research36 min studyLesson 687 in Mathematics

Free-boundary problems and smooth fit

Solve a discounted Brownian stopping threshold and verify where derivative matching is legitimate.

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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 Vt+LV−rV=0V_t+\mathcal LV-rV=0. 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.

max⁡{g−V, Vt+LV−rV}=0.\max\{g-V,\ V_t+\mathcal LV-rV\}=0.
The obstacle is g−V≤0; the generator expression vanishes where waiting is optimal and is nonpositive where stopping is optimal.↔ Scroll wide equations horizontally.

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.

V(b)=g(b),V′(b−)=g′(b)when the smooth-fit hypotheses apply.V(b)=g(b),\qquad V^{\prime}(b-)=g^{\prime}(b)\quad\text{when the smooth-fit hypotheses apply}.
The derivative condition needs its own regularity justification and does not assert second-derivative matching.↔ Scroll wide equations horizontally.

A discounted Brownian threshold can be checked globally

Let X=x+σWX=x+\sigma W on the real line, use positive discount r, and gain (x−K)+(x-K)^+. For a threshold b above K, the decaying continuation solution is proportional to an exponential with rate k=2r/σk=\sqrt{2r}/\sigma. Value matching and smooth fit yield b=K+1/kb=K+1/k. 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.

k=2rσ,b=K+1k,V(x)={1kek(x−b),x<b,x−K,x≥b.k=\frac{\sqrt{2r}}{\sigma},\quad b=K+\frac1k,\qquad V(x)=\begin{cases}\dfrac1k e^{k(x-b)},&x<b,\\x-K,&x\ge b.\end{cases}
The formula uses σ>0 and r>0; its boundary lies strictly above the payoff kink K.↔ Scroll wide equations horizontally.

A threshold with matching slope but different curvature

Take σ=1\sigma=1, r=1/2r=1/2, and K=2K=2 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.

  1. The exponential rate is one. Write the continuation candidate as AexAe^x. Value matching gives Aeb=b−2Ae^b=b-2, and matching its first derivative to the stopping slope one gives Aeb=1Ae^b=1. Therefore b equals three and A equals e−3e^{-3}.

  2. The candidate is ex−3e^{x-3} below three and x−2x-2 above. At x equal to two its value is e−1e^{-1}, although immediate gain is zero. Below the boundary, ex−3≥1+(x−3)=x−2e^{x-3}\ge1+(x-3)=x-2, and positivity also dominates the zero-gain part.

  3. 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 V′′/2−V/2=0V^{\prime\prime}/2-V/2=0 holds, and in the stopping region the same operator gives −(x−2)/2≤0-(x-2)/2\le0.

V(3)=1,V′(3−)=V′(3+)=1,V′′(3−)=1≠0=V′′(3+).V(3)=1,\quad V^{\prime}(3-)=V^{\prime}(3+)=1,\quad V^{\prime\prime}(3-)=1\ne0=V^{\prime\prime}(3+).
The curvature jump is compatible with the obstacle problem and does not create a first-derivative local-time term.↔ Scroll wide equations horizontally.
The result

The optimal threshold is three and V(2)=e⁻¹≈0.367879. Smooth fit holds for first derivatives, while second derivatives do not match.

Explore the model

Interactive lab

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.

Separate value and slope matching from curvatureThis 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. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

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.

Check your understanding

Which statement about smooth fit is supported by the Brownian threshold example and its assumptions?

Try it yourself

Change to K=1K=1, diffusion coefficient two, and discount r=1/2r=1/2. 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 k=1/2k=1/2, so the threshold is b=1+2=3b=1+2=3. The value at zero is 2e−3/2≈0.4462602e^{-3/2}\approx0.446260. The continuation derivative at b is (1/k)ke0=1(1/k)k e^0=1, 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.

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.

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