American Option Pricing with PSOR on the FD Grid

Hard finite-difference-methods · Module 3 · ~35 min

Cell 1 — Shared utilities: BS analytic, Thomas algorithm, PSOR solver

Foundation cell: Black-Scholes put/call for benchmarking, the Thomas algorithm from Module 2, and the PSOR solver for American puts. The PSOR solver accepts a relaxation parameter omega and a convergence tolerance tol, and returns the price, the final grid, the node array, and the total iteration count across all time steps.

import numpy as np
from scipy.stats import norm

# ── Black-Scholes analytic ────────────────────────────────────────────────
def bs_call(S, K, T, r, sigma):
    if T <= 0: return max(S - K, 0.0)
    d1 = (np.log(S / K) + (r + 0.5*sigma**2)*T) / (sigma*np.sqrt(T))
    d2 = d1 - sigma*np.sqrt(T)
    return S*norm.cdf(d1) - K*np.exp(-r*T)*norm.cdf(d2)

def bs_put(S, K, T, r, sigma):
    return bs_call(S, K, T, r, sigma) - S + K*np.exp(-r*T)

# ── Thomas algorithm ──────────────────────────────────────────────────────
def thomas_solve(lower, main, upper, rhs):
    n = len(main)
    w = main.copy().astype(float)
    d = rhs.copy().astype(float)
    for i in range(1, n):
        m = lower[i] / w[i-1]
        w[i] -= m * upper[i-1]
        d[i] -= m * d[i-1]
    x = np.zeros(n)
    x[-1] = d[-1] / w[-1]
    for i in range(n-2, -1, -1):
        x[i] = (d[i] - upper[i]*x[i+1]) / w[i]
    return x

# ── PSOR solver for American put (fully implicit scheme) ─────────────────
def fd_american_put_psor(S0, K, T, r, sigma, M=150, N=150, x_width=4.0,
                          omega=1.5, tol=1e-8, max_iter=500):
    """
    Price an American put via PSOR on the log-price implicit FD grid.

Parameters
    ----------
    omega     : PSOR relaxation parameter (0 < omega < 2); typical 1.2–1.5
    tol       : convergence criterion: max|V_new - V_old| < tol
    max_iter  : maximum PSOR iterations per time step

Returns
    -------
    price      : float — interpolated V(0, S0)
    V_final    : ndarray (M+1,) — option value at t=0 for all grid nodes
    x_nodes    : ndarray (M+1,)
    boundary   : list of (tau, S*) pairs — early exercise boundary
    total_iters: int — total PSOR iterations across all time steps
    """
    D = 0.5 * sigma**2
    c = r - D

x_min = np.log(K) - x_width * sigma * np.sqrt(T)
    x_max = np.log(K) + x_width * sigma * np.sqrt(T)
    dx = (x_max - x_min) / M
    x_nodes = np.linspace(x_min, x_max, M + 1)
    S_nodes = np.exp(x_nodes)
    dt = T / N

# Coupling coefficients (fully implicit, theta=1)
    r_D = D / dx**2
    r_c = c / (2.0 * dx)
    a_alpha = dt * (r_D - r_c)
    a_gamma = dt * (r_D + r_c)
    a_beta  = dt * (2.0 * r_D + r)
    d_ii    = 1.0 + a_beta              # main diagonal of A

# Payoff: American put
    h = np.maximum(K - S_nodes, 0.0)

# Terminal condition (tau=0): put payoff
    V = h.copy()

boundary    = []
    total_iters = 0

for n_step in range(N):
        tau_next = (n_step + 1) * dt

# Boundary conditions for the American put
        # Left (S→0): exercise immediately → full intrinsic value
        V_bc_left  = K - S_nodes[0]          # deep ITM, always exercise
        # Right (S→∞): put worthless
        V_bc_right = 0.0

# RHS for the fully implicit scheme: just the old values
        rhs = V[1:-1].copy()
        rhs[0]  += a_alpha * V_bc_left
        rhs[-1] += a_gamma * V_bc_right

# PSOR iteration
        V_int = V[1:-1].copy()           # initial guess = old interior values
        h_int = h[1:-1]
        n_int = M - 1

for iteration in range(max_iter):
            V_prev = V_int.copy()

for i in range(n_int):
                # Gauss-Seidel update (uses latest V_int[j] for j < i)
                gs = rhs[i]
                if i > 0:
                    gs += a_alpha * V_int[i - 1]
                if i < n_int - 1:
                    gs += a_gamma * V_int[i + 1]

V_hat = gs / d_ii

# Over-relaxation
                V_sor = (1.0 - omega) * V_int[i] + omega * V_hat

# Projection: enforce early exercise floor
                V_int[i] = max(V_sor, h_int[i])

if np.max(np.abs(V_int - V_prev)) < tol:
                total_iters += iteration + 1
                break
        else:
            total_iters += max_iter

# Reconstruct full solution
        V = np.empty(M + 1)
        V[0]    = V_bc_left
        V[1:-1] = V_int
        V[-1]   = V_bc_right

# Extract early exercise boundary: last node where V == h
        exercise_mask = (V[1:-1] <= h_int + 1e-10)
        if exercise_mask.any():
            last_exercise = np.where(exercise_mask)[0][-1] + 1  # +1 for offset
            # Linear interpolation for smoother boundary estimate
            if last_exercise < M:
                S_star = float(S_nodes[last_exercise])
            else:
                S_star = float(S_nodes[-1])
            boundary.append((tau_next, S_star))

price = float(np.interp(np.log(S0), x_nodes, V))
    return price, V, x_nodes, boundary, total_iters

# ── Smoke test ────────────────────────────────────────────────────────────
S0, K, T, r, sigma = 100.0, 100.0, 1.0, 0.05, 0.20

eu_put  = bs_put(S0, K, T, r, sigma)
am_put, _, _, _, iters = fd_american_put_psor(S0, K, T, r, sigma)

print(f"European put (BS analytic) : {eu_put:.6f}")
print(f"American put (PSOR FD)     : {am_put:.6f}")
print(f"Early exercise premium     : {am_put - eu_put:.6f}")
print(f"Total PSOR iterations      : {iters}")
print()
print("Expected: American put > European put (premium > 0).")
print("Setup complete.")

Run cell to see output.

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Cell 2 — Early exercise premium: American vs European put over a spot ladder

Price both the American and European put across S ∈ {70, 80, 90, 100, 110}. Compute the early exercise premium at each spot. The premium should be largest for deep ITM puts (where exercising to earn interest on K is most attractive) and zero for deep OTM puts.

Expected: Premium is ~0 at S=110 (OTM, worthless to exercise), positive and growing as S decreases. At S=70 the put is deep ITM and the premium is significant.

Run cell to see output.

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Cell 3 — Early exercise boundary: S*(τ) from τ=0 to τ=T

Extract and plot the early exercise boundary $S^*(\tau)$ from the PSOR solver. The boundary should: start near $K = 100$ at $\tau = 0$, decrease monotonically as $\tau$ increases (longer time means more time value → higher threshold for exercise), and approach a limiting value $S^*(\infty) < K$ at long maturities.

Expected: S*(0) ≈ K = 100, then decreasing. Near τ = 0, the boundary approaches K with square-root speed (steep slope). At T=1, S*(T) is typically around 75–85 for these parameters.

# Early exercise boundary S*(τ)
K, T, r, sigma = 100.0, 1.0, 0.05, 0.20
M, N = 200, 200

_, _, _, boundary, _ = fd_american_put_psor(
    100.0, K, T, r, sigma, M=M, N=N, omega=1.5
)

taus  = np.array([b[0] for b in boundary])
S_star = np.array([b[1] for b in boundary])

# Print a subset of the boundary
print(f"Early exercise boundary S*(τ) for K={K}, r={r}, σ={sigma}, T={T}\n")
print(f"{'τ':>8}  {'S*(τ)':>10}  {'S*(τ)/K':>10}")
print("-" * 34)
step = max(1, len(taus) // 20)
for i in range(0, len(taus), step):
    print(f"{taus[i]:>8.4f}  {S_star[i]:>10.4f}  {S_star[i]/K:>10.4f}")

# Key checks
monotone = bool(np.all(np.diff(S_star) <= 0.5))  # allow small grid noise
starts_near_K = S_star[0] >= 0.90 * K
ends_below_K  = S_star[-1] < K

print()
print(f"Boundary starts near K (S*(0) ≥ 0.9K): {'PASS' if starts_near_K else 'FAIL'}")
print(f"Boundary ends below K at T=1:           {'PASS' if ends_below_K  else 'FAIL'}")
print(f"Boundary is (broadly) decreasing:       {'PASS' if monotone      else 'FAIL'}")
print()
print(f"Final boundary value: S*({T}) ≈ {S_star[-1]:.2f}")
print(f"(theoretical: typically 75-85 for these parameters)")

Run cell to see output.

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Cell 4 — PSOR convergence: iteration count vs relaxation parameter ω

Sweep $\omega \in \{1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9\}$ and record the total PSOR iteration count across all time steps. The optimal $\omega$ minimises the iteration count. $\omega = 1$ (Gauss-Seidel) should require the most iterations; $\omega \in [1.4, 1.7]$ should be near-optimal for typical grids.

Expected: Iteration count is convex in ω, minimised somewhere in [1.4, 1.7]. ω=1 (Gauss-Seidel) requires roughly 1.5–3× more iterations than the near-optimal ω.

# PSOR convergence: total iterations vs omega
S0, K, T, r, sigma = 100.0, 100.0, 1.0, 0.05, 0.20
M, N = 100, 100

omega_values = [1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9]
iter_counts  = []

print(f"PSOR iteration count vs ω  (M={M}, N={N})\n")
print(f"{'ω':>6}  {'Total iters':>14}  {'Iters/step':>12}")
print("-" * 38)

for omega in omega_values:
    _, _, _, _, iters = fd_american_put_psor(
        S0, K, T, r, sigma, M=M, N=N, omega=omega, tol=1e-8
    )
    iter_counts.append(iters)
    print(f"{omega:>6.1f}  {iters:>14}  {iters/N:>12.2f}")

iter_counts = np.array(iter_counts)
best_idx = int(np.argmin(iter_counts))
best_omega = omega_values[best_idx]
gs_iters   = iter_counts[0]   # omega=1.0 (Gauss-Seidel)
best_iters = iter_counts[best_idx]

print()
print(f"Optimal ω: {best_omega:.1f}  (total iters: {best_iters})")
print(f"Gauss-Seidel (ω=1): {gs_iters} iters")
print(f"Speedup over Gauss-Seidel: {gs_iters / best_iters:.2f}×")
print()

omega_pass = 1.3 <= best_omega <= 1.8
speedup_pass = gs_iters / best_iters >= 1.2
print(f"Optimal ω in [1.3, 1.8]:        {'PASS' if omega_pass   else 'FAIL'}")
print(f"Over-relaxation speedup ≥ 1.2×: {'PASS' if speedup_pass else 'FAIL'}")

Run cell to see output.

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Cell 5 — American call on non-dividend stock: PSOR returns European price

Implement a PSOR pricer for an American call (payoff $h = (S-K)^+$). Verify Merton's theorem: the PSOR projection never binds (the call LCP solution equals the unconstrained European solution), so the American call price equals the European call price exactly.

# American call on non-dividend stock: PSOR should return the European price
def fd_american_call_psor(S0, K, T, r, sigma, M=150, N=150, x_width=4.0,
                           omega=1.5, tol=1e-8, max_iter=500):
    """
    Same structure as the put solver but for a call.
    Payoff h(S) = max(S - K, 0).
    Boundary conditions: left (S→0): V→0; right (S→∞): V→S - K·exp(-rτ).
    """
    D = 0.5 * sigma**2
    c = r - D
    x_min = np.log(K) - x_width * sigma * np.sqrt(T)
    x_max = np.log(K) + x_width * sigma * np.sqrt(T)
    dx = (x_max - x_min) / M
    x_nodes = np.linspace(x_min, x_max, M + 1)
    S_nodes = np.exp(x_nodes)
    dt = T / N

r_D = D / dx**2
    r_c = c / (2.0 * dx)
    a_alpha = dt * (r_D - r_c)
    a_gamma = dt * (r_D + r_c)
    a_beta  = dt * (2.0 * r_D + r)
    d_ii    = 1.0 + a_beta

h = np.maximum(S_nodes - K, 0.0)
    V = h.copy()

n_projections = 0   # count how many times projection actually binds

for n_step in range(N):
        tau_next = (n_step + 1) * dt
        V_bc_left  = 0.0
        V_bc_right = S_nodes[-1] - K * np.exp(-r * tau_next)

rhs = V[1:-1].copy()
        rhs[0]  += a_alpha * V_bc_left
        rhs[-1] += a_gamma * V_bc_right

V_int = V[1:-1].copy()
        h_int = h[1:-1]
        n_int = M - 1

for iteration in range(max_iter):
            V_prev = V_int.copy()
            for i in range(n_int):
                gs = rhs[i]
                if i > 0:       gs += a_alpha * V_int[i-1]
                if i < n_int-1: gs += a_gamma * V_int[i+1]
                V_hat = gs / d_ii
                V_sor = (1.0 - omega) * V_int[i] + omega * V_hat
                before = V_sor
                V_int[i] = max(V_sor, h_int[i])
                if V_int[i] > before + 1e-12:
                    n_projections += 1
            if np.max(np.abs(V_int - V_prev)) < tol:
                break

V = np.empty(M + 1)
        V[0]    = V_bc_left
        V[1:-1] = V_int
        V[-1]   = V_bc_right

price = float(np.interp(np.log(S0), x_nodes, V))
    return price, n_projections

# Test across a spot ladder
K, T, r, sigma = 100.0, 1.0, 0.05, 0.20
spots = [80, 90, 100, 110, 120]

print(f"American call = European call (Merton's theorem)")
print(f"K={K}, T={T}, r={r}, σ={sigma}\n")
print(f"{'Spot':>6}  {'EU call (BS)':>14}  {'AM call (PSOR)':>16}  {'Diff':>10}")
print("-" * 54)

merton_pass = True
for S in spots:
    eu = bs_call(S, K, T, r, sigma)
    am, n_proj = fd_american_call_psor(S, K, T, r, sigma)
    diff = abs(am - eu)
    if diff > 0.005:
        merton_pass = False
    print(f"{S:>6.0f}  {eu:>14.6f}  {am:>16.6f}  {diff:>10.2e}")

print()
print(f"American call ≈ European call (max diff < 0.005): {'PASS' if merton_pass else 'FAIL'}")
print()
print("Note: small residual difference is FD discretisation error, not early exercise premium.")
print("The projection step never binds for a call on a non-dividend paying stock,")
print("confirming that it is never optimal to exercise early.")

Run cell to see output.

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Cell 6 — Validation summary

Aggregate all checks from this notebook into a final pass/fail report. Run all preceding cells first.

# Final validation summary — run all cells first
print("=" * 60)
print("  VALIDATION SUMMARY: American Options with PSOR")
print("=" * 60)

checks = {
    "American put > European put (all spots)"  : premium_positive_for_itm,
    "OTM put premium near zero"                : premium_near_zero_for_otm,
    "Free boundary starts near K"              : starts_near_K,
    "Free boundary is monotonically decreasing": monotone,
    "Optimal ω in [1.3, 1.8]"                 : omega_pass,
    "Over-relaxation speedup ≥ 1.2×"          : speedup_pass,
    "American call = European call (Merton)"   : merton_pass,
}

all_pass = True
for label, result in checks.items():
    status = "PASS" if result else "FAIL"
    icon   = "✓" if result else "✗"
    if not result:
        all_pass = False
    print(f"  {icon}  {label:<44} {status}")

print("-" * 60)
print(f"  Overall: {'ALL CHECKS PASSED' if all_pass else 'SOME CHECKS FAILED'}")
print("=" * 60)
print()
print("Key facts confirmed by this notebook:")
print("  • American put > European put — early exercise premium is real.")
print("  • Free boundary S*(τ) starts at K and decreases with τ.")
print("  • Over-relaxation (ω > 1) reduces PSOR iteration count.")
print("  • Merton's theorem: American call = European call (no dividends).")
print("  • PSOR projection never binds for American calls.")

Run cell to see output.

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