I have a basic fundamental question on pricing an American option in the Black-Scholes (BS) framework: I seem to confuse two different approaches to price any early exercise,

  1. Write down a linear complimentary problem and use SOR to solve it;
  2. Solve the Black-Scholes PDE, but at every time step choose the maximum between the intrinsic value and the value from BS solution numerically.

Are these approaches equivalent or what's the difference between the two?


1 Answer 1


When you use a fully explicit finite difference scheme, you can simply apply the backward induction step and afterwards ensure that the option price at each node is at least equal to the intrinsic value. This is possible as any value $V_{i + 1, j}$ at time $\tau_{i + 1}$ only depends on the values $V_{i, j}$ at time $\tau_i$.

When you use a scheme that is at least partially implicit, then the values $V_{i + 1, j}$ at $\tau_{i + 1}$ additionally also depend on the other values at the same time step. In this case you use e.g. projected successive over-relaxation.

This issue is explicitly discussed in Chapter 78.9 of Wilmott (2006), pp. 1244ff.


Wilmott, Paul (2006) "Paul Wilmott on Quantiative Finance", John Wiley & Sons

  • $\begingroup$ I am not sure why I can't use implicit schemes. Assume I use Crank-Nicolson, solve for the value u^{n+1}, then I set u_i^{n+1}=max(u_i^{n+1}, (S_i-K)) for each i and use CN again, repeat. $\endgroup$
    – Medan
    Sep 10, 2016 at 17:57
  • $\begingroup$ As I wrote, this ignores that every value at time $\tau_{i + 1}$ depends on every other value on this time step. You can do this but it reduces the accuracy from $\mathcal{O} \left( \delta t^2 \right)$ to $\mathcal{O} \left( \delta t \right)$ which defeats the purpose of using Crank-Nicolson in the first place. See the reference that I provided - it discusses exactly this. $\endgroup$ Sep 10, 2016 at 18:01
  • $\begingroup$ I see, so basically it is simpler but loses the accuracy. Are there any downside of using SOR? Expensive or anything like that? I am curious as the approach I described is mentioned in many books so I assume it is a common practice even though it lowers the accuracy...or there are cases where I can't use SOR and have to use this FD method? $\endgroup$
    – Medan
    Sep 10, 2016 at 18:03
  • $\begingroup$ Compared to solving the linear system in the backward induction via LU decomposition, SOR is slightly slower. The books where the approach you mentioned is described probably use it in an explicit scheme or equivalently in a tree model. Regarding limitations - I am not aware of them.. $\endgroup$ Sep 10, 2016 at 18:08
  • $\begingroup$ @LocalVolatility: Can you elaborate why it lowers the accuracy to $O(\delta t)$? Is there a proof for it? Why not lowering any more or less than one order? $\endgroup$
    – Medan
    Sep 12, 2016 at 16:09

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