American Put Valuation by Backward Induction
Summary
The document explains the continuation-value step in a discrete-time recursion for valuing an American put. At each time step, the option value is the larger of its immediate exercise payoff and the expected discounted value of holding it to the next step. The question focuses on why the latter expression can be written using the option value at that next time.
The answer applies the tower property of conditional expectation: condition first on information at the next date, then on information at the current date. Since the discount factor over the single interval does not depend on the later exercise time, it can be taken outside the supremum over eligible stopping times. The conditional expectation inside then corresponds to the definition of the next-step American option value. A second answer gives the same economic interpretation as a choice between exercising now and continuing. The derivation assumes the stated filtration and discounting setup; it does not discuss numerical implementation or continuous-time limits.
Key ideas
- At each step, an American put is worth the maximum of immediate exercise value and continuation value.
- The tower property connects expectations conditioned at consecutive dates.
- A discount factor fixed over the current interval does not depend on the later stopping time.
- The continuation value is the conditional discounted expectation of the option value at the next step.
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Full text
# American Option Valuation - Induction algorithm
# 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?
## Answer by Sebastian (score 3)
https://quant.stackexchange.com/a/68812
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 by KT8 (score 0)
https://quant.stackexchange.com/a/68805
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 $t_{k-1}$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.