American Option Value, Optimal Stopping, and Discounted Martingales

# What is the true "value" process of American derivatives?


# What is the true "value" process of American derivatives?












Consider a continuous-time market where LOOP (law of one price) holds. The first fundamental theorem of asset pricing states explicitly that in the absence of arbitrage, the risk-neutral measure exists, i.e., for any asset price process $V_t$, the discounted process $V_t/B_t$ is a martingale, where $B_t$ is the risk-free money market account process.

However, there seems to be some ambiguity about American derivatives. The so-called "value process" $V_t$ of an American derivative written on an asset $S_t$ with the intrinsic value (the immediate-exercise value) process $Z_t$ is defined as $$V_t=\sup_{t\le\tau\le T} \Bbb E\left(\frac{B_t}{B_\tau}Z_\tau\mid \mathcal F_t\right)$$ And the first property we learn about $V_t$ is that $V_t/B_t$ is a supermartingale, not necessarily a martingale.

But when we consider the "stopped value process" $V_{\tau(t)\wedge t}$ (in which $\tau(t)$ is the optimal stopping time as seen at time $t$), it is not hard to prove that $V_{\tau(t)\wedge t}/B_{\tau(t)\wedge t}$ is a martingale.

My belief is that First FTAP still applies to American derivatives. It's just that we have not chosen the correct value process ("correct" as seen from the perspective of FTAP) for them. For example, I believe that the "stopped value process" $V_{\tau(t)\wedge t}$ can better characterise the value process of American derivatives because, well, who wouldn't exercise an American derivative when it's optimal to do so? (although $V_{\tau(t)\wedge t}$ still fails the FTAP because $V_{\tau(t)\wedge t}/B_t$ is still not a martingale.)

The so-called "value process" $V_t$ is worse in that it reflects the possibility of missed exercise opportunities i.e. the process still continues even after the optimal exercise boundary, which I think is not indicative of the derivative's true value in a rational market where no optimal exercise opportunities would be let go.

Could anybody kindly clarify?

It seems I have found something which is truly a martingale when discounted, and which is also indicative of the true value of American derivatives: $$V_{\tau(t)\wedge t}\cdot\frac{B_t}{B_{\tau(t)\wedge t}}$$ That it is a m.g. is obvious. And the intuition of this expression is that: along the path $\omega\in\Omega$, the value should be $V_t$ before exercise ($t<\tau$); if already exercised ($t\ge \tau$), we would have got $V_{\tau}$ at $\tau$, and this sum grows in the money market account to $V_{\tau}B_t / B_{\tau}$ at time $t$. I think this should be the "true" value process of American derivatives instead.