Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose ${X_t}$ is a random walk with mean zero. (either discrete or continuous time) Fix a time $T$. What is: $P[X_t < 0 \text{ for all } t \leq T]$?

In words, what's the probability the random walk from $0$ to $T$ has never been above $0$? Thank you.

share|improve this question

1 Answer 1

up vote 5 down vote accepted

Here is a solution for the discrete time, simple (steps are $\pm 1$), symmetric random walk $(X_t)$ that uses the reflection principle. You didn't specify your starting point, but presumably it is negative, say $-b$ for some $b>0$.

I'm going to consider the equivalent problem where the random walk starts at zero, and investigate when it hits $b>0$, that is, $T_b:= \inf (t>0: X_t =b )$. Then you should be able to use equation $(*)$ to find what you are looking for.

Let $t,b$ have opposite parity. Then $$P(X_t > b) = P(X_t>b \mid T_b < t) P(T_b < t)={1\over 2} P(T_b < t).$$ The conditional probability is exactly $1/2$ because $X_t=b$ is impossible. Given $T_b < t$, the sample paths satisfy $X_{T_b}=b$ at some time $T_b$ prior to $t$. The reflection principle shows that these paths divide equally into those with $X_t > b$ and those with $X_t < b$.

This gives us the formula $P(T_b < t)=2P(X_t > b)$ or $P(T_b\geq t)=P(-b\leq X_t\leq b)$. Since the random walk is symmetric around the origin, the random variables $T_b$ and $T_{-b}$ have the same distribution. Thus, we can combine the two cases as follows, taking into account that $X_t=\pm b$ is impossible, $$P(T_b\geq t)=P(-|b|< X_t < |b|),\quad b\in {\mathbb Z}\setminus \lbrace 0\rbrace ,\ t+b\mbox{ odd}.\tag{$\ast$} $$

For instance, in the special case where $b=1$, with the even time point $2t$, we get $$P(T_1\geq 2t)=P(-1< X_{2t} < 1)=P(X_{2t}=0)={2t\choose t}\left({1\over 2}\right)^{2t}.$$

share|improve this answer
    
This is such a brilliant solution. Thank you Byron. But what if the steps are normal (0,1), i.e. Xt+1 = Xt + et+1, where et+1 ~N(0,1)? And yes, b=0, and X0=0 as you guessed. I should have stated things clearer sorry. –  Toan Sep 8 '10 at 6:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.