Why Longstaff–Schwartz Values Exercise at Time Zero Separately

Summary

This note asks why an analysis of the Longstaff–Schwartz method defines its approximate initial option value as the maximum of immediate exercise value and the expected payoff from a stopping rule that begins at the next time step. It contrasts this with taking the expected payoff under a stopping time that also allows exercise at time zero.

The distinction matters in Monte Carlo implementation: maximizing the immediate payoff against the sample average differs from averaging the pathwise maximum. The note points out that the former can introduce a high bias through the convexity of the maximum operation. It poses the question but supplies no answer, derivation, or numerical evidence, so it does not establish why the cited analysis adopts that convention or quantify the bias. The discussion is specifically about American option valuation and the final step in the Longstaff–Schwartz procedure.

Key ideas

• The note compares two ways of estimating the initial value in Longstaff–Schwartz.
• One estimate takes a maximum between immediate exercise and the average continuation payoff.
• The alternative averages each path's maximum of immediate exercise and its later payoff.
• The author suggests that the first form can create upward bias through the maximum operation.
• The document poses the rationale as an open question and does not resolve it.

Tags

# The last step of the Longstaff-Schwartz method


# The last step of the Longstaff-Schwartz method












I'm reading An analysis of the Longstaff-Schwartz algorithm for American option pricing, by Clement, Lamberton and Protter.

They define the stopping times (top of page 4)

$$ \tau_j^{[m]} = \begin{cases} L & \text{ if } j = L\\ j \mathbf{1}_{\left\{ Z_j \ge P_j^m(Z_{\tau_{j+1}^{[m]}})\right\}} + \tau_{j+1}^{[m]} \mathbf{1}_{\left\{ Z_j < P_j^m(Z_{\tau_{j+1}^{[m]}})\right\}} & \text{ if } j \le L-1 \end{cases} $$

And then obtain the approximate value function (equation 2.1) $$ U_0^{[m]} = \max \left( Z_0, \mathbb{E} Z_{\tau_1^{[m]}} \right)$$

My question is, why didn't they take the more natural $$ U_0^{[m]} = \mathbb{E} Z_{\tau_0^{[m]}}$$ instead?

When implementing Longstaff-Schwartz (see equation 2.2 and above), this choice actually makes a difference, as they introduce some high-bias (Jensen) $$ U_0^{m, N} = \max \left(Z_0, \frac{1}{N} \sum_{n=1}^N Z^{(n)}_{\tau_1^{n, m, N}}\right) $$ instead of $$ \frac{1}{N} \sum_{n=1}^N Z^{(n)}_{\tau_0^{n, m, N}} = \frac{1}{N} \sum_{n=1}^N \max \left(Z_0, Z^{(n)}_{\tau_1^{n, m, N}}\right) $$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.