Expected Ruin Time With Infinite Capital

The Walk and Its Ruin Time

In the infinite-capital model, a gambler begins with a fixed positive integer fortune and has no upper boundary. Each play changes the fortune by one unit: a win adds one, and a loss subtracts one. The direction of the drift determines whether the expected time to reach zero is finite. In particular, the fair case is not obtained by substituting zero drift into the formula for negative drift.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space carrying independent increments \((Y_n)_{n\ge1}\), with \(\mathbb{P}(Y_n=1)=p\) and \(\mathbb{P}(Y_n=-1)=q=1-p\). Assume first that \(0<p<1\). For initial fortune \(x\in\{1,2,\ldots\}\), define \(X_0=x\) and, until ruin, \(X_n=x+\sum_{k=1}^nY_k\). We use the absorbed version of the process: once it reaches zero, it stays there. Its state space is \(S=\{0,1,2,\ldots\}\), its time index is discrete \(n\ge0\), and its filtration is \(\mathcal{F}_n=\sigma(Y_1,\ldots,Y_n)\), with \(\mathcal{F}_0=\{\varnothing,\Omega\}\). The ruin time is \(T=\inf\{n\ge0:X_n=0\}\), where the infimum of the empty set is \(\infty\). We write \(\mathbb{P}_x,\mathbb{E}_x\) when the starting fortune is \(x\).

For every positive state \(i\), the transition probabilities are \(p_{i,i+1}=p\) and \(p_{i,i-1}=q\); at zero, \(p_{00}=1\). Each positive-state row sums to \(p+q=1\), and the row at zero sums to \(1\). The expected time is interpreted as an extended expectation, so it equals \(\infty\) if \(T=\infty\) with positive probability or if \(T\) is finite almost surely but has no finite mean.

Negative Drift: Deriving the Finite Mean

Suppose \(p<q\), so one increment has mean \(\mathbb{E}[Y_k]=p-q<0\). We first stop at the bounded time \(T\wedge n\). For each \(k\le n\), the event \(\{T\ge k\}\) is determined by the first \(k-1\) increments and is therefore in \(\mathcal{F}_{k-1}\). Independence gives \(\mathbb{E}_x[Y_k\mathbf{1}_{\{T\ge k\}}]=(p-q)\mathbb{P}_x(T\ge k)\). Since the stopped fortune is nonnegative, these identities give a bound on the truncated expected time.

Proof. For every integer \(n\ge1\), the fortune at the stopped time is \(X_{T\wedge n}=x+\sum_{k=1}^nY_k\mathbf{1}_{\{T\ge k\}}\). Also, \(T\wedge n=\sum_{k=1}^n\mathbf{1}_{\{T\ge k\}}\). Taking expectations and using the conditional mean of each increment yields \(\mathbb{E}_x[X_{T\wedge n}]=x+(p-q)\mathbb{E}_x[T\wedge n]\). Because \(X_{T\wedge n}\ge0\) and \(q-p>0\), it follows that \(\mathbb{E}_x[T\wedge n]\le x/(q-p)\). As \(n\) increases, \(T\wedge n\) increases to \(T\); monotone convergence therefore gives \(\mathbb{E}_x[T]\le x/(q-p)<\infty\). In particular \(T<\infty\) almost surely.

To obtain equality, we may now sum the increments all the way to ruin. The sum is integrable: its absolute value is at most \(T\), which has finite expectation. Moreover, \(\mathbb{E}_x[Y_k\mathbf{1}_{\{T\ge k\}}]=(p-q)\mathbb{P}_x(T\ge k)\) for every \(k\), by the same independence argument. Summing these expectations is justified because \(\sum_{k\ge1}\mathbb{E}_x[|Y_k|\mathbf{1}_{\{T\ge k\}}]=\mathbb{E}_x[T]<\infty\). At ruin \(X_T=0\), so \(0=x+(p-q)\mathbb{E}_x[T]\). Rearranging gives \(\mathbb{E}_x[T]=x/(q-p)\). \(\square\)

Zero Drift: Ruin Does Not Have a Finite Mean

For a fair walk, the drift is zero, but this does not make the expected ruin time zero or finite. The key distinction is that a stopped-sum expectation argument is immediately valid at bounded times, whereas using it directly at \(T\) would require control of the unbounded stopping time. The following contradiction proves that the mean cannot be finite.

Proof. Let \(p=q=\tfrac12\), and suppose for contradiction that \(\mathbb{E}_x[T]<\infty\). Then \(T<\infty\) almost surely, and \(X_T=0\). As before, \(\{T\ge k\}\in\mathcal{F}_{k-1}\), so independence and zero mean give \(\mathbb{E}_x[Y_k\mathbf{1}_{\{T\ge k\}}]=0\) for each \(k\). The stopped sum is absolutely integrable because \(\sum_{k\ge1}\mathbb{E}_x[|Y_k|\mathbf{1}_{\{T\ge k\}}]=\mathbb{E}_x[T]<\infty\). Thus its expectation is the sum of the expectations, which is zero. But pathwise, \(0=X_T=x+\sum_{k\ge1}Y_k\mathbf{1}_{\{T\ge k\}}\). Taking expectations would give \(0=x\), contradicting \(x\ge1\). Hence \(\mathbb{E}_x[T]=\infty\). \(\square\)

Positive Drift: Positive Probability of No Ruin

When \(p>q\), the walk has positive drift. The earlier ruin-probability result for the infinite-capital model gives \(\mathbb{P}_x(T<\infty)=(q/p)^x\). Since \(0<q/p<1\), this probability is strictly less than one. The expected time is consequently infinite, for a different reason from the fair case: there is positive probability that \(T=\infty\). More explicitly, for every integer \(m\ge1\), \(\mathbb{E}_x[T]\ge m\mathbb{P}_x(T=\infty)\). The right side grows without bound as \(m\) increases.

Interpreting the Regimes and Avoiding a Common Error

The three cases separate two questions: whether ruin occurs and how long it takes on average. Negative drift gives both almost-sure ruin and a finite mean. In the fair case, the mean is infinite even though ruin occurs almost surely, as established in the earlier material. With positive drift, ruin has probability less than one, which alone forces the unconditional expected ruin time to be infinite. These are different mechanisms for an infinite expectation.

Common mistake. Applying the negative-drift formula \(\mathbb{E}_x[T]=x/(q-p)\) at \(p=q=\tfrac12\) produces \(x/0\), and treating that expression as a valid derivation is incorrect. For example, at \(x=2\), the formula has denominator \(\tfrac12-\tfrac12=0\); it does not give a finite value or prove anything by itself. The formula was derived under the strict hypothesis \(p<q\), where a bounded-stopping argument first proves \(\mathbb{E}_x[T]<\infty\). In the fair case, the separate contradiction proof shows that this finiteness hypothesis fails.

There are two boundary cases worth recording. If the initial fortune is \(x=0\), then \(T=0\) and \(\mathbb{E}_0[T]=0\), regardless of \(p\) and \(q\). If \(x\ge1\) and \(p=0\), then every play is a loss and ruin occurs after exactly \(x\) plays; the negative-drift formula still gives \(x/(1-0)=x\). If \(p=1\) and \(x\ge1\), the fortune increases forever, so \(T=\infty\) and \(\mathbb{E}_x[T]=\infty\).