I am trying to implement a LSMC to value an american-style real option with an underlying project value that is exposed to several risk factors.

In the paper of Longstaff & Schwartz, they use the first three Laguerre polynomials to execute their regressions. Unfortunately, they don't provide any explanation about what these functions are and how they are used.

My question is, can we always use these same functions irrespective of the dynamics of the underlying/model setup? How do we choose these basis functions and what do they actually mean?

Obviously I am new to this, so besides the technical stuff, I would very much appreciate some intuitive explanation.

Thank you for your support!

  • 3
    $\begingroup$ Traditionally, LS is indeed implemented using weighted Laguerre polynomials. This is because this set of functions forms a complete orthogonal system of the $L^2(0,\infty)$ space (see math.stackexchange.com/questions/100461/…). Intuitively, this means that any function $f : \Omega \to \Bbb{R}$ whose square is integrable over $\Omega=[0,\infty[$ can be expressed as a linear combinations of weighted laguerre polynomials, which is precisely your goal in LS (estimate the continuation value as a combo of elementary functions). $\endgroup$
    – Quantuple
    Commented Dec 21, 2017 at 11:50
  • 1
    $\begingroup$ See also this paper which seems to focus on some of the questions you ask concerning the choice of basis functions for real options' pricing (fep.up.pt/conferencias/pfn2006/Conference%20Papers/540.pdf) $\endgroup$
    – Quantuple
    Commented Dec 21, 2017 at 11:53
  • $\begingroup$ @Quantuple, from the LS paper, is not clear to me that all the payoffs they consider (americans, Asians, Bermudans) belong to $\mathcal L^2$ . They quickly direct us to some references that don't cover all their cases. Moreover, they never show how the stock prices they use as regressors form a closed subspace of the payoff space (which is the necessary condition for ortho projection). I believe LS should just be proved on finite spaces, not for infinite dimensional ones. $\endgroup$
    – Toofreak
    Commented Feb 27, 2020 at 13:47
  • $\begingroup$ I fail to see how this is relevant. The idea is to say that if you can approach the regression function using a subset of functions (regardless of finite space or infinite space), then the algorithm will converge. Now the question of what function space the true continuation value belongs to compared to the set of functions you are actually able to generate from your basis functions + regressors can be a research topic on its own indeed. It's really more of a technical discussion than a pragmatic one: you could definitely replace this step by some advanced ML if you feel like it! $\endgroup$
    – Quantuple
    Commented Mar 13, 2020 at 11:21

1 Answer 1


No, obviously LS chose 3 simple basis functions to illustrate their method initially. These will work poorly in general, even for a simple vanilla BS you probably need 5-6 of those for a good fit to the continuation value function (and thus the American price). But perhaps you haven't read the paper carefully. If I remember well they have like 7 different examples there where they use progressively more basis functions, products between them (when there are more that one stochastic factors) etc. In general the answer is no, you don't use what's in the paper, but you have to experiment to see what works better for your particular problem.


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