Suppose we are given a simple random walk starting in $0$, i.e. $(X_k)_{k\in\mathbb{N}}$ with $P[X_k=+1]=P[X_k=-1]=\frac{1}{2}$. What is the probability of hitting the level $a$ before hitting the level $b$, where we assume $b<0<a$ and $|a|\le |b|$. Let's define

$$T_a:=\inf\{n|S_n=a\}$$ and similarly $$ T_b:=\inf\{n|S_n=b\}$$ where $S_n:=\sum_{i=1}^nX_i$. Therefore we are interested in the probability $$P[T_a< T_b]$$ I think one needs the reflection principle at some point, but I'm not sure how it is exactly applied.


2 Answers 2


Hint: Since $(S_n)_{n \in \mathbb N}$ is a martingale, you can use the fact that $$\begin{align*}E[S_T]&=aP(S_T=a)+bP(S_T=b)=\\&=aP(S_T=a)+b(1-P(S_T=a))=(a-b)(P(S_T=a))+b=\\&=(a-b)P(T_a<T_b)+b\end{align*}$$ (where $S_T=\inf\{n|S_n \in \{a,b\}\}$) and the optional stopping theorem, i.e. that $$E[S_T]=E[S_0]=0$$ to deduce that $$P(S_T=a)=P(T_a<T_b)=\frac{-b}{a-b}$$

  • $\begingroup$ why one should take -b instead of +b in the first line of the equality? I notice that b<0 but i still don't understand why i shouldn't maintain the plus sign. $\endgroup$
    – Buddy_
    Oct 26, 2020 at 19:41
  • $\begingroup$ @Buddy_ You are right, it should be +b because b is already assumed to be negative. $\endgroup$
    – Jimmy R.
    Oct 27, 2020 at 7:07
  • 1
    $\begingroup$ i think that also your definition of $S_T$ is wrong. Maybe what you meant is the definition of T $\endgroup$
    – Buddy_
    Oct 27, 2020 at 8:30

@Jimmy R. Wondering how important is the original assumption: $b<0<a$ and especially $ \lvert a \rvert \leq \lvert b\rvert$. I didn't see where the above answer using this information.

I'm having a different assumption. What is the probability of hitting the level b before hitting the level a, which $b<0<a$ but $\lvert a \rvert \geq \lvert b\rvert$.

Here's the similar rewrite based on my assumption here.

$S_T=inf\{n|S_n \in \{a,b \} \}$

Since $S_n$ is a martingale, based on optimal stopping theorem, $E[S_T]=E[S_0]$ and $E[S_0]=0$.

$E[S_T]=a*P(S_T=a)+(-b)*P(S_T=b)=a[1-P(S_T=b)]+(-b)*P(S_T=b)=a-(a+b)*P(S_T=b)=a-(a+b)*(T_b < T_a)$

Hence, we can derive that $(T_b < T_a)=\frac{a}{a+b}$


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