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I am going through the paper by Longstaff and Schwartz (2001) on American-options pricing, and something got me confused.

There, in equation $(1)$ the continuation value at time $t_k$, $F(\omega; t_k)$, is defined as follows $$F(\omega; t_k) = E_Q\left[ \sum_{j = k+1}^{K} \exp \left( - \int_{t_k}^{t_j} r(\omega, s ) \, ds \right) C(\omega, t_j ; t_k, T) \; \Bigg\vert \; \mathcal{F}_k \right].$$

However, I think that it should instead be rewritten as $$F(\omega; t_k) = \sum_{j = k+1}^{K} D(t_k, t_j) \cdot E_Q\left[ C(\omega, t_j ; t_k, T) \; \Bigg\vert \; \mathcal{F}_k \right],$$ where $D(t_k, t_j) $ are the different discount factors at $t_k$, which are $\mathcal{F}_k$-measurable. My idea is nothing different to the martingale condition under the risk-free probability measure $Q$.

So my question is, is there something that I am missing here?

I appreciate every comment or discussion on the subject.

Edit

What I really mean is, isn't the price of a derivative at time $t$, chosen a numerary $\mathcal{N}$, given by

$$\dfrac{V(t,T)}{\mathcal{N}(t,T)} = E_{Q} \Bigg[ \dfrac{V(T,T)}{\mathcal{N}(T,T)} \Bigg\vert \mathcal{F}_t \Bigg]$$

where if $\mathcal{N}$ is chosen to be a zero coupon bond, then $\mathcal{N}(t,T) = D(t,T)$ and $\mathcal{N}(T,T) = 1$.


Longstaff, Schwartz - Valuing American Options by Simulation: A Simple Least-Squares Approach (2001)

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    $\begingroup$ I believe your rewrite does in general require a change of measure from the risk neutral measure to the t_j forward measure(s), which is not indicated in your formula. As mentioned already by Sebastian this is not necessary/has no effect if rates are deterministic. $\endgroup$ Commented Nov 17, 2021 at 14:27
  • $\begingroup$ Thanks for your comment @Robert, could you give me a hint on how to do so? Could you also check the edit on the question to see what I am missing here. Thanks! $\endgroup$
    – KT8
    Commented Nov 17, 2021 at 14:34
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    $\begingroup$ Your last formula will become valid when you change it to $$\dfrac{V(t,T)}{\mathcal{N}(t,T)} = E_{Q_{T}} \Bigg[ \dfrac{V(T,T)}{\mathcal{N}(T,T)} \Bigg\vert \mathcal{F}_t \Bigg]$$ where -as @Robert wrote- $Q_T$ is the $T$-forward measure. $\endgroup$
    – Kurt G.
    Commented Nov 17, 2021 at 15:03
  • $\begingroup$ Thanks @KurtG. do you know any references (books, articles, whatever) where I can find a good discussion on this? $\endgroup$
    – KT8
    Commented Nov 17, 2021 at 16:04
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    $\begingroup$ @castella08 : the book I have (it is surely not the only one) is Musiela & Rutkowski, Martingale Methods in Financial Modelling. The question how $Q_T$ is defined given $Q$ is not a big mystery: start from $V(t,T)=E_Q[\exp(-\int_t^T r_s\,ds)V(T,T)|{\cal F}_t]$ and set $dQ_T/dQ=\exp(-\int_0^T r_s\,ds)/{\cal N}(0,T)$ - the most obvious Radon-Nikodym density that comes to mind doing that job. $\endgroup$
    – Kurt G.
    Commented Nov 17, 2021 at 17:54

2 Answers 2

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I haven't read the paper but I would say that the formula in the paper seems more general. It accounts for the possibility of stochastic interest rates. If we have deterministic discount factors, then your formula and the paper's formula agree.

Answer to your edit: We know for sure, in the paper they are using the money market account process as numerarie since they are also using the risk-neutral probability measure $Q$. The money market account process is worth 1 at the initial time. So the denominator on the LHS of your formula in your edit, is worth 1, which means that the formula in the paper is correct. The numerarie can not be a zero coupon bond if we are using the risk-neutral probability $Q$. If you chose a zero coupon bond as numerarie, then you need to change the probablity measure from the risk neutral $Q$ to an equivalent probability measure $Q'$ (note that the probability in the paper is $Q$ and not $Q'$).

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  • $\begingroup$ Thank you for your comment and the edit Sebastian! $\endgroup$
    – KT8
    Commented Nov 17, 2021 at 16:15
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By the general no arbitrage pricing theory the price at time $t_k$ of a payoff $C$ that is paid at time $t_j\ge t_k$ is $$ E_Q\left[\exp\left(-\int_{t_k}^{t_j}r(s)\,ds\right)C\,\Bigg|\,{\cal F}_{t_k}\right]\,. $$ For $t\in[t_k,t_j]$ you can use a property of iterated conditional expectations to write this as $$ E_Q\left[E_Q\left[\exp\left(-\int_{t_k}^{t_j}r(s)\,ds\right)C\,\Bigg|\,{\cal F}_{t}\right]\Bigg|\,{\cal F}_{t_k}\right]\,. $$ If $C$ were known at $t$ (an assumption that Longstaff-Schwartz do not make (correct me if I am wrong) you could pull $\exp(-\int_{t_k}^tr(s)\,ds)\,C$ out of the inner expectation and get a formula similar to yours: $$ E_Q\left[\exp\left(-\int_{t_k}^tr(s)\,ds\right)\,C\,\underbrace{E_Q\left[\exp\left(-\int_{t}^{t_j}r(s)\,ds\right)\,\Bigg|\,{\cal F}_{t}\right]}_{D(t,t_j)}\Bigg|\,{\cal F}_{t_k}\right]\,. $$

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