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?



1 Answer 1


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.

  • $\begingroup$ Thank you very much for your answer. It is hard for me to prove that the payoffs they use are $\mathcal L^2$. Think of their example of a portfolio of an American plus a Bermudan plus an Asian. We don't even know the properties of this new function (i.e. the payoff of this portfolio). $\endgroup$
    – Toofreak
    Feb 28, 2020 at 15:43
  • $\begingroup$ Regarding the last question, to define a projection you need to define 3 things: (1) The space you start with (2) the space you are projecting onto and (3) the inner product you want to use. For the typical orthogonal projections in $\mathcal L^2$ the conditions are: (1) the starting space should be a Hilbert space (2) the space you are projecting onto should be a closed subspace of this space (it is not guaranteed that every subspace of a closed space is closed) . $\endgroup$
    – Toofreak
    Feb 28, 2020 at 15:47
  • $\begingroup$ The space they are projecting onto is chosen by them arbitrarily. In typical linear regression, the choice of regressors is more art than science. But in theory, the regressors should be a closed subspace of the original space. IMHO This needs to be proved: that (1) it's a subspace of the original space and (2) it's closed. $\endgroup$
    – Toofreak
    Feb 28, 2020 at 15:50

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