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I don't understand what's wrong in the following argument.

Assume that we have a no-arbitrage market where the following products are traded:

  • a risky asset $S$,
  • a risk-free bond $B$,
  • an American put option $P$ with finite maturity $T$ and payoff $K$. Its underlying is $S$.

Now, according to the first fundamental theorem of asset pricing, there exists an equivalent probability measure $\mathbb Q$, under which the two-dimensional process $(\frac{S}{B},\frac{P}{B})$ is a martingale. But then, $P$ would have the same fair price as an European put option, and this is (as far as I know) false.

Where is the mistake? Have I misunderstood the statement of the theorem?

EDIT: Perhaps the key point is that when we price an American option, we do not assume that, when exercised early, another one is available (and buyable) in the market. In my argument, on the contrary, we do assume that. Does this make some sense?

I don't understand what's wrong in the following argument.

Assume that we have a no-arbitrage market where the following products are traded:

  • a risky asset $S$,
  • a risk-free bond $B$,
  • an American put option $P$ with finite maturity $T$ and payoff $K$. Its underlying is $S$.

Now, according to the first fundamental theorem of asset pricing, there exists an equivalent probability measure $\mathbb Q$, under which the two-dimensional process $(\frac{S}{B},\frac{P}{B})$ is a martingale. But then, $P$ would have the same fair price as an European put option, and this is (as far as I know) false.

Where is the mistake? Have I misunderstood the statement of the theorem?

I don't understand what's wrong in the following argument.

Assume that we have a no-arbitrage market where the following products are traded:

  • a risky asset $S$,
  • a risk-free bond $B$,
  • an American put option $P$ with finite maturity $T$ and payoff $K$. Its underlying is $S$.

Now, according to the first fundamental theorem of asset pricing, there exists an equivalent probability measure $\mathbb Q$, under which the two-dimensional process $(\frac{S}{B},\frac{P}{B})$ is a martingale. But then, $P$ would have the same fair price as an European put option, and this is (as far as I know) false.

Where is the mistake? Have I misunderstood the statement of the theorem?

EDIT: Perhaps the key point is that when we price an American option, we do not assume that, when exercised early, another one is available (and buyable) in the market. In my argument, on the contrary, we do assume that. Does this make some sense?

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chalk
chalk
  • 61
  • 3

Why aren't american put options martingales?

I don't understand what's wrong in the following argument.

Assume that we have a no-arbitrage market where the following products are traded:

  • a risky asset $S$,
  • a risk-free bond $B$,
  • an American put option $P$ with finite maturity $T$ and payoff $K$. Its underlying is $S$.

Now, according to the first fundamental theorem of asset pricing, there exists an equivalent probability measure $\mathbb Q$, under which the two-dimensional process $(\frac{S}{B},\frac{P}{B})$ is a martingale. But then, $P$ would have the same fair price as an European put option, and this is (as far as I know) false.

Where is the mistake? Have I misunderstood the statement of the theorem?