From Reflection to a First-Passage Distribution
A path can reach a level and then finish on either side of it. That distinction matters when asking for the probability that Brownian motion has reached a positive level by a deadline: the endpoint alone does not record whether the path crossed the level earlier. Reflection converts paths that cross and finish below the level into paths with endpoints above it. This gives the first-passage distribution, and differentiating that distribution gives the density of the passage time.
Throughout, let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space carrying a standard Brownian motion \((B_t)_{t\ge0}\) with continuous paths, \(B_0=0\), and state space \(S=\mathbb{R}\). Time is continuous. We use its natural filtration \(\mathcal{F}_t=\sigma(B_s:0\le s\le t)\); no conditioning or stopping-time theorem is needed below. For a fixed level \(a>0\), define the first-passage time
with \(\inf\varnothing=\infty\). For \(t>0\), continuity of the paths implies that \(\{T_a\le t\}\) is the event that the path reaches or exceeds \(a\) at some time in \([0,t]\). Write \(M_t=\sup_{0\le s\le t}B_s\), so this event is \(\{M_t\ge a\}\).
Let \(\Phi(x)=\int_{-\infty}^x\phi(y)\,dy\), where \(\phi(y)=(2\pi)^{-1/2}e^{-y^2/2}\) is the standard normal density. The reflection principle, established in the earlier material on Brownian reflection, states that for \(a>0\) and \(x<a\), the probability of reaching \(a\) by time \(t\) and ending in an infinitesimal interval at \(x\) has density equal to the density of ending at \(2a-x\). This is obtained by reflecting the part of the path after its first visit to \(a\). We now use that identity to derive the passage law.
Proof. Fix \(a>0\) and \(t>0\). If \(B_t\ge a\), path continuity and \(B_0=0<a\) imply that the path has reached \(a\) by time \(t\). The endpoint event therefore contributes \(\mathbb{P}(B_t\ge a)\) to \(\mathbb{P}(T_a\le t)\). For paths that reach \(a\) but end below \(a\), the reflection principle identifies their endpoint density at \(x<a\) with the ordinary endpoint density at \(2a-x\). Consequently, their total probability is $$ \int_{-\infty}^{a}\frac{1}{\sqrt{2\pi t}}\exp\left(-\frac{(2a-x)^2}{2t}\right)\,dx =\int_a^\infty\frac{1}{\sqrt{2\pi t}}\exp\left(-\frac{y^2}{2t}\right)\,dy =\mathbb{P}(B_t>a), $$ where the first equality uses \(y=2a-x\), and the last uses the symmetry of the centered Gaussian density. Since \(B_t\) has a continuous distribution, \(\mathbb{P}(B_t=a)=0\). Adding the endpoint and reflected-path contributions gives $$ \mathbb{P}(T_a\le t)=\mathbb{P}(B_t\ge a)+\mathbb{P}(B_t>a) =2\mathbb{P}(B_t\ge a). $$ Since \(B_t/\sqrt{t}\) has the standard normal distribution, this is \(2[1-\Phi(a/\sqrt{t})]\), as claimed.
To obtain the density, differentiate the distribution function for \(t>0\). Set \(u=a/\sqrt{t}\). Then \(du/dt=-a/(2t^{3/2})\), so the chain rule gives $$ \frac{d}{dt}\left(2[1-\Phi(u)]\right) =-2\phi(u)\frac{du}{dt} =\frac{a}{t^{3/2}}\phi\left(\frac{a}{\sqrt{t}}\right) =\frac{a}{\sqrt{2\pi}\,t^{3/2}}\exp\left(-\frac{a^2}{2t}\right). $$ This derivative is nonnegative, so the distribution function is increasing. As \(t\) decreases to zero, \(a/\sqrt{t}\) increases without bound and the distribution function tends to zero. As \(t\) increases without bound, \(a/\sqrt{t}\) tends to zero and the distribution function tends to \(2[1-\Phi(0)]=1\). Thus the total probability over finite positive times is one: \(T_a<\infty\) almost surely. The distribution function is continuous for \(t>0\), and its limit at zero is zero, so there is no atom at a finite time. \(\square\)
Using the Distribution and Density
The formula for \(\mathbb{P}(T_a\le t)\) answers a deadline question directly. It also gives probabilities over intervals: for \(0\le s<t\), continuity of the distribution implies $$ \mathbb{P}(s<T_a\le t)=\mathbb{P}(T_a\le t)-\mathbb{P}(T_a\le s), $$ where the distribution function at \(s=0\) is taken to be zero. Equivalently, this probability is the integral of the density from \(s\) to \(t\). The density itself is not a probability at a single time; probabilities are obtained by integrating it over intervals.
Worked Example: Reaching Level 3 by Time 2
Find the probability that standard Brownian motion reaches \(a=3\) by time \(t=2\), and calculate the density of \(T_3\) at \(t=2\).
Substitute \(a=3\) and \(t=2\) into the distribution formula. The standardized level is \(a/\sqrt{t}=3/\sqrt{2}\), so $$ \mathbb{P}(T_3\le2)=2\left[1-\Phi\left(\frac{3}{\sqrt{2}}\right)\right]. $$ This expression is the probability; it includes paths that reached \(3\) and later ended below \(3\).
For the density at time \(2\), substitute the same parameter values into the density formula: $$ f_{T_3}(2)=\frac{3}{\sqrt{2\pi}\,2^{3/2}}\exp\left(-\frac{3^2}{2\cdot2}\right) =\frac{3}{4\sqrt{\pi}}e^{-9/4}. $$ Here \(2^{3/2}=2\sqrt{2}\), so the denominator \(\sqrt{2\pi}\,2\sqrt{2}\) is \(4\sqrt{\pi}\).
Check: Differentiating \(2[1-\Phi(3/\sqrt{t})]\) gives \(3t^{-3/2}\phi(3/\sqrt{t})\). At \(t=2\), this is \(3/(4\sqrt{\pi})e^{-9/4}\), agreeing with the direct substitution. The density value is not the probability of passage at exactly time \(2\); that probability is zero because the distribution has no atom there.
Worked Example: Passage During a Time Interval
For the level \(a=1\), find the probability that the first passage occurs after time \(1\) and by time \(4\), that is, \(\mathbb{P}(1<T_1\le4)\).
Evaluate the distribution function at each endpoint: $$ \mathbb{P}(T_1\le4)=2[1-\Phi(1/2)],\qquad \mathbb{P}(T_1\le1)=2[1-\Phi(1)]. $$ Subtracting the second probability from the first gives $$ \mathbb{P}(1<T_1\le4)=2[\Phi(1)-\Phi(1/2)]. $$ The strict inequality at the left endpoint is valid because \(\mathbb{P}(T_1=1)=0\).
Check by integrating the density: For \(a=1\), the density is \(t^{-3/2}\phi(1/\sqrt{t})\). Use \(u=1/\sqrt{t}\), so \(dt=-2u^{-3}\,du\) and \(t^{-3/2}=u^3\). When \(t=1\), \(u=1\); when \(t=4\), \(u=1/2\). Therefore $$ \int_1^4t^{-3/2}\phi(1/\sqrt{t})\,dt =2\int_{1/2}^{1}\phi(u)\,du =2[\Phi(1)-\Phi(1/2)], $$ which verifies the interval calculation.
Worked Example: Choosing a Deadline for a Target Probability
For a level \(a=2\), choose a deadline \(t\) so that the probability of reaching the level by that deadline is \(1/2\).
Start with the formula and set it equal to the target probability: $$ 2\left[1-\Phi\left(\frac{2}{\sqrt{t}}\right)\right]=\frac12. $$ Dividing by \(2\) and rearranging gives \(\Phi(2/\sqrt{t})=3/4\). The function \(\Phi\) is strictly increasing, so its inverse is defined at \(3/4\); let \(z=\Phi^{-1}(3/4)>0\). Then \(2/\sqrt{t}=z\), and solving for \(t\) gives $$ t=\frac{4}{[\Phi^{-1}(3/4)]^2}. $$
Check: At this deadline, \(2/\sqrt{t}=\Phi^{-1}(3/4)\). Substitution into the passage formula gives \(2[1-\Phi(\Phi^{-1}(3/4))]=2(1-3/4)=1/2\), as required. The inverse-normal expression specifies the deadline exactly without requiring a decimal approximation.
Interpretation and a Common Mistake
The level and time enter the passage probability through the ratio \(a/\sqrt{t}\). For example, replacing \(a\) by \(ca\) and \(t\) by \(c^2t\), for \(c>0\), leaves that ratio unchanged. The distribution also tends to zero as the deadline decreases to zero and to one as the deadline grows without bound. These limits express two facts: a positive level cannot be reached instantaneously from the starting point \(0\), but Brownian motion reaches each fixed positive level eventually with probability one.
The density describes how passage probability is distributed over time. For a short interval \((t,t+h]\), with \(t>0\) and small \(h>0\), the probability is approximately \(f_{T_a}(t)h\). This is an approximation to an interval probability, not an equality for a single time. In particular, a density value can have units of inverse time and need not be at most one.
Common mistake: counting only paths that end above the level. For \(a=1\) and \(t=1\), using \(\mathbb{P}(B_1\ge1)\) alone gives \(1-\Phi(1)\). That counts paths whose endpoint is above \(1\), but misses paths that reached \(1\) earlier and ended below it. Reflection shows that the missed paths have the same probability \(1-\Phi(1)\), so the correct calculation is $$ \mathbb{P}(T_1\le1)=2[1-\Phi(1)]. $$ The factor of \(2\) is essential. The endpoint event is contained in the passage event, but the two events are not equal.
The reflection argument here concerns a single positive level. It should not be confused with a probability involving both an upper and a lower barrier, or with the maximum of the absolute value of Brownian motion; those require different event descriptions. For this one-sided first-passage question, the reflection identity, the normal endpoint distribution, and differentiation give the complete distributional calculation.
Check Your Understanding
Use the distribution function and density above, keeping separate probabilities over intervals from density values at individual times.
- Write the formula for \(\mathbb{P}(T_a\le t)\) when \(a>0\) and \(t>0\), and explain why the endpoint event alone is insufficient.
- For \(a=2\) and \(t=8\), simplify the standardized level in the passage-probability formula.
- For \(a=1\), express \(\mathbb{P}(2<T_1\le5)\) in terms of \(\Phi\).
- Differentiate the distribution function to recover \(f_{T_a}(t)\), showing the derivative of \(a/\sqrt{t}\).
- Why does the density value \(f_{T_a}(t)\) not equal \(\mathbb{P}(T_a=t)\) for a fixed \(t>0\)?