# American Option Valuation - Induction algorithm

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?

• Could you please share the reference? Nov 17, 2021 at 7:04
• To be honest I don't remember, I found it somewhere on the internet article and write to my notebook. I think that we can use here tower property of conditional expectation $\sup_{\tau\in\mathcal{T}, \tau\ge t_{k}} E\{D(t_{k-1},t_k)\times e^{-\int_{t_{k}}^\tau r_sds} (K-S_{\tau})^+|\mathcal{F}_{t_{k-1}}=\sup_{\tau\in\mathcal{T}, \tau\ge t_{k}} E\{E\{D(t_{k-1},t_k)\times e^{-\int_{t_{k}}^\tau r_sds} (K-S_{\tau})^+|\mathcal{F}_{t_{k-1}}|\mathcal{F}_{t_k}\}$ but then why we can put $\sup$ inside expectation and still have equality sign? Nov 17, 2021 at 7:16

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.

• So we can put $\sup$ inside the first expectation and still have equality sign? Why? Nov 17, 2021 at 21:27
• The truth is that I am not sure. Un-accept the answer and leave it open. If I have a proper answer, I will reach out to you. Nov 18, 2021 at 14:21
• It would probably be useful if you could remember the book or notes where you got this from. Nov 18, 2021 at 14:24

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 $$t_{k-1}$$.