Ruin Probability With Favorable Drift

Ruin with Favorable Drift

With unlimited capital, a gambler can continue playing after any win, so the finite-capital ruin formulas do not directly answer the question of whether ruin will ever occur. When a win is more likely than a loss, the probability of eventual ruin decreases as the initial fortune grows. To compute it, we use a first-step recurrence—but must also choose the solution that represents the probability. In this problem, a key subtlety is that boundedness alone does not make that choice.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((X_n)_{n\ge0}\) be the gambler’s fortune at discrete time \(n\), with state space \(S=\{0,1,2,\ldots\}\). State \(0\) is absorbing. From every \(i\ge1\), the fortune increases by \(1\) with probability \(p\) and decreases by \(1\) with probability \(q\), where \(p+q=1\) and \(p>q\ge0\). Thus every transition row sums to \(1\), including the row at \(0\), which assigns probability \(1\) to staying at \(0\). Let \(\mathcal{F}_n=\sigma(X_0,\ldots,X_n)\) be the natural filtration. Write \(\mathbb{P}_i\) for probability when \(X_0=i\), and define the ruin time \(T_0=\inf\{n\ge0:X_n=0\}\), with \(\inf\varnothing=\infty\).

For \(i\ge1\), condition on the first step. If the first move is a win, the new fortune is \(i+1\); if it is a loss, it is \(i-1\). The Markov property therefore gives \(h_i=p h_{i+1}+q h_{i-1}\). The recurrence describes the probability, but a boundary condition at large fortunes is also needed to identify its solution.

Proof. We first show that the actual ruin probabilities tend to zero. On the same probability space, let \(Y_1,Y_2,\ldots\) be the successive increments before absorption, with independent values \(+1\) and \(-1\) having probabilities \(p\) and \(q\). Consider the unrestricted walk \(i+\sum_{k=1}^nY_k\), continuing the increments even after it first reaches \(0\). The event of eventual ruin in the absorbed model is exactly the event that this unrestricted walk reaches \(0\). For a ruin at time \(n\), the unrestricted sum of increments must equal \(-i\).

Because \(p>q\), there is a \(\theta>0\) sufficiently small that \(a=p e^{-\theta}+q e^\theta<1\). Indeed, the function \(f(\theta)=p e^{-\theta}+q e^\theta\) is continuous, satisfies \(f(0)=1\), and has derivative \(f'(0)=q-p<0\); hence it is less than \(1\) for some positive \(\theta\). For each \(n\ge1\), Markov’s inequality applied to the nonnegative random variable \(e^{-\theta\sum_{k=1}^nY_k}\) gives $$ \mathbb{P}\left(\sum_{k=1}^nY_k\le-i\right) \le e^{-\theta i}\mathbb{E}\left[e^{-\theta\sum_{k=1}^nY_k}\right] =e^{-\theta i}a^n. $$ Here independence gives the last equality, since \(\mathbb{E}[e^{-\theta Y_k}]=a\). A union bound over possible ruin times now yields $$ 0\le h_i\le \sum_{n=1}^\infty e^{-\theta i}a^n =e^{-\theta i}\frac{a}{1-a}. $$ The right side tends to zero as \(i\to\infty\), proving the claimed boundary condition.

It remains to solve the recurrence with that boundary condition. First suppose \(q>0\), and set \(r=q/p\), so \(0<r<1\). For any sequence \((u_i)\) satisfying the recurrence, define \(d_i=u_i-u_{i-1}\) for \(i\ge1\). Rearranging the recurrence gives $$ p(u_{i+1}-u_i)=q(u_i-u_{i-1}),\qquad d_{i+1}=r d_i. $$ Consequently \(d_i=d_1r^{i-1}\), and summing the differences gives \(u_n=u_0+d_1(1-r^n)/(1-r)\). If \(u_0=1\) and \(u_n\to0\), taking the limit as \(n\to\infty\) forces \(1+d_1/(1-r)=0\), or \(d_1=-(1-r)\). Substitution then gives \(u_n=r^n\), proving uniqueness. This sequence satisfies the recurrence because \(p r+q/r=p(q/p)+q(p/q)=p+q=1\); equivalently, \(p r^{i+1}+q r^{i-1}=r^i\). It has \(u_0=1\) and tends to zero.

If \(q=0\), then \(p=1\) and the recurrence becomes \(u_i=u_{i+1}\) for \(i\ge1\). Thus all \(u_i\) for \(i\ge1\) are equal. The condition \(u_i\to0\) forces each of them to be zero, while \(u_0=1\). This is also the actual ruin probability: every play from a positive fortune is a win. In both cases the actual probabilities satisfy the recurrence and the boundary conditions, so uniqueness identifies them with the stated formula. \(\square\)

Using and Checking the Formula

Why Boundedness Alone Is Not Enough

For \(q>0\), the recurrence has two basic solutions: the constant sequence \(1\) and the sequence \((q/p)^i\). Both are bounded because \(0<q/p<1\). More generally, every recurrence solution with \(u_0=1\) can be written as $$ u_i=A+B\left(\frac{q}{p}\right)^i,\qquad A+B=1. $$ The condition \(u_0=1\) leaves many bounded choices. For instance, \(u_i=1\) is bounded and satisfies the recurrence at every \(i\ge1\), but it is not the ruin probability when \(q<p\): it predicts certain ruin from every positive fortune. The condition that selects the probabilistic answer is \(u_i\to0\), which we proved from the model itself.

Common mistake: concluding that the ruin probability is \(1\) because the constant sequence solves the first-step recurrence and has \(u_0=1\). This calculation checks the recurrence but omits the behavior at large fortunes. The correction is to establish the boundary condition \(h_i\to0\) and use it along with \(h_0=1\). For \(q>0\), these conditions give \(h_i=(q/p)^i\); for \(q=0\), they give \(h_i=0\) for every \(i\ge1\).

The result quantifies how favorable drift affects risk: the probability of eventual ruin is a geometric function of the initial fortune, with ratio \(q/p<1\). It is a probability of ever reaching zero, not an expected time to ruin and not a claim that ruin occurs with probability one. The assumption \(p>q\) matters: it is what allows an exponential bound tending to zero and makes the ratio \(q/p\) less than \(1\).