# Early exercising American put options

I have found a proof that an American put option without dividend will never be exercised early. However, I suspect that that is not true, so there should be a mistake in the proof. The proof is as follows:

Consider an American put option $$P$$ without dividend. Let the strike price be $$E$$ and let $$S$$ be the underlying stock. First we prove that the price of the option is at least $$E-S$$. We do this by contradiction, so suppose that the option price is smaller than $$E-S$$. Then the following arbitrage option would occur: someone could buy the put option and 1 times the stock $$S$$. Then, the person could immediately exercise the put option. This would give the person an immediate risk-free profit. Therefore this can't be the case, hence the option price is at least $$E-S$$.

Next, suppose that the holder of the option would exercise the option, which would yield him $$E-S$$. Then he could instead sell the option to someone else. Since the option price is always at least $$E-S$$, this would give him at least as much money. Therefore, we can assume that the option is never exercised early.

Could you help me find the mistake in this proof?

• When you say you buy the put at $P<E-S_t$ and exercise it to make a profit that amounts to exercising the put early, no ? This cannot be used as an argument against early exercise of Puts. Jan 24, 2022 at 19:53
• I don't understand your point. In the first paragraph of my proof, I use a proof by contradiction to show that the option price is at least $E-S$. The proof by contradiction uses that arbitrage is impossible. Jan 25, 2022 at 16:55
• The proof by contradicion shows that if you can exercise early and $P<E-S_t$ then $P\ge E-S_t\,.$ This proves that either you cannot exercise early or $P\ge E-S_t\,.$ You still have to handle the case $P\ge E-S_t$ which is not sufficient to show the impossiblity of early exercise (see my posted answer). Jan 25, 2022 at 18:22

Suppose the option is worth $$E - S$$ and $$S=0$$ (the company is bankrupt). If interest rates are positive it's better to have the money now than later.

• Thanks, I added the proof that the option price is at least $E-S$! Jan 24, 2022 at 18:18
• For your bankrupt argument: in the case the S=0 and the option price is E, selling the option gives the same profit E as exercising the option, and both are instantaneous. Therefore I think that this is not a counterexample to my proof. Jan 24, 2022 at 18:20
• Who would buy the option though? If interest rates are positive they are better of with cash. Jan 24, 2022 at 18:49
• That is a good point. Jan 25, 2022 at 16:56

Bob made a good point in finding a hole in your proof. But I'll add on some more points here.

If a stock is bankrupt and S=0, and the Put Price is K-S, both selling the Put in the market as well as exercising represent an instantaneous opportunity. You would much rather sell the Put rather than exercise, because selling takes less hassle and doesn't have to deal with the brokerage/clearing house and what not.

But you see this comes down to "real world" and "risk-neutral world" probabilities in a sense. Even though you would much rather sell because it's "easier", you have no other choice but to exercise because like Bob said who would want to buy the Option from you whilst they gain interest off the cash instead? Maybe under "real world" probabilities you could find some unsuspecting buyer, but we won't count on that.

If an American ITM Put Price is smaller than K-S, that is not probable to happen in an American Options market with early-exercise because that represents an immediate arbitrage opportunity as you already said. You would buy the Put below Parity, buy the Stock, and immediately exercise.

Playing devil's advocate, you could also buy the Put, then hold and hope that someone brings the price back in line then sell to the bid at Parity, but that wouldn't make sense when you could lock in an immediate profit by exercising it.

On the matter of the early exercise of American Options, whether a stock has a dividend or not, is generally important only when/if you have an ITM Call and the dividend value is larger than the time value. If the dividend value is larger than time value, you would exercise the Call before the Ex-Dividend date. In the case that it is not, you would sell the Call instead of exercising it.

And generally, you are always better off selling an Option rather than exercising it due to the remaining time value. As expiration approaches, time value becomes negligible and exercising an Option will be able to complete your goal with less hassle and amount of transaction as opposed to selling the Option in the market, then buying or shorting the stock.

Therefore since the time value decreases less and less as expiration approaches, you will definitely be better off selling the Option, up until the point where the time value is negligible and you would exercise it then. Otherwise you'd be leaving money on the table in both cases of owning an ITM Call or Put before expiration and not exercising.

Only exercise an Option when it is trading exactly at or slightly above Parity with little time left.

I start this answer with some consideration about the call option first:

When interest rates or volatility are not zero an American call option on a stock without dividend should never be exercised early. The proof is well known, easy and in fact similar to your proof for the put: the payoff the option holder gets from exercising the call option is \begin{align} S_t-E&\color{red}{\le}S_t-e^{-r(T-t)}E\le\underbrace{\max(S_t-e^{-r(T-t)}E,0)}_{\text{intrinsic option price}}\\ &\color{red}{\le}\text{European call price if you don't exercise}\\ &\le\text{American call price if you don't exercise}\,. \end{align} When $$r>0$$ or $$\sigma>0$$ at least one of the red $$\color{red}\le$$-signs is a strict inequality. This means never exercise early as you would gain $$S_t-E$$ in cash which is worth less than the remaining American option price.

To emphasize: even for the American call without dividends it may be optimal to exercise early, namely when $$r=0$$ and $$\sigma=0$$ so that $$\tag{1} S_t-E=\text{American call price if you don't exercise}.^1$$ In general ($$r\ge 0,\sigma\ge 0$$) this relationship is known as the rule exercise when $$S_t$$ reaches the exercise boundary.

What you have shown for the Put is

$$E-S_t\color{red}{\le}\text{American put price if you don't exercise}\,.$$ To make the proof complete you would need to show that you never have an equals sign in this inequality.

$$^1\quad$$ When $$r=\sigma=0$$ "optimal" is to be taken with a grain of salt because (1) holds either for all $$t$$ or for no $$t\,.$$ If (1) holds with $$r=\sigma=0$$ one could therefore exercise whenever one wants.

• there is also nowadays the case where r<0 to handle, in which early exercise may be optimal. See "Pricing American options under negative rates" by Jherek Healy published in the Journal of Computational Finance, with preprint on arxiv.org/abs/2109.15157 Jan 27, 2022 at 12:10