the reflection principle for Brownian motion
Here is a question that looks hard: what is the chance that a Brownian motion ever rises above some level a at any moment during [0, t]? Tracking the running maximum of a wildly wiggling path seems hopeless. The reflection principle is a gorgeous trick that turns this path-history question into a simple endpoint question you can answer with the Normal distribution. The idea is a mirror: every path that crosses level a can be paired with a reflected twin, and counting the twins is easy.
Here is the trick step by step. Suppose a path touches level a at some first time T (a stopping time) and ends below a. Reflect the part of the path AFTER T across the horizontal line at height a — flip it like a mirror image. By the strong Markov property, the post-T continuation is a fresh symmetric Brownian motion, so the reflected path is just as likely as the original. The reflection sends an original path ending at a height x (below a) to a twin ending at 2a - x (above a). This pairing shows: P(max over [0,t] of B exceeds a, and B(t) below a) = P(B(t) above a). Adding the paths that end above a (which obviously had a max above a) gives the clean formula P(max over [0,t] of B at least a) = 2 P(B(t) at least a) = 2 P(Normal(0,t) at least a). The running maximum is twice as likely to have exceeded a as the endpoint alone is to lie above a.
From this one identity flow the headline distributions of Brownian motion. The maximum over [0, t] has the same law as the absolute value |B(t)| (a 'folded Normal'). The first time the path hits level a has a known heavy-tailed density (its mean is actually infinite — the path always gets there eventually, but the waiting time has no finite average). These hitting-time and maximum results are the bread and butter of pricing barrier options and analysing first-passage problems. The principle is exact, but note it leans entirely on two ingredients: the symmetry of Brownian increments and the strong Markov restart at the hitting time.
What is the chance a standard Brownian motion ever exceeds level 1 during [0, 1]? By the reflection principle it is 2 P(B(1) at least 1) = 2 P(Normal(0,1) at least 1) = 2 times 0.159 = 0.317 — about a 32 percent chance, far higher than the 16 percent chance of merely ending above 1.
Reflecting the post-hitting path turns a hard 'did it ever cross?' question into an easy endpoint probability.
The first-hitting-time of a level has infinite expected value: the path reaches any level with probability one, yet the average waiting time is infinite. 'Sure to happen' and 'happens soon on average' are different things.