Free-boundary problems and smooth fit
Solve a discounted Brownian stopping threshold and verify where derivative matching is legitimate.
On this page
The question for today
What you’ll understand
- Write the continuation equation and obstacle inequality for reward maximization.
- Determine and verify a threshold using value matching and smooth fit.
- Identify why continuous paths or a differentiable gain alone do not guarantee universal smooth fit.
An optimal stopping boundary is unknown in advance because deciding where to stop is part of the problem. Inside the continuation region the value satisfies a differential equation; inside the stopping region it equals the payoff. Matching values connects the regions, and smooth fit can add a derivative condition when the diffusion and reward have the necessary regularity. A perpetual Brownian example yields an explicit threshold and also shows why verifying the resulting candidate matters more than merely solving two matching equations.
The obstacle and continuation equations have different roles
For a Markov diffusion and discounted reward, the value dominates the immediate gain because stopping now is allowed. In a continuation region, local waiting gives the equation . In a stopping region, the value equals the gain and waiting must not improve its expected discounted value. These statements combine into a variational inequality, with a maximum for this reward convention. The unknown interface between the two regions is the free boundary.
For a perpetual time-homogeneous problem, the time derivative disappears, but behavior at infinity and admissibility of the proposed hitting time still matter. Solving the continuation ordinary differential equation introduces constants that the growth condition and boundary matching can determine. A candidate must dominate the gain everywhere and have the correct supermartingale inequality in the stopping region. Matching conditions alone can select a false candidate that fails either of these global tests.
Smooth fit reflects boundary regularity
Value matching requires the limiting continuation value to equal the stopping gain at the boundary. Smooth fit additionally equates the relevant first derivatives. For ordinary one-dimensional Brownian motion and a smooth gain near a regular threshold, this matching can be justified by the path behavior and the associated stopping verification. It does not require equality of second derivatives: a valid value function can be continuously differentiable and only piecewise twice differentiable.
Smooth fit is not a universal consequence of continuity of paths. Peskir constructs regular diffusions and differentiable gains for which ordinary spatial smooth fit fails, and explains the role of differentiability of the scale function. Jump processes, irregular boundaries, or nonsmooth gains can require different matching principles. Here the threshold lies away from the gain's kink, and Brownian scale is smooth. We will verify the resulting piecewise function directly, rather than extend its boundary rule to all stopping problems.
A discounted Brownian threshold can be checked globally
Let on the real line, use positive discount r, and gain . For a threshold b above K, the decaying continuation solution is proportional to an exponential with rate . Value matching and smooth fit yield . The candidate exceeds the gain below b by the elementary exponential tangent inequality, and above b its generator-minus-discount expression is nonpositive because the gain is positive and affine there.
The candidate is continuously differentiable at b, so the local-time contribution associated with a first-derivative jump vanishes in the generalized Itô formula. Its bounded first derivative gives a true discounted martingale term. Discounted Brownian moment bounds justify the terminal tail limit, and the threshold hitting-time Laplace transform attains the candidate below b. These checks certify the value among stopping times, taking a zero payoff on an infinite stopping time if needed.
Work it through
A threshold with matching slope but different curvature
Take , , and in the perpetual Brownian stopping problem. Determine the boundary, value at x equal to two, and the first and second one-sided derivatives at the boundary.
The exponential rate is one. Write the continuation candidate as . Value matching gives , and matching its first derivative to the stopping slope one gives . Therefore b equals three and A equals .
The candidate is below three and above. At x equal to two its value is , although immediate gain is zero. Below the boundary, , and positivity also dominates the zero-gain part.
Both first derivatives at three are one. The left second derivative is one and the right second derivative is zero. In the continuation region the equation holds, and in the stopping region the same operator gives .
The optimal threshold is three and V(2)=e⁻¹≈0.367879. Smooth fit holds for first derivatives, while second derivatives do not match.
The living notebook
Explore the model
Interactive lab · function
Separate value and slope matching from curvature
Pause and Reset, then set coefficient a to 0.5 and position x to 1. Predict value 0.5 and slope 1 for the plotted parabola. On paper compare its tangent line y=x−0.5: it has the same value and slope at one, but zero curvature rather than the parabola's curvature one. This illustrates what first-order matching does and does not require. The instrument does not solve the exponential stopping problem or locate a free boundary. Reset and set both controls before a fresh comparison; edits keep elapsed time.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forEqual boundary values and first derivatives do not impose equal curvature or prove a stopping candidate is globally valid.
Pause & reason
Check your understanding
Your turn
Try it yourself
Change to , diffusion coefficient two, and discount . Compute the exponential rate, threshold, value at zero, and the continuation-side slope at the threshold.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The rate scales inversely with diffusion. Use the general value formula including its 1/k prefactor.
Compare with the solution
The rate is , so the threshold is . The value at zero is . The continuation derivative at b is , matching the stopping slope. The unchanged numerical boundary compared with the worked example comes from the changed strike and rate offsetting each other, not from diffusion being irrelevant.
Keep the idea
Take it with you
- An obstacle problem combines gain domination with a continuation generator equation.
- Smooth fit needs boundary and process regularity beyond mere continuity.
- Global domination, martingale control, and tail limits verify a matched threshold candidate.
Read further
Original sources and deeper explanations to follow at your own pace.
Optional progress, saved in this browser when available. Revisit any lesson at any time.
Lesson not marked complete.