One-Sided Ruin and the Limits of Optional Stopping

The One-Sided Ruin Setup

In a finite gambler’s-ruin problem, the walk is stopped when it reaches either of two barriers. With only a lower barrier, the walk may escape upward forever. This changes the question: even if the ruin time is almost surely finite, it does not follow that optional stopping can be applied at that unbounded time. The key is to analyze the stopped process at bounded times first, then justify any limiting step separately.

Let $(\Omega,\mathcal{F},\mathbb{P})$ support independent increments $(Y_n)_{n\ge1}$ with $\mathbb{P}(Y_n=1)=p$ and $\mathbb{P}(Y_n=-1)=q=1-p$, where $0<p<1$. For a fixed integer $i\ge1$, define $X_0=i$ and $X_n=i+\sum_{k=1}^n Y_k$. The state space is $\mathbb{Z}$, and the time index is discrete, $n\ge0$. This is a Markov chain: from any state $j\in\mathbb{Z}$, it moves to $j+1$ with probability $p$ and to $j-1$ with probability $q$; each transition row sums to $p+q=1$. Use the filtration $\mathcal{F}_n=\sigma(Y_1,\ldots,Y_n)$, with $\mathcal{F}_0=\{\varnothing,\Omega\}$. The one-sided ruin time is the stopping time $T_0=\inf\{n\ge0:X_n=0\}$, where $\inf\varnothing=\infty$. Because the steps have size one, a path starting above zero cannot cross below zero without first reaching zero.

Theorem: For this walk starting at $i\ge1$, if $p\le q$, then $\mathbb{P}_i(T_0<\infty)=1$. If $p>q$, then $$ \mathbb{P}_i(T_0<\infty)=\left(\frac{q}{p}\right)^i. $$ The probability in the second case is strictly between $0$ and $1$. These conclusions concern the probability of eventual ruin; they do not assert that the expected ruin time is finite.

The three drift regimes require different arguments. When $p<q$, the walk tends downward. When $p=q$, the drift is zero, so that argument does not apply; finite upper barriers are useful instead. When $p>q$, ruin is not certain, and a positive exponential martingale gives the probability of the event that ruin does occur.

Ruin with Zero or Negative Drift

For $p<q$, the standard strong law of large numbers for independent, identically distributed bounded random variables gives $X_n/n\to p-q<0$ almost surely. The proof of that standard law is not needed here. In particular, $X_n\to-\infty$ almost surely. Since the walk starts at $i\ge1$ and changes by one at each step, it must reach zero before it can become negative. Thus $T_0<\infty$ almost surely.

For $p=q=\tfrac12$, the strong law only gives $X_n/n\to0$; it does not imply that the walk hits zero. To prove ruin in this case, place an upper barrier at an integer $N>i$ and write $T_N=\inf\{n\ge0:X_n=N\}$ and $\sigma_N=T_0\wedge T_N$. The walk exits the finite interval $\{1,\ldots,N-1\}$ almost surely: from any state in that interval, $N$ consecutive downward steps, an event of probability $2^{-N}$, force an exit within the next $N$ steps. Conditional on still being in the interval at the start of a block of $N$ steps, the probability of surviving that block is at most $1-2^{-N}$. Therefore the probability of remaining in the interval for $kN$ steps is at most $(1-2^{-N})^k$, which tends to zero.

For the fair walk, $(X_n)$ is a martingale: its one-step conditional mean increment is $\tfrac12-\tfrac12=0$. For each deterministic $n$, $\sigma_N\wedge n$ is a bounded stopping time, so bounded optional stopping gives $\mathbb{E}_i[X_{\sigma_N\wedge n}]=i$. Until exit, and at exit, $X_{\sigma_N\wedge n}$ lies in $[0,N]$. Since $\sigma_N<\infty$ almost surely, $X_{\sigma_N\wedge n}\to X_{\sigma_N}$ almost surely. The variables are bounded by $N$, so dominated convergence gives $\mathbb{E}_i[X_{\sigma_N}]=i$. At exit the walk is either at $0$ or at $N$; hence $\mathbb{P}_i(T_N<T_0)=i/N$. If $T_0=\infty$, the walk must hit $N$ before zero, except on the event of never exiting the finite interval, which has probability zero. It follows that $\mathbb{P}_i(T_0=\infty)\le i/N$ for every $N>i$. Letting $N\to\infty$ proves that this probability is zero.

Worked Example: A Fair Walk from 2

Take $p=q=\tfrac12$ and $i=2$. Each transition row sums to $\tfrac12+\tfrac12=1$. For an upper barrier $N=8$, the fair-walk exit calculation gives $\mathbb{P}_2(T_8<T_0)=2/8=1/4$. For $N=20$, it gives $\mathbb{P}_2(T_{20}<T_0)=2/20=1/10$. If ruin never occurs, the walk must hit each such upper barrier before zero, so its probability is at most $1/4$, at most $1/10$, and, in general, at most $2/N$ for every $N>2$. Since $2/N\to0$, ruin occurs almost surely.

Check: At $N=8$, the exit position has expectation $8(1/4)+0(3/4)=2$, equal to the starting position. This agrees with the bounded-stopping calculation.

Positive Drift: A Martingale and a Justified Limit

Now suppose $p>q$, and put $r=q/p$, so $0<r<1$. Define $M_n=r^{X_n}$. Conditional on $\mathcal{F}_n$, the next step is upward with probability $p$ and downward with probability $q$, so $$ \mathbb{E}[M_{n+1}\mid\mathcal{F}_n] =r^{X_n}(pr+qr^{-1}) =r^{X_n}(q+p) =M_n. $$ Thus $(M_n)$ is a martingale. For every deterministic $n$, $T_0\wedge n$ is bounded. Bounded optional stopping therefore gives $\mathbb{E}_i[r^{X_{T_0\wedge n}}]=r^i$. Splitting according to whether ruin has occurred by time $n$ yields $$ r^i=\mathbb{P}_i(T_0\le n)+\mathbb{E}_i\!\left[r^{X_n}\mathbf{1}_{\{T_0>n\}}\right]. $$ To pass to the limit, the strong law gives $X_n/n\to p-q>0$ almost surely. On $\{T_0=\infty\}$, this implies $X_n\to\infty$, so $r^{X_n}\to0$. On $\{T_0<\infty\}$, the indicator $\mathbf{1}_{\{T_0>n\}}$ is eventually zero. The random variable inside the expectation is bounded between zero and one: on $\{T_0>n\}$, $X_n\ge1$, and $0<r<1$. Dominated convergence applies. Also, $\mathbb{P}_i(T_0\le n)\to\mathbb{P}_i(T_0<\infty)$, since the events increase to $\{T_0<\infty\}$. The displayed identity consequently gives $\mathbb{P}_i(T_0<\infty)=r^i$.

This argument does not apply optional stopping directly at $T_0$. It applies bounded optional stopping at $T_0\wedge n$, then uses a separately justified limit. The two steps matter: almost-sure finiteness of a stopping time alone does not establish that expectations of stopped variables converge to the expectation at that time.

Worked Example: Positive Drift from 2

Take $p=\tfrac23$, $q=\tfrac13$, and $i=2$. Each transition row sums to $\tfrac23+\tfrac13=1$, and $r=q/p=(\tfrac13)/(\tfrac23)=\tfrac12$. The theorem gives $$ \mathbb{P}_2(T_0<\infty)=r^2=\left(\frac12\right)^2=\frac14. $$ Thus $\mathbb{P}_2(T_0=\infty)=1-\tfrac14=\tfrac34$. In particular, the stopping time is not almost surely finite in this example.

Check by substitution: For $h_i=(\tfrac12)^i$, the one-step equation for $i\ge1$ is $h_i=\tfrac23 h_{i+1}+\tfrac13 h_{i-1}$, with $h_0=1$. Its right-hand side is $$ \frac23\left(\frac12\right)^{i+1}+\frac13\left(\frac12\right)^{i-1} =\left(\frac13+\frac23\right)\left(\frac12\right)^i =\left(\frac12\right)^i=h_i. $$ The candidate also tends to zero as $i\to\infty$, consistent with the exponential-martingale limit.

Worked Example: Negative Drift from 2

Take $p=\tfrac13$, $q=\tfrac23$, and $i=2$. Each transition row sums to $\tfrac13+\tfrac23=1$, and the mean increment is $\tfrac13-\tfrac23=-\tfrac13$. By the strong law, $X_n/n\to-\tfrac13$ almost surely. Hence $X_n\to-\infty$ almost surely, and the unit steps force the walk to visit zero first. Therefore $\mathbb{P}_2(T_0<\infty)=1$.

Check: This is consistent with the drift calculation: a path that never visited zero would remain positive for every $n$, which is incompatible with $X_n\to-\infty$. The positive-drift formula does not apply, because $q/p=2$ is not less than one.

Why Almost-Sure Stopping Is Not Enough

In the fair case, $(X_n)$ itself is a martingale and $T_0<\infty$ almost surely. It is tempting to write $\mathbb{E}_i[X_{T_0}]=i$ by optional stopping. But $X_{T_0}=0$ almost surely, so that conclusion would say $0=i$, a contradiction. The error is applying optional stopping at an unbounded time without a condition that justifies it.

Worked Example: A Failed Expectation Limit

Take the fair walk with $i=2$. Bounded optional stopping gives $\mathbb{E}_2[X_{T_0\wedge n}]=2$ for every $n$. Since $T_0<\infty$ almost surely, $X_{T_0\wedge n}\to X_{T_0}=0$ almost surely. The incorrect calculation is $$ \mathbb{E}_2[X_{T_0}] =\lim_{n\to\infty}\mathbb{E}_2[X_{T_0\wedge n}] =2. $$ Its first equality has not been justified. In fact, the limiting variable has expectation zero, whereas every approximating variable has expectation two.

Correction: Almost-sure convergence does not in general imply convergence of expectations. If the variables $X_{T_0\wedge n}$ were uniformly integrable, almost-sure convergence would imply convergence in $L^1$, and their expectations would converge to zero. Since all their expectations equal two, they are not uniformly integrable. The stopping time is almost surely finite, but that fact does not supply the missing expectation control.

The exponential-martingale argument for $p>q$ avoids this mistake for a different reason: the stopped exponential variables are bounded by one, and the limit is justified by dominated convergence. In the fair case, the stopped positions are not bounded uniformly in $n$, so the same domination is unavailable. These examples show why one must examine the particular stopped variables and their limiting behavior, rather than infer validity from the stopping time being finite almost surely.

There is also a useful distinction between a ruin probability and an expected duration. The calculations above determine whether ruin occurs and, in the positive-drift case, its probability. They do not establish that $\mathbb{E}_i[T_0]$ is finite. A stopping-time probability argument and an expectation-of-duration calculation are different questions and require their own justification.

Takeaway: For a one-sided nearest-neighbor walk, ruin is certain when $p\le q$, while for $p>q$ its probability is $(q/p)^i$. Derive unbounded-time conclusions by first stopping at $T_0\wedge n$, then proving that the relevant limit can pass through the expectation.

Check Your Understanding

For each question, identify which limit or stopping argument is justified before drawing a conclusion.

For $p=\tfrac34$, $q=\tfrac14$, and $i=2$, compute the probability of eventual ruin and the probability of never being ruined.
For a fair walk starting at $i=3$, what is the probability of hitting $N=9$ before zero? State why the walk exits $\{1,\ldots,8\}$ almost surely.
Explain why the strong law proves eventual ruin when $p<q$, but does not by itself prove eventual ruin when $p=q$.
In the positive-drift case, why is dominated convergence applicable to $r^{X_n}\mathbf{1}_{\{T_0>n\}}$?
For a fair walk starting at $i=1$, bounded optional stopping gives $\mathbb{E}_1[X_{T_0\wedge n}]=1$. Why does this not imply $\mathbb{E}_1[X_{T_0}]=1$?