First Passage Time of Brownian Motion: Density and Mean
Let $W(t)$ be a standard Wiener process (Brownian motion) with $W(0) = 0$, and fix a level $x > 0$. Define the first passage time
$$\tau_x = \inf\{t \ge 0 : W(t) = x\}.$$
(a) Find $P(\tau_x \le t)$ and the probability density function of $\tau_x$.
(b) Find $E[\tau_x]$.
Hints
- Reflection principle: for $x > 0$, $P(\tau_x \le t) = 2\,P(W(t) \ge x)$, because a path that has touched $x$ by time $t$ is equally likely to end above or below $x$.
- Write $P(W(t) \ge x) = 1 - \Phi(x/\sqrt{t})$, so $P(\tau_x \le t) = 2 - 2\Phi(x/\sqrt{t})$. Differentiate in $t$ to get the density.
- The density behaves like $t^{-3/2}$ for large $t$, so $\int t \cdot f(t)\,dt$ diverges: $E[\tau_x] = \infty$ even though $P(\tau_x < \infty) = 1$.
Worked Solution
How to Think About It: The distribution of the hitting time comes from a symmetry argument, not from an SDE. Once the process reaches $x$, by the strong Markov property and symmetry its future is equally likely to be above or below $x$; reflecting the path after $\tau_x$ maps "hit $x$ and end below" bijectively onto "end above." That doubles the tail probability of $W(t)$ into the CDF of $\tau_x$.
Quick Estimate: $W(t)$ has standard deviation $\sqrt{t}$, so for $t \gg x^2$ the level $x$ has almost surely been reached, while for $t \ll x^2$ it almost surely has not. $P(\tau_1 \le 1) = 2(1 - \Phi(1)) \approx 0.317$. The tail $P(\tau_x > t) \approx 2x/\sqrt{2\pi t}$ decays like $t^{-1/2}$, too slowly for a finite mean.
Formal Solution:
Part (a): Distribution and density
*Step 1 -- Reflection principle.* For $x > 0$,
$$P(\tau_x \le t) = P(\tau_x \le t, W(t) \ge x) + P(\tau_x \le t, W(t) < x).$$
The first term equals $P(W(t) \ge x)$ (if $W(t) \ge x$ then by continuity the path has crossed $x$). For the second, reflect the path about the level $x$ after time $\tau_x$; by the strong Markov property and the symmetry of Brownian increments this is a measure-preserving bijection onto $\{\tau_x \le t, W(t) > x\}$, so the second term also equals $P(W(t) \ge x)$. Therefore
$$P(\tau_x \le t) = 2\,P(W(t) \ge x) = 2\left[1 - \Phi\!\left(\frac{x}{\sqrt{t}}\right)\right] = 2 - 2\Phi\!\left(\frac{x}{\sqrt{t}}\right),$$
where $\Phi$ is the standard normal CDF. Equivalently $P(\tau_x \le t) = P(|W(t)| \ge x)$.
*Step 2 -- Density.* Differentiate with respect to $t$, using $\Phi' = \phi$ and $\frac{d}{dt}\big(x t^{-1/2}\big) = -\tfrac{x}{2}t^{-3/2}$:
$$f_{\tau_x}(t) = -2\,\phi\!\left(\frac{x}{\sqrt{t}}\right)\cdot\left(-\frac{x}{2}\,t^{-3/2}\right) = \frac{x}{\sqrt{2\pi\, t^3}}\,\exp\!\left(-\frac{x^2}{2t}\right) = \frac{x}{t\sqrt{2\pi t}}\,e^{-x^2/(2t)}, \qquad t > 0.$$
This is the Levy distribution (an inverse-gamma with shape $1/2$).
*Step 3 -- Reaches the level almost surely.* As $t \to \infty$, $\Phi(x/\sqrt{t}) \to \Phi(0) = 1/2$, so $P(\tau_x < \infty) = 2 - 2\cdot\tfrac12 = 1$.
Part (b): Expected first passage time
*Step 4 -- Divergent mean.* For large $t$, $e^{-x^2/(2t)} \to 1$, so $f_{\tau_x}(t) \sim \dfrac{x}{\sqrt{2\pi}}\,t^{-3/2}$ and
$$E[\tau_x] = \int_0^\infty t\, f_{\tau_x}(t)\,dt \ \ge\ \int_{t_0}^\infty \frac{c\,x}{\sqrt{t}}\,dt = \infty.$$
*Step 5 -- Martingale cross-check.* Treat $x$ as an upper barrier and $-\beta$ as a lower barrier; the two-barrier exit time has mean $x\beta$ (from the martingale $W(t)^2 - t$). Letting $\beta \to \infty$ recovers the one-sided hitting time, and $x\beta \to \infty$. So $\tau_x$ is finite almost surely but has infinite expectation.
Answer: (a) $P(\tau_x \le t) = 2 - 2\Phi(x/\sqrt{t})$, with density $f_{\tau_x}(t) = \dfrac{x}{t\sqrt{2\pi t}}\,e^{-x^2/(2t)}$ for $t > 0$. (b) $E[\tau_x] = \infty$, although $P(\tau_x < \infty) = 1$.
Intuition
The reflection principle says that among all paths that reach $x$ by time $t$, the continuation is symmetric, so "hit $x$ by $t$" is exactly twice as likely as "end above $x$ at $t$." That gives the Levy distribution for $\tau_x$, whose density decays only like $t^{-3/2}$: the level is reached with probability one, but the wait has infinite mean because of the heavy tail. Traders meet this as the sobering fact that a driftless price will eventually touch any target, yet the expected time to get there is infinite, and it is the reason single-barrier expected hitting times need drift (or a second barrier) to be finite.