The Reflection Principle and Brownian First Passage

From a Path Event to a Gaussian Tail

At a fixed time, Brownian motion has a simple Gaussian distribution. A first-passage event, however, depends on the entire path: it asks whether the path has reached a specified level at any earlier time. The reflection principle relates these two kinds of probabilities. Once the path first reaches a positive level, reflecting its subsequent increments produces a one-to-one correspondence between paths that hit the level and finish below it, and paths that finish above it.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, with continuous paths, $B_0=0$, and independent increments. For a fixed level $a>0$, define the first hitting time

\tau_a=\inf\{t\ge0:B_t=a\},

with $\inf\varnothing=\infty$. By path continuity, $\{\tau_a\le t\}=\{\sup_{0\le s\le t}B_s\ge a\}$. We use the standard strong Markov property of Brownian motion: after a stopping time, the process of future increments, on the event that the stopping time is finite, is a Brownian motion independent of the information available at that stopping time. No stochastic integral is needed here; whenever an Ito integral is used elsewhere, its integrand is taken to be progressively measurable and in $\mathcal{L}^2$ on the time interval under consideration.

The Reflection Principle

Theorem: Let $a>0$, $T>0$, and let $A$ be a Borel subset of $(-\infty,a)$. Then $$ \mathbb{P}(\tau_a\le T,\ B_T\in A)=\mathbb{P}(B_T\in 2a-A), $$ where $2a-A=\{2a-y:y\in A\}$. In particular, $$ \mathbb{P}(\tau_a\le T)=2\mathbb{P}(B_T\ge a). $$

Proof. Define the reflected process by $B'_t=B_t$ for $t\le\tau_a$, and $B'_t=2a-B_t$ for $t>\tau_a$ when $\tau_a<\infty$; if $\tau_a=\infty$, set $B'_t=B_t$ for every $t$. The paths agree up to the first hit, and after that time the increments from $a$ are negated. By the strong Markov property, those post-$\tau_a$ increments are Brownian and independent of the past; their negatives have the same distribution, by the symmetry of centered normal increments. Thus reflection preserves Brownian path probabilities after the hit. Also, the first hitting time of $a$ is unchanged by this operation, since the paths agree through that first hit.

If a path has $\tau_a\le T$ and $B_T=y\in A$, its reflected path has terminal value $2a-y\in2a-A$, which is greater than $a$. Conversely, any path with $B_T\in2a-A$ must have hit $a$ by time $T$, by continuity; reflecting it gives a path with terminal value in $A$. Reflection is its own inverse. Since it preserves Brownian path probabilities, the two events have equal probability. This proves the first identity.

For the second identity, split $\{\tau_a\le T\}$ into the disjoint events with $B_T<a$ and $B_T\ge a$. The event $B_T\ge a$ implies $\tau_a\le T$, again by continuity. The first identity, applied to $A=(-\infty,a)$, gives $\mathbb{P}(\tau_a\le T,B_T<a)=\mathbb{P}(B_T>a)$. Since $B_T$ has a continuous normal distribution, $\mathbb{P}(B_T=a)=0$, so $\mathbb{P}(B_T>a)=\mathbb{P}(B_T\ge a)$. Adding the two disjoint probabilities proves the claimed formula. $\square$

The endpoint version is useful when the hitting event must be combined with a terminal constraint. It says more than the formula for the probability of hitting: for endpoints below $a$, the reflected endpoint $2a-B_T$ has the same Gaussian probability as the original path event together with a hit. The endpoint $a$ itself causes no extra term in the distribution calculation because its probability at a fixed positive time is zero.

Distribution and Density of the First-Passage Time

Corollary: For $a>0$, the distribution function $F_{\tau_a}(t)=\mathbb{P}(\tau_a\le t)$ is $$ F_{\tau_a}(t)= \begin{cases} 0,&t\le0,\\ 2\left[1-\Phi\left(\dfrac{a}{\sqrt{t}}\right)\right],&t>0, \end{cases} $$ where $\Phi$ is the standard normal CDF. For $t>0$, the first-passage time has density $$ f_{\tau_a}(t)=\frac{a}{\sqrt{2\pi}\,t^{3/2}} \exp\left(-\frac{a^2}{2t}\right). $$

Proof. For $t>0$, the reflection principle gives $F_{\tau_a}(t)=2\mathbb{P}(B_t\ge a)$. Since $B_t/\sqrt{t}$ is standard normal, this equals $2[1-\Phi(a/\sqrt{t})]$. At $t=0$, continuity at the starting point and $a>0$ imply $\tau_a>0$ almost surely; for $t<0$, a hitting time cannot be at most $t$. Hence the stated CDF applies for all real $t$.

To differentiate for $t>0$, put $u(t)=a t^{-1/2}$. Then $u'(t)=-a/(2t^{3/2})$, and the standard normal density is $\phi(u)=e^{-u^2/2}/\sqrt{2\pi}$, with $\Phi'(u)=\phi(u)$. The chain rule gives

\frac{d}{dt}\left(2[1-\Phi(u(t))]\right) =-2\phi(u(t))u'(t) =\frac{a}{\sqrt{2\pi}\,t^{3/2}} \exp\left(-\frac{a^2}{2t}\right).

This derivative is nonnegative. Also, as $t\downarrow0$, $a/\sqrt{t}\to\infty$, so the CDF tends to $0$; as $t\to\infty$, $a/\sqrt{t}\to0$, so it tends to $2[1-\Phi(0)]=1$. Thus the density accounts for the full distribution, with no atom at zero or at infinity. $\square$

Worked Example: Probability of Hitting by a Fixed Time

Find the probability that Brownian motion reaches level $a=\sqrt{2}$ by time $T=2$, and verify it using the standardized endpoint.

The hitting probability is $2\mathbb{P}(B_2\ge\sqrt{2})$. Since $B_2/\sqrt{2}$ is standard normal, the threshold in standard normal units is $\sqrt{2}/\sqrt{2}=1$. Therefore

\mathbb{P}(\tau_{\sqrt{2}}\le2) =2[1-\Phi(1)].

For the check, split according to whether the endpoint is above or below $\sqrt{2}$. The endpoint-above probability is $1-\Phi(1)$. Reflection says the probability of having hit by time $2$ and ending below $\sqrt{2}$ is also $1-\Phi(1)$. Their sum is $2[1-\Phi(1)]$, as required.

Worked Example: Hitting and Ending in an Interval

For level $a=1$ and time $T=1$, compute the probability that Brownian motion hits level $1$ and ends in the interval $(0,1)$.

Take $A=(0,1)$, which lies strictly below $a=1$. Reflection maps this endpoint interval to $2-A=(2-1,2-0)=(1,2)$. Since $B_1$ is standard normal,

\mathbb{P}(\tau_1\le1,\ 0<B_1<1) =\mathbb{P}(1<B_1<2) =\Phi(2)-\Phi(1).

The inequalities are strict on both intervals; including either endpoint would not change the probability, because the normal distribution assigns probability zero to each single point. As a check, paths ending in $(1,2)$ necessarily hit $1$, and their reflected paths end in $(0,1)$ after having hit $1$. The reflection is reversible, so the two probabilities agree.

Worked Example: Probability of Hitting During a Time Window

For level $a=1$, find the probability that the first hit occurs after time $1$ and no later than time $4$.

The event is $\{1<\tau_1\le4\}$. Since the CDF is continuous for positive times, its probability is $F_{\tau_1}(4)-F_{\tau_1}(1)$. At time $4$, the standardized level is $1/\sqrt{4}=1/2$; at time $1$, it is $1/\sqrt{1}=1$. Substitution gives

\begin{aligned} \mathbb{P}(1<\tau_1\le4) &=2[1-\Phi(1/2)]-2[1-\Phi(1)]\\ &=2[\Phi(1)-\Phi(1/2)]. \end{aligned}

This is positive because $\Phi$ is increasing and $1>1/2$. As a check, it is the difference between the probability of having hit by time $4$ and the probability of having hit by time $1$; the events are nested, so subtracting their probabilities gives exactly the probability of the time window.

Reading the Formula Correctly

The first-passage distribution is not just the probability that $B_t$ is above $a$ at the observation time. A path can reach $a$, turn around, and finish below it. Reflection shows that these hit-and-finish-below paths contribute a second Gaussian tail of equal size. That is the reason for the factor of $2$.

A common mistake is to write $\mathbb{P}(\tau_a\le T)=\mathbb{P}(B_T\ge a)=1-\Phi(a/\sqrt{T})$. This counts only paths that finish at or above the level and misses paths that crossed it earlier but ended below. For example, when $a=1$ and $T=1$, the one-tail calculation gives $1-\Phi(1)$, while the reflection principle gives $2[1-\Phi(1)]$. The omitted contribution is exactly the probability of hitting by time $1$ and ending below $1$; reflection identifies it with another $1-\Phi(1)$.

The density and distribution answer different questions. The CDF gives a probability accumulated up to time $t$; the density gives probability per unit time near $t$, so a density value itself is not a probability of hitting at exactly that time. The density integrates to one because the CDF starts at zero and tends to one. These formulas concern standard Brownian motion starting at zero and a fixed positive level; changing the starting point, adding drift, or considering two-sided exits changes the problem and requires a separate derivation.

Check Your Understanding

Use the endpoint reflection identity and the derived distribution to answer each question.

For $a>0$ and $T>0$, why does $\{B_T\ge a\}$ imply $\{\tau_a\le T\}$? Which path property is needed?
Express $\mathbb{P}(\tau_a\le T,\ B_T\in(-\infty,0))$ as a Gaussian probability on a reflected endpoint interval.
For level $a=2$, write the exact expression for $\mathbb{P}(\tau_2\le8)$ using $\Phi$.
Differentiate $2[1-\Phi(a/\sqrt{t})]$ with respect to $t$, showing the derivative of $a/\sqrt{t}$ explicitly.
Explain why the one-tail probability $\mathbb{P}(B_T\ge a)$ does not by itself give the probability of hitting $a$ by time $T$.

The Reflection Principle and Brownian First Passage

What You'll Learn