Suppose a put option on a stock $S(t)$ following a Geometric Brownian motion is given, with strike $K$ and maturity $T$. Let us denote its price at time $t$ by $p(t,S(t))$. Now, by no-arbitrage consideration, we can easiy see that the price at time $t$ of this option is always at least the instrinsic value of the option itself, namely, the following inequality holds: $$p(t,S(t))\geq (K-S(t))^+$$ To see this, suppose for example that the opposite inequality holds, i.e., $p(t,S(t)) < (K-S(t))^+$. If so, then at time $t$ buy the put option, buy the stock and exercise the option. It results a riskless profit of $K-S(t)-p(t,S(t)) >0$.

So, from this simple no-arbitrage argument, we see that the price of the option must always be at least its intrisic value. However, at this point I realized something strange: if this is true, why in the world should I exercise my put option before expiry?? The inequality seems to indicate that it would be an unwise decision to ever exercise the American put option at time $t$, and so, the only right time to exercise an American put option would be at expiry, making the American put like an European put (the same as for the American call case).

And yet, we know that the American put option is not the same as the European one. So in my argument there should be something wrong.... but what, exactly??

• You forgot that with positive interest rates, a dollar now is worth a little more than a dollar at expiration. For this reason it is sometimes (not always) rational to exercise the American put early ("to cash in on my profits") if it is deep in the money. Commented Jun 27, 2016 at 11:01
• @Alex C: You completely missed the point. My question is about logic. Interest rates have nothing to do with that, here. Commented Jun 27, 2016 at 12:17
• Help me to understand. The paradox is that once it is optimal to exercise (even by a small amount $\epsilon$), all holders will exercise and the option will go out of existence. Is that it? That is true in an efficient market. Such an option literally can't exist without providing an arbitrage opportunity by immediate exercise. Commented Jun 27, 2016 at 13:01
• There is no paradox just a big confusion IMHO... @RandomGuy it would be nice if you could explain your reasoning: how in the world can $p(t,S_t) \geq (K-S_t)^+$ mean "the only right time to exercise is at $T$" to you? On a given path, how can you even compare $(K-S_t)^+$ with $(K-S_T)^+$ at time $t$ since the latter is a random variable? Commented Jun 27, 2016 at 13:13
• You never exercise put coz stock can’t go below zero, but you do exercise deep itm calls early Commented May 29, 2019 at 12:08

So, from this simple no-arbitrage argument, we see that the price of the option must always be at least its intrisic value.

Yes indeed

However, at this point I realized something strange: if this is true, why in the world should I exercise my put option before expiry?? The inequality seems to indicate that it would be an unwise decision to ever exercise the American put option at time $$t$$, and so, the only right time to exercise an American put option would be at expiry

How does the fact that $$P(t,S_t;K,T-t) \geq (K-S_t)^+$$ gets you to that conclusion? This is a completely fallacious reasoning IMHO.

The price of an American option is: $$P(t,S_t;K,T-t) = \text{sup}_{\tau \in [t,T]} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} (K-S_\tau)^+ \right]$$ the inequality you observe can be obtained by splitting up the family of stopping times $$\tau$$ with values in $$[t,T]$$ using the fact that $$[t,T] = \{t\}\ \cup\ ]t,T]$$ We then get, \begin{align} P(t,S_t;K,T-t) &= \text{sup}_{\tau \in [t,T]} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} (K-S_\tau)^+ \right] \\ &= \max\left( \underbrace{\text{sup}_{\tau=t} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} (K-S_\tau)^+ \right]}_{\text{immediate exercise}}, \underbrace{\sup_{\tau \in ]t,T]} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} (K-S_\tau)^+ \right]}_{\text{differed exercise}} \right)\\ &= \max\left( (K-S_t)^+, \sup_{\tau \in ]t,T]} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} (K-S_\tau)^+ \right] \right) \\ &= (K-S_t)^+ + \max\left(0, \sup_{\tau \in ]t,T]} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} (K-S_\tau)^+ \right] - (K-S_t)^+ \right) \\ &\geq (K-S_t)^+ \end{align}

When the holder has to choose whether or not to exercise at time $$t$$, he/she should compare the value of his option position $$P(t,S_t;K,T-t)$$ with the payoff he/she would get if he/she exercised immediately (intrinsic value) $$(K-S_t)^+$$. $$\underbrace{(K-S_t)^+}_{\text{immediate exercise}} - \underbrace{P(t,S_t;K,T-t)}_{\text{option value}} = \max\left( 0, (K-S_t)^+ - \sup_{\tau \in ]t,T]} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} (K-S_\tau)^+ \right] \right)$$ The holder would then exercise at $$t$$ if the RHS is positive, that is iff $$(K-S_t)^+ \geq \sup_{\tau \in ]t,T]} \mathbb{E}_t^\mathbb{Q}\left[ e^{-r(\tau-t)} (K-S_\tau)^+ \right]$$ or equivalently if the intrinsic value is greater than the continuation value. You see that it is not possible to make a general claim such as "the only right time to exercise would be at expiry" from the above.

Let $$\frac{dS_t}{S_t} = rdt + \sigma dW_t^\mathbb{Q}$$ Elaborating on @MJ73550's remark if $$(S_t)_{t\geq 0}$$ is a martingale, that is if $$r = 0$$, one can show that, for any convex function $$\phi$$, $$\mathbb{E}_t\left[ \phi(S_\tau) \right] \leq \mathbb{E}_t\left[ \phi(S_T) \right], \ \ \forall \tau: t \leq \tau \leq T$$ to see this, we can appeal to Jensen's inequality along with the optional stopping theorem. Indeed for all stopping time $$\tau \leq T$$ we can write: \begin{align} \mathbb{E}_t\left[ \phi(S_T) \right] &= \mathbb{E}_t\left[ \mathbb{E}\left[ \phi(S_T) \vert \mathcal{F}_\tau \right] \right]\ \ \text{(Tower property)}\\ &\geq \mathbb{E}_t\left[ \phi(\mathbb{E}[S_T \vert \mathcal{F}_\tau]) \right] \ \ \text{(Jensen's inequality)} \\ &= \mathbb{E}_t\left[ \phi(S_\tau) \right] \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(Optional sampling theorem)} \end{align}

Using this result the price of an American option, when $$S_t$$ is a martingale becomes: $$V^{AME}(t,S_t;K,T-t) = \sup_{\tau \in [t,T]} \mathbb{E}_t^\mathbb{Q}\left[ \phi(S_\tau) \right] = \mathbb{E}_t^\mathbb{Q}\left[\phi(S_T)\right] = V^{EUR}(t,S_t;K,T-t)$$

Now, if we consider the case $$r \leq 0$$, the stock process $$(S_t)_{t \geq 0}$$ becomes a sub-martingale since in that case: $$\mathbb{E}_t[S_T] = S_t \underbrace{e^{r(T-t)}}_{\leq 1} \leq S_t$$ Should we let $$\phi: x \rightarrow e^{-r(\tau-t)}(K-x)^+$$ which is still a convex function of $$x$$, we would see that for put options, where $$\phi$$ is a monotonically decreasing function of $$x$$, applying a similar reasoning as earlier shows that it is never optimal to exercise the American put before maturity $$T$$.

• I would add that in the particular case of interest rate being equal to 0, (i.e $r=0$ and $S$ being martingale), it is true by Jensen's inequality that exercise is at expiry. Commented Jun 28, 2016 at 8:48

I think the chain of logic should be as follows:
We have put value >= intrinsic.
Therefore either put value > intrinsic or put value= intrinsic.
If put value > intrinsic, then it is not optimal to exercise.
If put value = intrinsic , it may be optimal to exercise.. Hence there is no contradiction.

The logic is that $P_t\geq e^{-r(T-t)}K-S_t$, but ${\bf \text{NOT}}$ $P_t\geq K-S_t$
• Not quite. $P_t \geq K - S_t$ is a stronger condition than the one you presented, although both hold. (assuming $r$ is positive, off course). Commented May 29, 2019 at 9:35