# Random walk $< 0$

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.

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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 $$P(T_1\geq 2t)=P(-1< X_{2t} < 1)=P(X_{2t}=0)={2t\choose t}\left({1\over 2}\right)^{2t}.$$