Brownian Bridge™

1. Why does an American put on a non-dividend-paying stock have positive probability of early exercise, while an American call on the same stock should never be exercised early?

Calls have unlimited upside; puts do notA put can earn interest on the strike by exercising; a call would sacrifice time value with no dividend to offset itThe put's delta is negative; the call's delta is positiveBlack-Scholes assumes puts are always exercised early

2. The linear complementarity problem (LCP) for the American put requires three simultaneous conditions. At a point in the holding region ( $S > S^*(t)$ ), which conditions hold?

\mathcal{L}V = 0

and

V = g

and

\mathcal{L}V \cdot (V - g) = 0

\mathcal{L}V = 0

and

V > g

— the PDE holds exactly, option exceeds intrinsic value

\mathcal{L}V > 0

and

V = g

— early exercise is forced

\mathcal{L}V > 0

and

V > g

— both conditions active

3. In the PSOR update rule, what does the 'projected' step $V_i^{\text{new}} = \max(g_i, \tilde{V}_i)$ accomplish?

It eliminates numerical oscillations from Crank-NicolsonIt enforces the early exercise constraint

V_i \geq g_i

at each node during each iteration, without needing to know the free boundary locationIt over-relaxes the solution to speed up convergence by a factor of

\omega

It projects the solution onto the space of smooth functions

4. The PSOR algorithm for the American option LCP converges for all $\omega \in (0, 2)$ when the discretisation matrix $A$ is an M-matrix.

truefalse

5. Compared to a binomial tree at matched computational cost, a PSOR finite difference scheme with Crank-Nicolson achieves better accuracy for the American put primarily because:

The binomial tree cannot price American puts at allCN is second-order in time (

O(\Delta\tau^2)

) vs. the binomial tree's

O(1/\sqrt{N})

convergence due to the staircase approximation of the free boundaryPSOR avoids all numerical errors; the binomial tree is an approximationThe binomial tree requires more memory than a finite difference grid

6. The free boundary $S^*(t)$ for the American put satisfies $S^*(T) = ?$ and $\partial S^*/\partial t = ?$ near expiry.

S^*(T) = K

and

\partial S^*/\partial t \to 0

t \to T

S^*(T) = K

and

S^*(t) \to K

at rate

O(\sqrt{(T-t)|\ln(T-t)|})

t \to T

S^*(T) = 0

and

S^*(t)

increases monotonically from 0 to

K

S^*(T) = Ke^{-rT}

(the discounted strike)

Quiz: PSOR for American Options

Quick Quiz