In the Longstaff and Schwartz method of pricing American options, what is the point of the regressions at each step?

The goal is to approximate an optimal stopping time for each path. However, why approximate, when you can get an exact stopping time for each path by just ... you know ... using the values you've simulated to figure out when the best stopping time is?

For example, say the maturity is $T$, and at time $T - 1$, we want to estimate the continuation value for each path. But I already know what the continuation value is for each path, I know it exactly, I just use what the stock value for that particular path is, and then calculate what my option payoff will be in the next period. Why do I need to use regression? What is the point?

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    $\begingroup$ The method you propose involves insight into the future, which is not realistic and results in large overestimation of the value of the option (try it). The LS method forms an estimate using only past data, which is key to a fair valuation. $\endgroup$
    – Alex C
    Jun 16, 2018 at 16:39
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    $\begingroup$ @AlexC that's not entirely true since the dependent variable in the training data for the cross sectional regression is the discounted one-step-ahead cashflow. The algorithm does indeed look ahead since it uses that path's realized continuation value, but most of the data are from independent simulations. $\endgroup$
    – user217285
    Jun 16, 2018 at 17:34
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    $\begingroup$ The answer is: Read the L&S paper carefully, This is explained pertty well in there. $\endgroup$
    – Yian Pap
    Jun 16, 2018 at 17:36
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    $\begingroup$ people.math.ethz.ch/~hjfurrer/teaching/… $\endgroup$
    – Alex C
    Jun 16, 2018 at 17:38
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    $\begingroup$ This is a classic paper that is very easy to follow, so it's a must read for anyone interested in these things. I think that it will definitely help you more if you read the paper, than getting an easy shortcut to the exact page from me. Of course be free to seek lazy shortcuts and insulting other people. But take it from me, no one will ever hire someone with your personality, so there's no point really for you being here. $\endgroup$
    – Yian Pap
    Jun 16, 2018 at 23:02

1 Answer 1


You are mixing up the realization of a random variable with its expected value at a certain stage.

Let's say you are at path $i$ and time step $t_j$, what you want is not the realization of the stock at $t_{j+1}$ but rather its expected value at $t_{j+1}$ conditional on the info you have up to $t_j$.

The brute force approach here would be to do a (nested) Monte Carlo starting at $t_j$ to get this expectation, which is very costly in terms of computing power.

Longstaff and Schwarz' approach uses a regression to kind of extract this info from all realizations $t_{j+1}$ across all paths.

I won't go further into the details. As everybody said in the comments, the paper is a must read and is very well explained using a simple example. If anything is unclear for you in the paper then everybody here will be happy to help.

  • $\begingroup$ Actually it's not clear at all, and I think it's a legit question. Why do we need regression to infer a value we already have in the simulation? We want to fit a polynomial with the minimum least square error. If the error is 0, we are just using the exact data as if we were not having a polynomial. So why do we need the error? $\endgroup$
    – asdf
    Feb 1, 2021 at 2:40
  • $\begingroup$ Of course it's a legit question :) But, I don't agree with you when you say that we already have the value in the simulation. What we want is the conditional expectation of $X(t_{j+1})$ knowing that the value of $X(t_j)$ is say $x_j$. We don't have this in the simulation, we simply have the realization along each Monte Carlo path. Other numerical methods are better suited to american options. For example, with trees, you have the conditional probas at each node (cond. on the value at that node). This is not the case of MC. MC is useful for american options only when you have many underlyings. $\endgroup$
    – byouness
    Feb 9, 2021 at 10:40
  • $\begingroup$ Exactly. But I feel the right answer to this question should involve Snell envelopes. $\endgroup$
    – asdf
    Feb 9, 2021 at 14:11
  • $\begingroup$ The aim of my answer wasn't to get into the technical detail, but rather to give an intuition about what the regression aims to do and why. Please feel free to add a new, more technically involved answer if you think it'll help clarify things. $\endgroup$
    – byouness
    Feb 9, 2021 at 14:31
  • $\begingroup$ Hehe, if I'm here is because I don't know the details :p $\endgroup$
    – asdf
    Feb 9, 2021 at 17:57

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