Laplace Transform of a One-Sided Hitting Time

The One-Sided Hitting-Time Transform

For Brownian motion, the time to reach a level above the starting point is finite almost surely, yet its expected value is infinite. A Laplace transform captures this first-passage time in a form that is both explicit and useful. The key is to choose an exponential martingale whose time decay matches the transform parameter, then stop it at the hitting time truncated at a fixed time. That truncation makes the stopping argument valid and leaves a residual term that can be bounded.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ carrying standard Brownian motion $(B_t)_{t\ge0}$ with $B_0=x$, where $x<a$. Use its natural filtration $(\mathcal{F}_t)_{t\ge0}$, with the usual completion and right-continuous modification. Write $\mathbb{P}_x$ and $\mathbb{E}_x$ for probability and expectation under this starting point. The state space is $S=\mathbb{R}$, and time is continuous. Define the first hitting time of the fixed level $a$ by $\tau_a=\inf\{t\ge0:B_t=a\}$, with $\inf\varnothing=\infty$. This is a stopping time for the Brownian filtration. For $s>0$, use the convention $e^{-s\infty}=0$ and define the Laplace transform as $\mathbb{E}_x[e^{-s\tau_a}]$.

Theorem: If $B_t$ is standard Brownian motion started at $x<a$, then for every $s>0$, $$ \mathbb{E}_x[e^{-s\tau_a}]=\exp\!\left(-(a-x)\sqrt{2s}\right). $$ Moreover, $\mathbb{P}_x(\tau_a<\infty)=1$, so the transform at $s=0$, interpreted as $\mathbb{E}_x[1]$, equals $1$.

The Exponential Martingale

For a real number $\theta$, define $M_t=\exp(\theta(B_t-x)-\tfrac12\theta^2t)$. This is adapted to $(\mathcal{F}_t)$, is integrable, and starts at $M_0=1$. To verify the martingale property, let $0\le u\le t$. Brownian increments $B_t-B_u$ are independent of $\mathcal{F}_u$ and are normally distributed with mean $0$ and variance $t-u$. The normal moment-generating function gives $\mathbb{E}_x[e^{\theta(B_t-B_u)}\mid\mathcal{F}_u]=e^{\theta^2(t-u)/2}$. Consequently,

\mathbb{E}_x[M_t\mid\mathcal{F}_u] =M_u e^{-\theta^2(t-u)/2} \mathbb{E}_x[e^{\theta(B_t-B_u)}\mid\mathcal{F}_u] =M_u.

Thus $(M_t)$ is a martingale. The parameter that will be useful is $\theta=\sqrt{2s}$: it makes the time factor in $M_t$ exactly $e^{-st}$. The choice of the positive square root is important because the stopped Brownian position at the target is $a$, above the starting point.

Stopping principle used here: For a martingale and a bounded stopping time $\sigma$, the bounded optional stopping theorem gives $\mathbb{E}_x[M_\sigma]=\mathbb{E}_x[M_0]$, provided the martingale is integrable on each fixed finite time horizon. We apply it only to $\sigma=\tau_a\wedge T$, where $T<\infty$. We do not assume optional stopping at the possibly unbounded time $\tau_a$.

Deriving the Transform

Fix $s>0$, set $\theta=\sqrt{2s}>0$, and choose a deterministic truncation time $T>0$. The time $\tau_a\wedge T$ is bounded, so the stopping principle applies to $M_t=\exp(\theta(B_t-x)-st)$: $\mathbb{E}_x[M_{\tau_a\wedge T}]=1$. Split this expectation according to whether the level has been reached by time $T$. On $\{\tau_a\le T\}$, continuity gives $B_{\tau_a}=a$, and hence $M_{\tau_a}=\exp(\theta(a-x)-s\tau_a)$. On $\{\tau_a>T\}$, the stopped value is $M_T$. Therefore

1=e^{\theta(a-x)}\mathbb{E}_x[e^{-s\tau_a}\mathbf{1}_{\{\tau_a\le T\}}] +\mathbb{E}_x[M_T\mathbf{1}_{\{\tau_a>T\}}].

The second term is the contribution from paths that have not hit $a$ by time $T$. On that event, $B_T<a$; otherwise continuity would imply that the path had already reached $a$. Since $\theta>0$, it follows that $M_T\le e^{\theta(a-x)-sT}$ there. Thus the residual term is nonnegative and at most $e^{\theta(a-x)-sT}$, which tends to zero as $T$ tends to infinity. Meanwhile, $e^{-s\tau_a}\mathbf{1}_{\{\tau_a\le T\}}$ increases to $e^{-s\tau_a}$ under the convention $e^{-s\infty}=0$. By monotone convergence,

1=e^{\theta(a-x)}\mathbb{E}_x[e^{-s\tau_a}], \qquad \mathbb{E}_x[e^{-s\tau_a}] =e^{-\theta(a-x)} =\exp\!\left(-(a-x)\sqrt{2s}\right).

This proves the transform for $s>0$ without first assuming that $\tau_a$ is finite. To handle $s=0$, take any sequence of positive parameters decreasing to zero. For each path, $e^{-s\tau_a}$ increases to $\mathbf{1}_{\{\tau_a<\infty\}}$, including when $\tau_a=\infty$. Monotone convergence and the formula just proved give

\mathbb{P}_x(\tau_a<\infty) =\lim_{s\downarrow0}\mathbb{E}_x[e^{-s\tau_a}] =\lim_{s\downarrow0}e^{-(a-x)\sqrt{2s}} =1.

So the level is reached almost surely, and the transform at zero is $1$. This limiting argument also specifies why the $s=0$ case is not obtained by casually substituting into an expression involving $e^{-s\infty}$.

Worked Applications

Worked Example: Starting at Zero and Hitting Level Two

Let $x=0$, $a=2$, and $s=\tfrac12$. The distance to the target is $a-x=2-0=2$, and $\sqrt{2s}=\sqrt{2\cdot\tfrac12}=1$. Substitution into the transform gives

\mathbb{E}_0[e^{-\tau_2/2}]=e^{-2\cdot1}=e^{-2}.

Check: The stopping identity has multiplier $e^{\theta(a-x)}=e^{1\cdot2}=e^2$, so the proposed transform gives $e^2e^{-2}=1$, as required by the martingale calculation.

Worked Example: A Larger Discount Rate

Let $x=-1$, $a=2$, and $s=2$. The distance is $a-x=2-(-1)=3$, and the chosen martingale parameter is $\theta=\sqrt{2s}=\sqrt{4}=2$. Therefore

\mathbb{E}_{-1}[e^{-2\tau_2}]=e^{-2\cdot3}=e^{-6}.

Check: At the target, the spatial factor in the martingale is $e^{\theta(a-x)}=e^{2\cdot3}=e^6$. Multiplying it by the computed transform gives $e^6e^{-6}=1$. Increasing the discount parameter from the previous example decreases the transform, consistent with the fact that $e^{-s\tau_a}$ decreases as $s$ increases for each nonnegative $\tau_a$.

Worked Example: A Half-Unit Martingale Parameter

Let $x=1$, $a=3$, and $s=\tfrac18$. Then $a-x=3-1=2$, and $\theta=\sqrt{2s}=\sqrt{2\cdot\tfrac18}=\sqrt{\tfrac14}=\tfrac12$. The transform is

\mathbb{E}_1[e^{-\tau_3/8}] =e^{-(1/2)\cdot2} =e^{-1}.

Check: The stopped martingale identity uses the factor $e^{\theta(a-x)}=e^{(1/2)\cdot2}=e$. The product $e\cdot e^{-1}=1$ verifies the result. The calculation also shows how the transform depends on the gap: for fixed $s$, doubling $a-x$ doubles the magnitude of the negative exponent.

A Consequence: The Mean Hitting Time Is Infinite

Almost-sure finiteness does not imply finite expectation. In this case the transform proves that the mean is infinite. For $s>0$ and $y\ge0$, the identity $(1-e^{-sy})/s=\int_0^y e^{-su}\,du$ shows that this quantity increases to $y$ as $s\downarrow0$. Since $\tau_a<\infty$ almost surely, monotone convergence yields

\mathbb{E}_x[\tau_a] =\lim_{s\downarrow0} \frac{1-\mathbb{E}_x[e^{-s\tau_a}]}{s} =\lim_{s\downarrow0} \frac{1-e^{-(a-x)\sqrt{2s}}}{s}.

Put $d=a-x>0$. As $s\downarrow0$, the numerator is asymptotic to $d\sqrt{2s}$, so the ratio is asymptotic to $d\sqrt{2}/\sqrt{s}$, which tends to infinity. Therefore $\mathbb{E}_x[\tau_a]=\infty$. The limit step is justified by the integral identity and monotone convergence; it is not an assumption that the hitting time has a density or a finite mean.

Proof. For each finite $y\ge0$, $(1-e^{-sy})/s=\int_0^y e^{-su}\,du$ increases to $y$ as $s\downarrow0$. Apply this pathwise with $y=\tau_a$, which is finite almost surely by the hitting-probability result. Monotone convergence identifies the limit of the expectations with $\mathbb{E}_x[\tau_a]$. With $d=a-x>0$, differentiability of $z\mapsto e^{-z}$ at zero gives $1-e^{-d\sqrt{2s}}\sim d\sqrt{2s}$. Dividing by $s$ gives a quantity asymptotic to $d\sqrt{2}/\sqrt{s}$, which diverges. Hence the expectation is infinite. $\square$

Why the Truncation Matters

The bounded truncation separates the paths that have hit from those that have not. Its residual term is small for a specific reason: on $\{\tau_a>T\}$, the Brownian position remains below $a$ at time $T$, while the martingale includes the decaying factor $e^{-sT}$. Both the sign of $\theta$ and the assumption $s>0$ matter in the bound. A bare appeal to optional stopping at an unbounded time would skip this control and would not establish the result.

Common mistake. A tempting but incorrect computation is to stop at $\tau_a$, write $M_{\tau_a}=e^{\theta(a-x)-s\tau_a}$, and immediately assert $\mathbb{E}_x[M_{\tau_a}]=1$ by optional stopping, without checking the stopping conditions. The resulting transform happens to match the correct answer, but that argument has not justified the unbounded-time stopping step. The correction is to stop first at $\tau_a\wedge T$, use bounded optional stopping, and show that the remaining contribution is at most $e^{\theta(a-x)-sT}$. Only after sending $T$ to infinity does the transform follow. A separate algebraic error is to drop the factor $e^{-s\tau_a}$ on the hitting event; the stopped value there is $e^{\theta(a-x)-s\tau_a}$, not just $e^{\theta(a-x)}$.

The formula gives a compact description of the one-sided hitting time without requiring its probability density. Its dependence on the starting point and target is only through the gap $a-x$, and it reveals the infinite mean despite almost-sure hitting. The bounded stopping argument is also a useful model for other problems: identify a martingale adapted to the relevant filtration, truncate an unbounded stopping time, and explicitly control what remains on paths that have not yet reached the target.

Check Your Understanding

Use the transform and the stopping argument to answer the following questions. Assume standard Brownian motion starts at $x<a$, and use $s>0$ whenever a Laplace transform is requested.

For $x=0$, $a=1$, and $s=2$, compute $\mathbb{E}_0[e^{-2\tau_1}]$ and verify it using the multiplier at the hitting point.
Why is the choice $\theta=\sqrt{2s}$ the one that makes the exponential martingale's time factor agree with the Laplace discount?
On the event $\{\tau_a>T\}$, explain why $M_T\le e^{\theta(a-x)-sT}$ when $\theta>0$.
How does taking $s\downarrow0$ in the transform establish that $\tau_a$ is finite almost surely?
Use the small-$s$ behavior of the transform to explain why the mean hitting time is infinite even though the level is reached almost surely.