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