, conclude that $e^{-\tau S_0}=P(ruin)E(e^{-\tau R})$ where $R\le 0$ are your assets upon ruin. From here you get $P(ruin)\le e^{-\tau S_0}$. If you knew that $\xi\ge -A$ (not the case here) then you could also an easy lower bound for $P(ruin)$. $\endgroup$ – A.S. Feb 24 '16 at 4:23
1
$\begingroup$

Note that $$\mathbb E\left[e^{-\left(\frac{2(c-\mu)}{\sigma^2}\right)\xi_1}\right] = e^{-\left(\frac{2(c-\mu)}{\sigma^2}\right)(c-\mu) -\frac12 \left(\frac{2(c-\mu)}{\sigma^2}\right)^2\sigma^2}=e^0=1.$$ Set $\theta=\frac{-2(c-\mu)}{\sigma^2}$ and $X_n=S_0-S_n$, then $\{e^{-\theta X_n}\}$ is a martingale. Let $B>S_0$ and define $$N=\inf\{n>0: X_n>S_0\text{ or } X_n<-B\}. $$ Then by optional stopping, $$\mathbb E\left[e^{-\theta X_N}\right] = 1, $$ and so $$1 = \mathbb E\left[e^{-\theta X_N}\mid X_N>S_0\right]\mathbb P(X_N>S_0) + \mathbb E\left[e^{-\theta X_N}\mid X_N<-B\right]\mathbb P(X_N<-B).$$ Since $-\theta>0$, we have $$1\geqslant e^{-\theta S_0}\mathbb P(X_N>S_0),$$ and letting $B\to\infty$ yields $$\mathbb P(\mathrm{ruin})\leqslant e^{-\left(\frac{2(c-\mu)}{\sigma^2}\right)S_0}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.