Square Integrability and Projection Spaces in Longstaff–Schwartz
Summary
The document examines why the Longstaff–Schwartz method assumes American option payoffs are square-integrable, and asks how that assumption relates to finite Monte Carlo samples, regression bases, and path-dependent payoffs. The answer distinguishes the finite-dimensional span produced by a particular sample from the broader payoff space needed to discuss convergence across possible samples and sample sizes. It also notes that square integrability is a distinct condition from bounded variation, and suggests that payoffs bounded on compact sets may suffice in practical settings where payouts have finite prices.
For regression, the response argues that a finite collection of polynomial basis functions spans a finite-dimensional, hence closed, subspace, and that projection remains within the original space. However, it does not give a rigorous proof that the relevant option payoffs are square-integrable, nor does it fully resolve how stock-price regressors relate to the payoff space. Its explanation is a brief conceptual answer rather than a complete treatment of the functional analysis or convergence assumptions.
Key ideas
- A finite Monte Carlo sample spans a finite-dimensional space, but that alone does not establish convergence for all possible samples.
- Square integrability and bounded variation are different mathematical conditions.
- The answer suggests that payoffs bounded on compact sets may meet the square-integrability requirement in practical cases.
- A finite set of polynomial basis functions spans a closed finite-dimensional subspace for regression.
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# L2 Assumptions of the Longstaff Schwartz method
# L2 Assumptions of the Longstaff Schwartz method
In page 121 of the original LS Paper they use the fact that the space of functions they are dealing with (payoffs of American options), belong to the $\mathcal L^2$ space.
They use this assumption to allow the following: (1) a unique orthogonal projection of this space of Americans Payoffs exist and (2) the orthogonal projection can be decomposed as a finite combination of bases.
So in the end, they come with a polynomial representation of the conditional expectation.
I have some questions regarding the $\mathcal L^2$ assumption:
1) Do we need to deal with infinite dimensional spaces?
1.1.) Our Monte-Carlo simulation gives us a finite number of vectors, therefore aren't we on finite spaces?
1.2) If we are on finite spaces, I want to confirm that in the finite space case, we will always find an orthogonal projection and we will always find a decomposition on countable number of bases.
2) In the case that we want to insist with infinite-dimension spaces:
2.1) we know that Americans are convex functions, but how do we prove that they are they of bounded variance (square integrable)? I read the papers that they mention Karatzas and Bensoussan but it's still not clear to me how to prove that Americans payoff can form an $\mathcal L^2$ space
2.3) Then the authors continue to work on a portfolio of many path dependent options (as their example of American-Bermudan-Asian)? how do we prove they are of bounded variation?
2.2) Now, a question regarding the subspace used in the regression (their "Xs" i.e. the stock prices).
In the case of orthogonal projections, we not only need to show that we are starting with a Hilbert space, but also, that the space we are projecting onto, is a subspace of it (i.e. it's a closed subspace of the "Ys" - the american prices). My understanding is that the subspace is the observed stock prices, how do we know they form a subspace of the original Hilbert space?
Thanks!
## Answer by g g (score 3)
https://quant.stackexchange.com/a/51402
> 1) Do we need to deal with infinite dimensional spaces?
Yes, I think you need an infinite dimensional pay-off space. Your remark that a finite sample spans a finite dimensional space of pay-offs is true. But you would like to prove convergence of the method for any pay-off, i.e. for all possible samples of all sizes.
> 2) In the case that we want to insist with infinite-dimension spaces
I think $\mathcal{L}^2$ should mean square integrable here and not bounded variation. And again you are right this is something which requires proof. But being in $\mathcal{L}^2$ given the pay-outs are in $\mathcal{L}^1$ (since they have a price) is not a strong restriction in practice. For example being bounded on compact sets should be enough.
I do not entirely understand your last question. If you are projecting you remain always in the original space by definition of projection. And the space is closed since it is finite dimensional being spanned by polynomials of finite degree.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
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