Free Boundary Problems and American Options¶

An American option holder can exercise at any moment, so at every $(S,t)$ they compare "keep waiting" against "cash $\Phi(S)$ now" and pick whichever is larger. The "keep waiting" value follows the Black-Scholes PDE, the "cash now" value is $\Phi(S)$, and the actual price $V$ is the upper envelope of the two, glued along an unknown curve $S^*(t)$. That curve -- chosen by the option, not specified by us -- is the free boundary, and the resulting variational inequality (obstacle problem) is what every numerical method in this folder discretizes.

The American Option Problem¶

Recall (see § American Options): under risk-neutral dynamics (see § Risk-neutral measure) the price

\[ V(t,S) = \sup_{\tau \in \mathcal{T}_{t,T}} \mathbb{E}^{\mathbb{Q}}\!\left[e^{-r(\tau-t)}\Phi(S_\tau) \mid S_t = S\right] \]

satisfies the variational inequality $\min(-\partial_t V - \mathcal{L}V + rV,\; V - \Phi) = 0$ with $\mathcal{L}V = \tfrac12\sigma^2 S^2 V_{SS} + rS V_S$, splitting the domain into a continuation region $\{V>\Phi\}$ (PDE holds) and an exercise region $\{V=\Phi\}$, separated by the free boundary $S^*(t)$ with smooth-pasting $V=\Phi$, $V_S=\Phi'$. The early-exercise premium $V_\text{Am}-V_\text{Eu}\ge 0$ is the value of this right.

Numerical Methods: Projection¶

Simple Projection Method¶

After each time step, project onto the constraint:

\[ \boxed{ u_j^{n+1} \leftarrow \max(u_j^{n+1}, \Phi_j) } \]

Algorithm (Implicit + Projection)¶

1. Solve (I - Δτ A)ũ^{n+1} = u^n (unconstrained) 2. Project: u_j^{n+1} = max(ũ_j^{n+1}, Φ_j) for all j 3. Repeat for next time step

Issues¶

Simple but may lose accuracy near the free boundary
First-order in time (even with Crank-Nicolson base scheme)
Smooth pasting not explicitly enforced

Linear Complementarity Problem (LCP)¶

Recall (see § Linear Complementarity Formulation): after implicit time stepping with $L = I + \Delta\tau A$ and $\mathbf{f} = \mathbf{u}^n$, the time step becomes

\[ L\mathbf{u}\ge\mathbf{f},\quad \mathbf{u}\ge\boldsymbol{\Phi},\quad (L\mathbf{u}-\mathbf{f})^T(\mathbf{u}-\boldsymbol{\Phi})=0, \]

with the continuation / exercise dichotomy described there.

PSOR Algorithm¶

Recall (see § PSOR Algorithm): Projected SOR solves the LCP by combining an SOR sweep with a componentwise projection $u_j \leftarrow \max(\tilde u_j, \Phi_j)$. With $\omega\in(1,2)$ (often $\omega\approx 1.2$), convergence is linear at typically 5-20 iterations per time step.

Penalty Method¶

Recall (see § Penalty Method): replacing the hard constraint by a soft penalty $\rho(V-\Phi)^-$ yields the nonlinear PDE

\[ V_t + \mathcal{L}V - rV + \rho(V-\Phi)^- = 0, \]

with error $O(1/\rho)$ and typical $\rho = 10^6$-$10^8$.

Brennan-Schwartz Algorithm¶

For American puts specifically, there's an elegant direct method.

Key Observation¶

The exercise boundary $S^*(t)$ increases as $t \to T$ (from below).

Algorithm¶

Work backward from maturity:

Find the grid point where exercise first becomes optimal
Set $u_j = \Phi_j$ for $j \leq j^*$
Solve PDE only for $j > j^*$

Advantage¶

No iteration required—direct solve at each time step.

Front-Fixing Methods¶

Transform coordinates to fix the free boundary.

Coordinate Transform¶

Let $\xi = S/S^*(t)$ so that the boundary is always at $\xi = 1$.

Transformed PDE¶

Additional terms appear involving $\dot{S}^*$, which must be determined as part of the solution.

Advantage¶

High accuracy near the free boundary.

Disadvantage¶

More complex implementation; boundary must be tracked.

Convergence and Accuracy¶

Order of Convergence¶

Method	Time Order	Space Order
Projection + Implicit	1	2
PSOR + C-N	1-1.5	2
Penalty	1	2
Front-fixing	2	2

Note: The free boundary introduces non-smoothness that typically limits time accuracy to first-order.

Grid Considerations¶

Place grid points near expected $S^*$
Adaptive grids can improve efficiency
Smoothing at maturity helps (Rannacher)

Comparison of Methods¶

Method	Ease	Accuracy	Speed
Projection	Easy	Low	Fast
PSOR	Moderate	Good	Moderate
Penalty	Easy	Good	Fast
Front-fixing	Hard	High	Moderate

Summary¶

\[ \boxed{ \min\left(-\frac{\partial V}{\partial t} - \mathcal{L}V + rV, \; V - \Phi\right) = 0 } \]

Aspect	Description
Problem type	Free boundary / obstacle problem
Constraint	$V \geq \Phi$ (can exercise anytime)
Free boundary	$S^*(t)$ separates exercise/continuation
Smooth pasting	$V$ and $V_S$ continuous at boundary

Method	Key Idea
Projection	Enforce $u \geq \Phi$ after each step
PSOR	Iteratively solve LCP
Penalty	Soft constraint via large $\rho$
Front-fixing	Track boundary explicitly

American option pricing requires specialized numerical techniques to handle the free boundary between continuation and exercise regions.

Exercises¶

Exercise 1. For an American put with $K = 100$, $r = 0.05$, $\sigma = 0.20$, explain why the exercise boundary $S^*(t)$ satisfies $S^*(t) < K$ for $t < T$ and $S^*(T) = K$. Why is early exercise never optimal for an American call on a non-dividend-paying stock?

Solution to Exercise 1

Why $S^*(t) < K$ for $t < T$:

At the exercise boundary $S^*(t)$, the holder exercises the put and receives $K - S^*(t)$. If $S^*(t) = K$, the exercise payoff would be zero, so there would be no reason to exercise. Hence $S^*(t) < K$.

More precisely, the holder exercises when the expected gain from holding (continuation value) equals the intrinsic value. Since $r > 0$, holding the put when the stock is very low ties up the option and delays receiving the strike cash $K$. The interest lost on $K$ makes early exercise optimal once $S$ falls below some critical level $S^*(t)$ that is strictly less than $K$.

Why $S^*(T) = K$:

At expiration, the put is exercised whenever $S < K$ (the payoff $(K-S)^+$ is positive). There is no remaining time value, so $S^*(T) = K$.

Why early exercise of an American call on a non-dividend-paying stock is never optimal:

For a non-dividend-paying stock, the European call price satisfies $C_E \geq S - Ke^{-r(T-t)}$. Since $e^{-r(T-t)} < 1$ for $t < T$, we have

\[ C_E \geq S - Ke^{-r(T-t)} > S - K = \Phi(S) \]

for in-the-money options. The call value always exceeds the immediate exercise payoff, so early exercise is never optimal. Intuitively, exercising early means paying $K$ now rather than later, forfeiting interest on $K$, and there are no dividends to offset this cost.

Exercise 2. State the smooth pasting conditions at the free boundary $S = S^*(t)$ for an American put. Explain geometrically why both $V(t, S^*) = K - S^*$ and $V_S(t, S^*) = -1$ must hold simultaneously.

Solution to Exercise 2

The smooth pasting conditions at $S = S^*(t)$ for an American put with payoff $\Phi(S) = (K - S)^+$ are:

Value matching: $V(t, S^*(t)) = K - S^*(t)$

Slope matching: $V_S(t, S^*(t)) = -1$

Geometric interpretation:

The value matching condition says that the option value curve meets the intrinsic value line $K - S$ at the boundary. This is simply the requirement that the option value is continuous.

The slope matching condition ($V_S = -1$) requires that the option value curve is tangent to the intrinsic value line at the boundary. Geometrically, the option value curve "kisses" the payoff line from above rather than meeting it at a corner.

If the slope were discontinuous (say $V_S(t, S^{*+}) \neq -1$), then a small perturbation of the boundary location would either increase the option value (contradicting optimality) or create an arbitrage opportunity. Specifically:

If $|V_S| < 1$ at the boundary, the holder could improve the stopping strategy by exercising slightly earlier (lower $S^*$).
If $|V_S| > 1$ at the boundary, the holder could improve by exercising slightly later (higher $S^*$).

The optimal boundary is the one where no local improvement is possible, which requires the tangency condition.

Exercise 3. The simple projection method enforces $u_j^{n+1} \leftarrow \max(u_j^{n+1}, \Phi_j)$ after each time step. Suppose at a given time step, the unconstrained implicit solve yields $\tilde{u} = (48, 12, 3, 0.5)^T$ on a grid with $S = (40, 80, 120, 160)$ and the put payoff is $\Phi = (60, 20, 0, 0)^T$. Apply the projection and identify the exercise and continuation regions.

Solution to Exercise 3

Given the unconstrained solution $\tilde{u} = (48, 12, 3, 0.5)^T$ on the grid $S = (40, 80, 120, 160)$ and put payoff $\Phi = (60, 20, 0, 0)^T$:

Apply projection $u_j = \max(\tilde{u}_j, \Phi_j)$ componentwise:

$j = 1$: $u_1 = \max(48, 60) = 60$ (constraint active)
$j = 2$: $u_2 = \max(12, 20) = 20$ (constraint active)
$j = 3$: $u_3 = \max(3, 0) = 3$ (unconstrained)
$j = 4$: $u_4 = \max(0.5, 0) = 0.5$ (unconstrained)

The projected solution is $\mathbf{u} = (60, 20, 3, 0.5)^T$.

Exercise region: $\{S = 40, S = 80\}$ where $u_j = \Phi_j$ (the constraint binds and immediate exercise is optimal).

Continuation region: $\{S = 120, S = 160\}$ where $u_j > \Phi_j$ (the PDE solution exceeds the payoff and holding is optimal).

The free boundary lies between $S = 80$ and $S = 120$, which is consistent with the exercise boundary being below the strike $K = 100$.

Exercise 4. Write out the linear complementarity problem (LCP) for a single time step of the implicit scheme: $(I - \Delta\tau A)\mathbf{u}^{n+1} \geq \mathbf{u}^n$, $\mathbf{u}^{n+1} \geq \boldsymbol{\Phi}$, $(L\mathbf{u}^{n+1} - \mathbf{u}^n)^T(\mathbf{u}^{n+1} - \boldsymbol{\Phi}) = 0$. Explain the complementarity condition in terms of the exercise and continuation regions.

Solution to Exercise 4

The linear complementarity problem for a single time step with $L = I - \Delta\tau A$ (or more precisely $L = I + \Delta\tau A$ in the time-to-maturity formulation) is:

Condition 1 (PDE residual non-negative):

\[ L\mathbf{u}^{n+1} - \mathbf{u}^n \geq \mathbf{0} \]

This means the discrete PDE residual is non-negative at every grid point.

Condition 2 (Early exercise constraint):

\[ \mathbf{u}^{n+1} \geq \boldsymbol{\Phi} \]

The option value must be at or above the payoff at every grid point.

Condition 3 (Complementarity):

\[ (L\mathbf{u}^{n+1} - \mathbf{u}^n)^T(\mathbf{u}^{n+1} - \boldsymbol{\Phi}) = 0 \]

Interpretation: The complementarity condition requires that for each grid point $j$, at least one of the two inequalities holds with equality:

Continuation region ($u_j^{n+1} > \Phi_j$): Since the factor $(u_j^{n+1} - \Phi_j) > 0$, the complementarity condition forces $(L\mathbf{u}^{n+1} - \mathbf{u}^n)_j = 0$. The discrete PDE is satisfied exactly. The option is worth more than immediate exercise, so the holder continues.
Exercise region ($u_j^{n+1} = \Phi_j$): The factor $(u_j^{n+1} - \Phi_j) = 0$, so the product is zero regardless of the residual. The PDE residual satisfies $(L\mathbf{u}^{n+1} - \mathbf{u}^n)_j \geq 0$, meaning the option would be "overpriced" by the PDE alone --- the continuation value is below the payoff, so exercising is optimal.

The two conditions cannot simultaneously have strict inequality: we cannot have both $u_j > \Phi_j$ and $(L\mathbf{u})_j > q_j$, because the first implies the PDE holds with equality.

Exercise 5. The penalty method replaces the hard constraint with $\rho(V - \Phi)^-$ for large $\rho$. If $\rho = 10^6$, estimate the penalty approximation error. What value of $\rho$ would be needed to make the penalty error smaller than the discretization error on a grid with $\Delta S = 1$ and $\Delta\tau = 0.01$?

Solution to Exercise 5

Penalty approximation error with $\rho = 10^6$:

The error bound is $|V - V^\rho| \leq C/\rho$, so for $\rho = 10^6$, the penalty error is of order $10^{-6}$ (times a constant $C$ that depends on the problem parameters). In dollar terms, this is negligible for option prices of order $\$1$-$\$100$.

Balancing penalty and discretization errors:

The discretization error for a second-order scheme in space and first-order in time (typical near the free boundary) is:

\[ E_{\text{disc}} \sim O(\Delta\tau) + O(\Delta S^2) \]

With $\Delta S = 1$ and $\Delta\tau = 0.01$:

\[ E_{\text{disc}} \sim 0.01 + 1 = O(1) \]

The dominant discretization error is $O(\Delta S^2) = O(1)$. To make the penalty error smaller:

\[ \frac{C}{\rho} < \Delta S^2 = 1 \]

So $\rho > C$ suffices. Even $\rho = 10^2$ would make the penalty error negligible relative to the spatial discretization error.

However, for the time discretization error $O(\Delta\tau) = O(0.01)$, we need $\rho > C/0.01 = 100C$, so $\rho \approx 10^3$ to $10^4$ is sufficient (depending on the constant $C$, typically $C \sim O(1)$ to $O(10)$).

In practice, $\rho = 10^4$ is more than adequate for this grid resolution.

Exercise 6. Compare the four methods for American option pricing (projection, PSOR, penalty, front-fixing) in terms of implementation complexity, accuracy near the free boundary, and computational cost per time step. Which method would you recommend for a production implementation pricing American puts on a single underlying?

Solution to Exercise 6

Projection method:

Implementation: Trivial --- solve the unconstrained system, then apply $u_j \leftarrow \max(u_j, \Phi_j)$.
Accuracy: Low near the free boundary due to splitting error (the PDE and constraint are enforced sequentially, not simultaneously).
Cost: One tridiagonal solve per time step (cheapest).

PSOR:

Implementation: Moderate --- requires the LCP formulation, the sweep loop, and tuning of $\omega$.
Accuracy: Good --- solves the LCP exactly (to iteration tolerance), so the constraint and PDE are coupled.
Cost: 5-20 sweeps per time step, each costing $O(M)$. Total cost is moderate.

Penalty method:

Implementation: Easy --- add a penalty term to the existing PDE solver and use Newton iteration.
Accuracy: Good --- error $O(1/\rho)$ from penalization, but no splitting error. With $\rho = 10^6$, the penalty error is negligible.
Cost: 2-4 Newton iterations per time step, each requiring one tridiagonal solve. Comparable to or slightly more than projection.

Front-fixing:

Implementation: Hard --- requires a coordinate transformation, tracking the boundary $S^*(t)$, and solving the transformed PDE with additional terms.
Accuracy: High --- the free boundary is resolved exactly (by construction), and smooth pasting is enforced.
Cost: One PDE solve per time step plus boundary update. Moderate total cost.

Recommendation for production: The penalty method offers the best balance of ease of implementation, accuracy, and speed for a single-asset American put. It integrates seamlessly into existing finite difference infrastructure, has negligible approximation error for reasonable $\rho$, and requires only a few Newton iterations. PSOR is a close second and is preferred if exact LCP solutions (without penalty approximation) are required.

Exercise 7. The early exercise premium is defined as $V_{\text{American}} - V_{\text{European}}$. For an at-the-money put with $K = 100$, $r = 0.05$, $\sigma = 0.20$, $T = 1$, the European put price is approximately $\$5.57$. If a numerical solver gives an American put price of $\$6.08$, what is the early exercise premium? Is this magnitude reasonable?

Solution to Exercise 7

The early exercise premium is:

\[ V_{\text{American}} - V_{\text{European}} = \$6.08 - \$5.57 = \$0.51 \]

As a percentage of the European price:

\[ \frac{0.51}{5.57} \approx 9.2\% \]

Is this magnitude reasonable?

Yes, this is a reasonable magnitude. For an at-the-money put with these parameters:

The interest rate $r = 0.05$ creates an incentive for early exercise (receiving $K$ earlier earns interest).
The volatility $\sigma = 0.20$ creates an incentive to wait (the option might become more valuable).
For $T = 1$ year, the early exercise premium is typically 5-15% of the European price for moderate interest rates and volatilities.

The premium of approximately 9% of the European price falls squarely within the expected range. As a rough check, the maximum possible early exercise premium for a put is bounded by the interest savings from receiving $K$ immediately:

\[ K(1 - e^{-rT}) = 100(1 - e^{-0.05}) \approx \$4.88 \]

The observed premium of $\$0.51$ is well below this upper bound, which is consistent since only deep in-the-money options would capture the full interest benefit.

Method	Key Idea
Projection	Enforce \(u \geq \Phi\) after each step
PSOR	Iteratively solve LCP
Penalty	Soft constraint via large \(\rho\)
Front-fixing	Track boundary explicitly