Finite-Target Limits for Infinite-Capital Ruin

From a Finite Target to Infinite Capital

In a finite gambler’s ruin problem, the game ends when the gambler reaches either zero or an upper target $N$. With infinite capital, there is no upper boundary to stop the walk. One way to understand the infinite-capital ruin probability is therefore to solve the finite problem for each $N$, then let $N$ grow. This limit makes the effect of drift explicit: a walk with nonpositive drift is eventually ruined with probability one, while positive drift leaves a nonzero chance of escaping ruin.

Let $(X_n)_{n\ge0}$ be a Markov chain on $S=\{0,1,2,\ldots\}$, on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Time is discrete, and we use the natural filtration $\mathcal{F}_n=\sigma(X_0,\ldots,X_n)$. At each positive state $k$, the chain moves to $k+1$ with probability $p$ and to $k-1$ with probability $q=1-p$; state $0$ is absorbing. Thus the two transition probabilities from every positive state sum to $p+q=1$. Write $\mathbb{P}_i$ for probabilities when $X_0=i$. For $i\ge1$, define $T_0=\inf\{n\ge0:X_n=0\}$; for a target $N>i$, define $T_N=\inf\{n\ge0:X_n=N\}$.

The finite-target ruin probability is $h_i^{(N)}=\mathbb{P}_i(T_0<T_N)$. The strict inequality matters: the event is that ruin occurs before the upper target is reached. The boundary values are $h_0^{(N)}=1$ and $h_N^{(N)}=0$.

Theorem (Finite-target ruin formula): Suppose $N\ge2$ and $0<p,q<1$, with $p+q=1$. For $0\le i\le N$, the probability of hitting $0$ before $N$ is $$ h_i^{(N)}= \begin{cases} \dfrac{(q/p)^i-(q/p)^N}{1-(q/p)^N},&p\ne q,\\[6pt] \dfrac{N-i}{N},&p=q=\tfrac12. \end{cases} $$ In particular, the formula gives the specified boundary values at $i=0,N$.

Proof. First, absorption at one of the two boundaries occurs with probability one. From any interior state, a run of $N$ downward moves reaches $0$ within $N$ steps; its probability is at least $m^N>0$, where $m=\min(p,q)$. Conditional on being unabsorbed at the start of any block of $N$ steps, the Markov property gives the same lower bound for absorption during that block. Hence the probability of surviving $r$ such blocks is at most $(1-m^N)^r$, which tends to zero.

Conditioning on the first move from an interior state $k$ gives $$ h_k^{(N)}=p\,h_{k+1}^{(N)}+q\,h_{k-1}^{(N)},\qquad 1\le k\le N-1, $$ with $h_0^{(N)}=1$ and $h_N^{(N)}=0$. If $p\ne q$, set $r=q/p$. The constant sequence $1$ satisfies the recurrence, and so does $r^k$, since $$ p r^{k+1}+q r^{k-1}=r^k(pr+q/r)=r^k(q+p)=r^k. $$ The proposed expression $(r^k-r^N)/(1-r^N)$ is therefore a solution and has the required boundary values. If $p=q=\tfrac12$, the recurrence is $h_k=(h_{k+1}+h_{k-1})/2$; the linear sequence $(N-k)/N$ satisfies it and has the same boundaries.

To check uniqueness, subtract any two solutions and call their difference $u_k$. It has zero boundary values and satisfies $u_k=p u_{k+1}+q u_{k-1}$. If $\max_{0\le k\le N}|u_k|=M>0$, the maximum absolute value is attained at an interior state $k$. The recurrence and $p,q>0$ imply $$ M=|u_k|\le p|u_{k+1}|+q|u_{k-1}|\le pM+qM=M. $$ Equality forces both neighboring values to have absolute value $M$ and the same sign as $u_k$. Repeating this implication toward state $0$ forces $|u_0|=M$, contradicting $u_0=0$. Thus $M=0$, proving uniqueness and hence the formula. $\square$

Why the Finite-Target Events Increase

To pass from the formula to the infinite-capital problem, it is important to compare the events themselves, not just their probabilities. For $N>i$, let $A_N=\{T_0<T_N\}$. If a path hits $0$ before $N$, it must also hit $0$ before $N+1$: to reach $N+1$ from below, the path would first have to visit $N$. Thus $A_N\subseteq A_{N+1}$.

Also, every path that is eventually ruined belongs to some $A_N$. If $T_0<\infty$, only finitely many states are visited before time $T_0$, so their maximum is finite. Choose $N$ larger than this maximum and larger than $i$. Then $T_0<T_N$. Conversely, membership in any $A_N$ implies $T_0<\infty$. Therefore $$ \bigcup_{N>i} A_N=\{T_0<\infty\}. $$ By continuity of probability for an increasing sequence of events, $\mathbb{P}_i(T_0<\infty)=\lim_{N\to\infty}h_i^{(N)}$.

Theorem (Infinite-capital ruin probability): For an initial state $i\ge1$, the probability of eventual ruin is $$ \mathbb{P}_i(T_0<\infty)= \begin{cases} 1,&0\le p\le q,\\[2pt] (q/p)^i,&p>q. \end{cases} $$ Here $p+q=1$. The cases $p=0$ and $q=0$ are included: the walk moves down deterministically in the first case and up deterministically in the second.

Proof. For $0<p,q<1$ with $p\ne q$, put $r=q/p$. If $p<q$, then $r>1$. Dividing the numerator and denominator of the finite formula by $r^N$ gives $$ h_i^{(N)}=\frac{r^{i-N}-1}{r^{-N}-1}\longrightarrow\frac{-1}{-1}=1. $$ If $p>q$, then $0<r<1$, so $r^N\to0$ directly in the formula and $h_i^{(N)}\to r^i$. If $p=q=\tfrac12$, then $(N-i)/N=1-i/N\to1$. The event argument above identifies each of these limits with $\mathbb{P}_i(T_0<\infty)$. When $p=0$, the walk reaches $0$ after $i$ downward steps, so the probability is $1$; when $q=0$, it moves strictly upward and never reaches $0$, so the probability is $0$. These agree with the stated cases. $\square$

Worked Limits in the Three Drift Regimes

Worked Example: Unfavorable Drift

Take $p=\tfrac13$, $q=\tfrac23$, initial capital $i=2$, and upper target $N=5$. The move probabilities sum to $\tfrac13+\tfrac23=1$, as required. Here $r=q/p=2$, so $$ h_2^{(5)}=\frac{2^2-2^5}{1-2^5} =\frac{4-32}{1-32} =\frac{-28}{-31} =\frac{28}{31}. $$ For a recurrence check, the formula gives $h_1^{(5)}=\tfrac{30}{31}$, $h_2^{(5)}=\tfrac{28}{31}$, and $h_3^{(5)}=\tfrac{24}{31}$. At state $2$, $$ \tfrac13 h_3^{(5)}+\tfrac23 h_1^{(5)} =\tfrac13\cdot\tfrac{24}{31}+\tfrac23\cdot\tfrac{30}{31} =\tfrac{24+60}{93} =\tfrac{28}{31} =h_2^{(5)}. $$ As $N$ grows, $r>1$, so the finite-target probabilities tend to $1$. This does not mean every finite game has ruin probability one: at $N=5$, the probability is $\tfrac{28}{31}$. It is the limit as the upper target recedes that equals one.

Worked Example: Favorable Drift

Take $p=\tfrac23$, $q=\tfrac13$, initial capital $i=2$, and target $N=6$. Again $\tfrac23+\tfrac13=1$. Now $r=q/p=\tfrac12$, and the finite probability is $$ h_2^{(6)} =\frac{(1/2)^2-(1/2)^6}{1-(1/2)^6} =\frac{1/4-1/64}{1-1/64} =\frac{15/64}{63/64} =\frac{15}{63} =\frac{5}{21}. $$ The formula also gives $h_1^{(6)}=\tfrac{31}{63}$ and $h_3^{(6)}=\tfrac19$. Substitution into the first-step equation at state $2$ yields $$ \tfrac23 h_3^{(6)}+\tfrac13 h_1^{(6)} =\tfrac23\cdot\tfrac19+\tfrac13\cdot\tfrac{31}{63} =\tfrac{14}{189}+\tfrac{31}{189} =\tfrac{45}{189} =\tfrac{5}{21}. $$ As $N\to\infty$, the finite formula tends to $r^i=(\tfrac12)^2=\tfrac14$. The limiting probability is positive, but smaller than the finite-target probability $\tfrac{5}{21}$; the latter includes paths that eventually hit $0$ before $6$, whereas some paths counted as eventual ruin may reach $6$ first.

Worked Example: Fair Play

Let $p=q=\tfrac12$, start at $i=2$, and set $N=5$. The probabilities sum to $1$. The fair-game formula gives $$ h_2^{(5)}=\frac{5-2}{5}=\frac35. $$ Indeed, the values at states $1,2,3$ are respectively $\tfrac45,\tfrac35,\tfrac25$, and the recurrence check at $2$ is $$ \tfrac12\cdot\tfrac25+\tfrac12\cdot\tfrac45 =\tfrac15+\tfrac25 =\tfrac35. $$ For general $N>2$, $h_2^{(N)}=(N-2)/N=1-2/N$, which tends to $1$. Fair play has zero average one-step change, but that does not leave a positive probability of escaping all ruin: the increasing finite-target events show that eventual ruin still has probability one.

Interpreting the Limit and Avoiding a Ratio Error

The ratio $q/p$ determines the limiting behavior. When downward moves are more likely, the finite formula approaches one because $q/p>1$. When the game is fair, the linear formula also approaches one, though at a different rate. When upward moves are more likely, $q/p<1$, and the probability of ruin from capital $i$ is $(q/p)^i$. In this last case, increasing the starting capital decreases the eventual ruin probability geometrically.

Common mistake. The ratio in the biased formula is $q/p$, not $p/q$. For the favorable example above, using $p/q=2$ would give $$ \frac{2^2-2^6}{1-2^6} =\frac{4-64}{1-64} =\frac{60}{63} =\frac{20}{21}, $$ which is not the finite-target ruin probability. The correction is to use $q/p=\tfrac12$, giving $\tfrac{5}{21}$. One direct check is the first-step recurrence: the correctly computed neighboring values give $\tfrac23\cdot\tfrac19+\tfrac13\cdot\tfrac{31}{63}=\tfrac{5}{21}$, while the erroneous ratio does not supply the solution to that recurrence.

A second point to keep straight is that a finite-target probability and an infinite-capital probability answer different questions. Reaching the upper target ends the finite game without ruin; in the infinite-capital model, reaching a particular finite level does not end the process. The increasing-event argument is what justifies taking the limit: it shows that eventual ruin is exactly the union of the events of ruin before some sufficiently large finite target. The limit would not follow merely from observing that the formula has a limit; the event identification is essential.

Check Your Understanding

For each calculation, use the strict event $T_0<T_N$ and verify the result from the finite formula or its recurrence.

For $p=\tfrac14$, $q=\tfrac34$, $i=1$, and $N=4$, calculate the finite-target ruin probability and its limit as $N$ grows.
For $p=\tfrac34$, $q=\tfrac14$, and $i=3$, find the eventual ruin probability.
For a fair game starting at $i=4$, write the finite-target ruin probability for $N>4$, then take its limit.
Explain why $\{T_0<T_N\}\subseteq\{T_0<T_{N+1}\}$.
In the biased formula, what fails if $q/p$ is replaced by $p/q$? Give a recurrence-based check.

Finite-Target Limits for Infinite-Capital Ruin

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