Stochastic Rates and Discounting in Longstaff–Schwartz Continuation Values

# Continuation value definition in Longstaff and Schwartz


# Continuation value definition in Longstaff and Schwartz












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)

## Answer by Sebastian (score 2, accepted)

https://quant.stackexchange.com/a/68810

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 $Q'$).

## Answer by Kurt G. (score 0)

https://quant.stackexchange.com/a/68809

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: $$ E_Q\left[\exp\left(-\int_{t_k}^tr(s)\,ds\right)\,C\,\underbrace{E_Q\left[\exp\left(-\int_{t}^{t_j}r(s)\,ds\right)\,\Bigg|\,{\cal F}_{t}\right]}_{D(t,t_j)}\Bigg|\,{\cal F}_{t_k}\right]\,. $$