American Option Valuation - Induction algorithm

Question

The price of an American put option is given by

$$V_k = \sup_{\tau\in\mathcal{T}, \tau\ge t_K} E\{e^{-\int_{t_k}^\tau r_sds} (K-S_{\tau})^+|\mathcal{F}_{t_k}\}$$

I found in one book the following: $$\begin{aligned} V_{k-1} & = \sup_{\tau\in\mathcal{T}, \tau\ge t_{k-1}} E\{e^{-\int_{t_{k-1}}^\tau r_sds} (K-S_{\tau})^+|\mathcal{F}_{t_{k-1}}\} \\ & =\max\{(K-S_{t_{k-1}})^+, \sup_{\tau\in\mathcal{T}, \tau\ge t_{k}} E\big[D(t_{k-1},t_k)\times e^{-\int_{t_{k}}^\tau r_sds} (K-S_{\tau})^+|\mathcal{F}_{t_{k-1}}\big]\} \\ & = \max\{(K-S_{t_{k-1}})^+, E\big[D(t_{k-1},t_k)V_k|\mathcal{F}_{t_{k-1}}\big] \} \end{aligned}$$

and I don't understand the last equality. Can anyone explain it to me?

Answer 1

By the tower property of the conditional expectation first and the definition of the American put later (first equation in the question), we obtain

\begin{align} \sup_{\tau\in\mathcal{T}, \tau\ge t_{k}} &E\big[D(t_{k-1},t_k)\times e^{-\int_{t_{k}}^\tau r_sds} (K-S_{\tau})^+|\mathcal{F}_{t_{k-1}}\big] \\ &= \sup_{\tau\in\mathcal{T}, \tau\ge t_{k}} E\left[E\big[D(t_{k-1},t_k) e^{-\int_{t_{k}}^\tau r_sds} (K-S_{\tau})^+|\mathcal{F}_{t_{k}}\big]|\mathcal{F}_{t_{k-1}}\right] \\ &=E\left[D(t_{k-1},t_k) V_k | \mathcal{F}_{t_{k-1}}\right]. \end{align} Note that the term $D(t_{k-1},t_k)$ doesn't depend on $\tau$ so it can come out of the supremum. Also note that the $\sigma$-algebras in your comment above are swapped.

Answer 2

If I just focus on the last term of your last formula, what you have is

$$ V_{k-1} = \max\{(K-S_{t_{k-1}})^+, D(t_{k-1},t_k) E\big[V_k|\mathcal{F}_{t_{k-1}}\big] \}.$$

The idea behind that equality is that the value of an american option at time $t_{k-1}$ should be the most convenient one (therefore the maximum of) between

• exercising the option at that time, i.e. $$(K-S_{t_{k-1}})^+$$
• the continuation value $$E\big[D(t_{k-1},t_k)V_k|\mathcal{F}_{t_{k-1}}\big],$$ which you can understand there as the discounted expectation value, where the discounting goes from time $t_k$ up to $t_{k-1}$.