Comparing Bounded and Unbounded Ruin Stopping Rules

Two Ruin Rules, Two Different Stopping Arguments

For a random walk, changing a stopping rule from “leave a finite interval” to “hit one boundary on an unbounded state space” can change whether an optional-stopping calculation is valid. With two finite barriers, the stopped position stays in a bounded interval. With only a lower boundary, the walk may wander arbitrarily far before ruin, and the stopped positions need not be uniformly integrable. The equations can look nearly identical, but the limit argument is not.

Work on a probability space \((\Omega,\mathcal F,\mathbb P)\) carrying independent increments \((\xi_n)_{n\ge1}\). Let \(\mathcal F_n=\sigma(\xi_1,\ldots,\xi_n)\), with \(\mathcal F_0\) containing only events of probability zero or one. For a starting integer \(x\), define \(X_0=x\) and \(X_n=x+\sum_{k=1}^n\xi_k\). In the fair walk, \(\xi_k=1\) and \(\xi_k=-1\), each with probability \(\tfrac12\), so \((X_n,\mathcal F_n)\) is a martingale. A stopping time \(\tau\) is a random time for which \(\{\tau\le n\}\in\mathcal F_n\) for every \(n\). The finite-barrier exit time and the one-sided ruin time are, respectively, \(\tau_{0,N}=\inf\{n\ge0:X_n\in\{0,N\}\}\) and \(\tau_0=\inf\{n\ge0:X_n=0\}\), with \(\inf\varnothing=\infty\).

Finite Barriers: The Stopped Walk Is Bounded

First consider the fair walk with integer starting position \(x\) satisfying \(0<x<N\). Each increment has mean zero, so \(X_n\) is a martingale. The exit time \(\tau_{0,N}\) is finite almost surely. To see this, from any interior state, \(N\) consecutive increments all equal to \(1\) would force exit through \(N\) within the next \(N\) steps; this event has probability \(2^{-N}\). Conditional on the past at the start of any block of \(N\) steps, its probability is still \(2^{-N}\). Thus the probability of remaining inside for \(m\) such blocks is at most \((1-2^{-N})^m\), which tends to zero.

Proof. For each deterministic \(n\), the time \(n\wedge\tau_{0,N}\) is bounded. Optional stopping for a martingale at a bounded stopping time therefore gives \(\mathbb E_x[X_{n\wedge\tau_{0,N}}]=x\). Almost surely, \(\tau_{0,N}<\infty\), so \(X_{n\wedge\tau_{0,N}}\to X_{\tau_{0,N}}\). These random variables all lie in \([0,N]\), and dominated convergence gives \(\mathbb E_x[X_{\tau_{0,N}}]=x\). The terminal position is \(N\) if the walk exits at the upper barrier and \(0\) otherwise. If \(u=\mathbb P_x(X_{\tau_{0,N}}=N)\), then \(x=Nu+0(1-u)\), so \(u=x/N\). \(\square\)

The essential point is not that \(\tau_{0,N}\) is bounded by a fixed deterministic time; it is not. The stopped values are bounded by the barriers, so their expectations pass to the almost-sure terminal limit by dominated convergence. Equivalently, the family of stopped values is uniformly integrable. This distinction is useful whenever applying optional stopping at an almost surely finite but not deterministically bounded time.

One-Sided Fair Ruin: The Limit Can Lose Expectation

Now keep the fair walk but remove the upper barrier. The position process is still a martingale, and \(\tau_0\wedge n\) is bounded for each \(n\), so optional stopping gives \(\mathbb E_x[X_{n\wedge\tau_0}]=x\). However, this identity does not by itself allow \(n\) to tend to infinity inside the expectation. The stopped positions are nonnegative but not uniformly bounded: if ruin has not occurred by time \(n\), the walk can be far above zero.

Proof. Fix an integer \(N>x\). The finite-barrier result gives \(\mathbb P_x(X_{\tau_{0,N}}=N)=x/N\). Also, exit from \(\{1,\ldots,N-1\}\) occurs almost surely: in each block of \(N\) steps, \(N\) consecutive upward increments force exit, with probability \(2^{-N}\) conditional on the past. If the walk never hits \(0\), it must therefore hit \(N\) before \(0\). Hence \(\mathbb P_x(\tau_0=\infty)\le x/N\) for every \(N>x\). Letting \(N\) increase without bound gives \(\mathbb P_x(\tau_0=\infty)=0\).

For every \(n\), bounded-time optional stopping yields \(\mathbb E_x[X_{n\wedge\tau_0}]=x\). Since \(\tau_0<\infty\) almost surely and the walk is stopped at zero, \(X_{n\wedge\tau_0}\to0\) almost surely. If this family were uniformly integrable, almost-sure convergence together with uniform integrability would imply convergence in \(L^1\), and therefore convergence of expectations to \(0\). That contradicts \(\mathbb E_x[X_{n\wedge\tau_0}]=x>0\) for every \(n\). Thus the family is not uniformly integrable, and passing the expectation through this limit is not justified. \(\square\)

One-Sided Biased Ruin: A Bounded Exponential Martingale

Failure of the position martingale in the fair one-sided example does not mean that every one-sided stopping calculation fails. For a biased walk, an exponential martingale can remain bounded after stopping at ruin. Let each increment equal \(1\) with probability \(p=\tfrac34\) and \(-1\) with probability \(q=\tfrac14\). The mean increment is \(p-q=\tfrac12\), so the position itself is not a martingale. Set \(r=q/p=\tfrac13\). Since \(p r+q/r=\tfrac34\cdot\tfrac13+\tfrac14\cdot3=\tfrac14+\tfrac34=1\), the process \(r^{X_n}\) is a martingale.

Proof. Define \(Y_n=r^{X_{n\wedge\tau_0}}\). For each \(n\), this is the exponential martingale stopped at the bounded time \(n\wedge\tau_0\), so \(\mathbb E_x[Y_n]=r^x\). Before ruin, \(X_k\ge1\); at ruin, \(X_{\tau_0}=0\). Since \(0<r<1\), it follows that \(0\le Y_n\le1\) for all \(n\), so dominated convergence applies.

On \(\{\tau_0<\infty\}\), \(Y_n\to1\). On \(\{\tau_0=\infty\}\), the strong law of large numbers for the independent increments gives \(X_n/n\to p-q=\tfrac12\) almost surely. Thus \(X_n\to\infty\) on this event and \(Y_n=r^{X_n}\to0\). Therefore \(Y_n\to\mathbf 1_{\{\tau_0<\infty\}}\) almost surely. Taking expectations by dominated convergence gives \(r^x=\mathbb P_x(\tau_0<\infty)\), as claimed. \(\square\)

What to Check Before Using Optional Stopping

These examples separate three questions that are easy to conflate: is the stopping time finite almost surely, is the stopped process bounded or uniformly integrable, and does the stopped process converge to the claimed terminal value? Almost-sure finiteness addresses only the last question’s event of eventual stopping. It does not ensure that expectations converge. In the finite-barrier fair walk, stopped positions are bounded by \(N\). In the one-sided fair walk, they are not uniformly integrable. In the one-sided biased example, the position can be unbounded, but the chosen exponential martingale, stopped at ruin, is bounded.

Common mistake. Applying optional stopping to \(X_{\tau_0}\) merely because \(\tau_0<\infty\) almost surely is not valid. For the fair walk from \(x=2\), the incorrect substitution gives \(\mathbb E_2[X_{\tau_0}]=2\), but direct evaluation gives \(X_{\tau_0}=0\) almost surely and hence expectation \(0\). The correction is to stop first at \(\tau_0\wedge n\), then check whether the resulting family permits passage to the limit. Here \(\mathbb E_2[X_{n\wedge\tau_0}]=2\), but the family is not uniformly integrable, so the expectations do not converge to the expectation of the almost-sure limit.