Gambler's Ruin Problem with Simple Solution - Different Approach

Question

What is the probability that symmetric simple random walk starting at the origin reaches $−1$ before it reaches $9$? Briefly explain your answer.

Solution: This is $p = 1$ gambler's ruin with states $0, 1, . . . , 10$ but the states have been relabelled $−1, 0, . . . , 9$. The answer is $\frac{9}{10}$ because state 0 is one tenth of the way from $−1$ to $9$.

Unfortunately, I used a different and rather clumsy way to approach this question.

My attempt was not successful but I was wondering if there is anyone who can tell me how to go further with it and arrive at the solution.

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My attempt: I drew it out like birth and death process with 11 states. With $-1$ and $9$ having the probability of returning to itself as 1 (absorbing state). Let's $P(X_i)$ be probability to reach $-1$ before reaching $9$, starting from $i$.

Then I listed out $$P(X_0) = \frac{1}{2} + \frac{1}{2}P(X_1)$$ $$P(X_i) = \frac{1}{2}\left(P(X_{i-1}) + P(X_{i+1})\right) \ \ \text{ for } i \in\{1,2,3,...7\}$$

$$P(X_8) = \frac{1}{2}P(X_7) + 0$$

There are 9 equations and 9 unknowns. We should be able to calculate the result. However, how to go from here?

Answer 1

Hint: The equations $$ P\left(X_{i+1}\right) - 2P\left(X_i\right) + P\left(X_{i-1}\right)=0\ \ \ \mathrm{for}\ i=1,2,\dots,7 $$ constitute a second-order linear recurrence whose general solution is $\ P\left(X_n\right) = a + bn\ $ for $\ n=0,1, \dots, 8\ $, and some constants $\ a\ $ and $\ b\ $. The boundary conditions, $\ P\left(X_0\right) = \frac{1}{2} + \frac{1}{2}P\left(X_1\right)\ $, and $\ P\left(X_8\right) = \frac{1}{2}P\left(X_7\right)\ $, give you two linear equations to solve for the values of $\ a\ $ and $\ b\ $.

Answer 2

The equation $$P(X_i) = \frac{1}{2}\left(P(X_{i-1}) + P(X_{i+1})\right) \: \: \: \; (*)$$ holds for $i \in\{0,2,3,...8\}$ since $P(X_{-1})=1$ and $P(X_9)=0$. Write $$Q_i=P(X_i)-P(X_{i-1})$$ for $i=0,1,\ldots,9$. Then by doubling (*) and subtracting $P(X_{i-1}) + P(X_{i})$ from both sides, we get that $$Q_i = Q_{i+1} $$ holds for $i \in\{0,2,3,...8\}$. Since $\sum_{i=0}^9 Q_i=-1$ we conclude that $Q_i=-1/10$ for all $i$. In particular $P(X_0)=1+Q_0=9/10$.