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$$ \newcommand{\cbkt}[1]{\left\{{#1}\right\}} \newcommand{\rbkt}[1]{\left({#1}\right)} \newcommand{\sqbkt}[1]{\left[{#1}\right]} $$

Shreve volume I, defines an American derivative security as follows:


Definition. An American derivative security is any contract that can be exercised at times $0,1,\ldots,N$. Let the random variable $G_n$ denote the intrinsic value (contract function) of the American derivative at time $n$. Thus, if the derivative is exercised at time $n$, it has a payoff $G_n$.

The price process $V_n$ for this contract is given by:

$$V_n =\max_{\tau \in \mathcal{S}_n} \tilde{\mathbb{E}}_n\left[\mathbb{I}_{\tau\leq N}\frac{1}{(1+r)^{\tau - n}}\right]$$


where we take the maximum over all exercise strategies.

Theorem 4.4.2 proves some important properties of the American derivative price process.


The American derivative security price process $(V_n)_{n=0}^{N}$ satisfies the following properties:

(1) $V_n \geq \max \{G_n,0\}$

(2) the discounted price process $\frac{V_n}{(1+r)^n}$ is a super-martingale.

(3) if $Y_n$ is any arbitrary process satisfying (1) and (2), then $Y_n \geq V_n$.


I tried to study the proof extremely carefully, but I did not follow a couple of arguments.

In the proof of claim (3), we proceed as follows.

Let $\tau$ be any arbitrary (valid) exercise rule in $\mathcal{S}_n$. And let $Y_n$ be another process satisfying (1) and (2).

Because $Y_k \geq \max \cbkt{G_k,0}$ for all $k$, we have:

\begin{align*} \mathbb{I}_{\cbkt{\tau \leq N}}G_\tau &\leq \mathbb{I}_{\cbkt{\tau \leq N}} \max \cbkt {G_\tau,0} \\ &\leq \mathbb{I}_{\cbkt{\tau \leq N}} \max \cbkt {G_{\tau \land N},0} + \mathbb{I}_{\cbkt{\tau = \infty}} \max \cbkt {G_{\tau \land N},0} \\ &= \max \cbkt {G_{\tau \land N},0}\\ &\leq Y_{\tau \land N} \end{align*}

[Question]. In the above inequality, what property are we alluding to in the step:

$$\mathbb{I}_{\cbkt{\tau \leq N}} \max \cbkt {G_\tau,0} \leq \mathbb{I}_{\cbkt{\tau \leq N}} \max \cbkt {G_{\tau \land N},0} + \mathbb{I}_{\cbkt{\tau = \infty}} \max \cbkt {G_{\tau \land N},0}$$

Now, we write:

\begin{align*} \tilde{\mathbb{E}}_n \sqbkt{\mathbb{I}_{\cbkt{\tau \leq N}} \frac{1}{(1+r)^\tau} G_\tau} &= \tilde{\mathbb{E}}_n \sqbkt{\mathbb{I}_{\cbkt{\tau \leq N}} \frac{1}{(1+r)^{\tau \land N}} G_\tau} \end{align*}

[Question] Why are we allowed to replace $\tau$ by $\min \cbkt{\tau,N}$? Is it because if we never exercise the option, the indicator random variable $\mathbb{I}_{\tau \leq N} = 0$?

In the next step, we give an upper bound:

\begin{align*} \tilde{\mathbb{E}}_n \sqbkt{\mathbb{I}_{\cbkt{\tau \leq N}} \frac{1}{(1+r)^{\tau \land N}} G_\tau} \leq \tilde{\mathbb{E}}_n \sqbkt{\frac{1}{(1+r)^{\tau \land N}} Y_{\tau \land N}} \end{align*}

[Question]. In the above step, what property are we alluding to?

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