# Properties of the American derivative security price process

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

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?