First-Passage Time of Brownian Motion

Worked Solution

How to Think About It: Before any formulas, think about what a first-passage time question is really asking: how long does a random walk take to reach a fixed level? The reflection principle is a beautiful symmetry trick -- once a Brownian path reaches $a$, its future is symmetric, so continuing above $a$ is equally likely as reflecting below. This immediately doubles probabilities in a clean way. For the expectation, your gut should already suspect something is off: Brownian motion is recurrent, so $\tau_a$ is finite with probability 1, but the path meanders so wildly that the average time is infinite. This is a counterintuitive but crucial fact that shows up repeatedly in exotic options pricing.

Quick Sanity Check: Before computing, ask: what does $P(\tau_a \leq t)$ look like qualitatively? At $t = 0$, it should be 0 (the path starts at 0, not $a > 0$). As $t \to \infty$, it should go to 1 (Brownian motion is recurrent and will eventually hit every level). And it should be an increasing function of $t$ and a decreasing function of $a$ (higher barriers take longer to hit). Whatever formula we derive had better satisfy these.

Part (i): The Distribution of $\tau_a$

Reflection Principle Derivation:

The event $\{\tau_a \leq t\}$ means the path touched level $a$ by time $t$. To compute this, note:

$$P(\tau_a \leq t) = P\left(\max_{0 \leq s \leq t} B_s \geq a\right).$$

Now apply the reflection principle. Condition on paths that hit $a$ by time $t$ and reflect them after the first hitting time:

• Any path with $\tau_a \leq t$ and $B_t \geq a$ contributes as-is.
• Any path with $\tau_a \leq t$ and $B_t < a$: reflect the portion after $\tau_a$ to get a path ending above $a$.

By the strong Markov property and symmetry of Brownian motion, these two sets are in bijection, so:

$$P(\tau_a \leq t) = P(\tau_a \leq t,\, B_t \geq a) + P(\tau_a \leq t,\, B_t < a) = 2\,P(B_t \geq a).$$

Since $B_t \sim N(0, t)$, we have $P(B_t \geq a) = P\left(Z \geq a/\sqrt{t}\right) = 1 - \Phi\left(a/\sqrt{t}\right) = \Phi\left(-a/\sqrt{t}\right)$, so:

$$\boxed{P(\tau_a \leq t) = 2\,\Phi\left(-\frac{a}{\sqrt{t}}\right) = 2\left(1 - \Phi\left(\frac{a}{\sqrt{t}}\right)\right).}$$

This satisfies all our qualitative checks: at $t = 0$, $\Phi(-\infty) = 0$; as $t \to \infty$, $\Phi(0) = 1/2$ so the expression goes to 1; larger $a$ means the argument $-a/\sqrt{t}$ is more negative, giving smaller probability. Equivalently, $P(\tau_a \leq t) = P(|B_t| \geq a)$.

For $a = t = 1$: $$P(\tau_1 \leq 1) = 2\,\Phi(-1) = 2(1 - 0.8413) \approx 0.3173,$$ double the terminal probability $P(B_1 \geq 1) \approx 0.1587$.

The density follows by differentiating with respect to $t$:

$$f_{\tau_a}(t) = \frac{a}{\sqrt{2\pi}\, t^{3/2}} \exp\!\left(-\frac{a^2}{2t}\right), \quad t > 0.$$

This is the Levy distribution (the driftless limit of the inverse Gaussian) with heavy tails -- specifically, $f_{\tau_a}(t) \sim t^{-3/2}$ for large $t$.

Part (ii): The Expected Hitting Time is Infinite

We need $$E[\tau_a] = \int_0^{\infty} t \cdot f_{\tau_a}(t)\, dt = \int_0^{\infty} \frac{a}{\sqrt{2\pi}}\, t^{-1/2} \exp\!\left(-\frac{a^2}{2t}\right) dt.$$

For $t \ge a^2$ the exponential factor is at least $e^{-1/2}$, so

$$\int_{a^2}^{\infty} t \cdot f_{\tau_a}(t)\, dt \ \ge\ \frac{a\, e^{-1/2}}{\sqrt{2\pi}} \int_{a^2}^{\infty} t^{-1/2}\, dt = +\infty.$$

So $E[\tau_a] = +\infty$. Numerically, $\int_0^T t f_{\tau_a}(t)\,dt$ grows like $a\sqrt{2T/\pi}$ (about $798$ at $T = 10^6$ for $a = 1$) and never levels off.

Martingale check (Wald): If $E[\tau_a]$ were finite, Wald's identity for Brownian motion would give $E[B_{\tau_a}] = E[B_0] = 0$. But $B_{\tau_a} = a > 0$ on every path, a contradiction. So $E[\tau_a] = +\infty$.

Two-barrier view: Add a lower barrier at $-b < 0$. The exit time $\tau_{-b,a}$ from $(-b, a)$ has finite mean, and optional stopping on the martingale $B_t^2 - t$ gives $E[\tau_{-b,a}] = ab$. As $b \to \infty$ the exit time increases to $\tau_a$, so by monotone convergence $E[\tau_a] = \lim_{b\to\infty} ab = \infty$. This also shows why the tempting answer $E[\tau_a] = a^2$ (optional stopping on $B_t^2 - t$ with only the upper barrier) is wrong: optional stopping needs exactly the integrability that fails here.

Answer (ii): $E[\tau_a] = +\infty$. The $t^{-3/2}$ tail of the density is not integrable when multiplied by $t$.

Part (iii): Barrier Option Intuition

For a barrier option that knocks out if $B_T$ (or a GBM analog) hits a barrier at level $a$ before maturity $T$, the relevant quantity is $P(\tau_a \leq T)$, NOT $P(B_T \geq a)$.

From our formula: - $P(B_T \geq a) = 1 - \Phi(a/\sqrt{T})$ -- probability the terminal value exceeds the barrier. - $P(\tau_a \leq T) = 2(1 - \Phi(a/\sqrt{T})) = 2 \cdot P(B_T \geq a)$ -- probability the path ever touches the barrier.

The hitting probability is exactly TWICE the terminal exceedance probability. This factor of 2 is the whole story: even if the barrier seems remote from the terminal distribution (say $a = 3\sqrt{T}$, so $P(B_T \geq a) \approx 0.135\%$), the hitting probability is still $\approx 0.27\%$ -- not negligible for exotic products.

In practice, for barrier options on stocks modeled with GBM, the analog involves the running maximum of a drifted Brownian motion, but the qualitative lesson is the same: paths explore a much wider range than the terminal snapshot suggests. A barrier that looks far out-of-the-money at expiry may have been touched many times during the life of the option. This is why barrier options are materially cheaper (knock-outs) or more expensive (knock-ins) than their vanilla counterparts, and why barrier risk is hard to hedge -- small moves in spot or vol can dramatically change the hitting probability.