6
$\begingroup$

It is often explained, that the rule of thumb for exercising American options is to check when the benefit from the interest rate (sell the stock earlier, get the cash, put in the bank) is higher than the time value of the option. This is all clear in the positive interest rate environment, but the question is then - would we exercise some put options in case $r = 0$, and why would we do that? I thought, that there's no reason for such exercise, however according to this paper it is even sometimes optimal to early exercise American puts when $r<0$. What am I missing?

$\endgroup$
4
$\begingroup$

The classic result is never early exercise an American call if $r \geq 0, d \leq 0.$ If we think in terms of FX, calls and puts are really the same thing and by switching currency, we get never early exercise an American put if $ r \leq 0, d \geq 0.$ If one of these is violated it may be worth early exercising.

$\endgroup$
  • 1
    $\begingroup$ Thanks for the answer, but I'm still not sure what am I missing. Apparently, puts can trade below intrinsic even if $r= d = 0$. Why such asymmetry with the case of calls, just because the stock price can't go negative, but it's not bounded from above? $\endgroup$ – Ulysses Jun 10 '15 at 7:10
  • $\begingroup$ I added an answer to show that early exercise of put is optimal if $r=0$ (so you should correct or delete your answer). $\endgroup$ – emcor Jun 10 '15 at 11:27
  • 2
    $\begingroup$ @emcor, in your answer the company becomes unbust after reaching $S=0$? Also please stop flagging answers you don't like. I've said this to you before. $\endgroup$ – Bob Jansen Jun 10 '15 at 12:02
3
$\begingroup$
  • Let $r=0$:

The maximum payoff ever from the put is when $S=0$ so $P= K$. So one would always exercise at this maximum because you cant get any better in the future and dont forego any interest. Based on @MarkJoshi 's comment, we have to assume that $S=0$ never recovers, so if $r=0$ you are essentially indifferent between exercising then or later because you gain/loose no interest and the payoff will never decrease. So it holds $P_t=p_t$.

If $S>0$ you can theoretically always have an infinitesimal increase in payoff when $S\to 0$ and wait, however I would expect that there is an exercise boundary at which point you would exercise because the probability of ending OTM vs. being close to maximum payoff already is too high and hence the AM put dominates at this point and $P_t\geq p_t$.

  • Let $r<0$:

If you exercise at the maximum payoff, your received cash will vanish over time through the negative interest. I expect it is then a question of the expected opportunity cost from not exercising at the maximum vs. the negative interest until maturity. If the volatility of the asset is very high, you have a high probability of moving away from the maximum in the future and would probably rather accept the negative interest. If the volatility is very low or even close to zero, you will likely not move away from the maximum much and rather exercise later to avoid negative interest. Hence the AM put fully dominates and $P_t>p_t$.

$\endgroup$
  • $\begingroup$ In case $r = 0$ can it still be the case that European put trades below intrinsic in the BS model? $\endgroup$ – Ulysses Jun 10 '15 at 11:33
  • $\begingroup$ @Ulysses The general lower bound for EU puts is $p_t>Ke^{-r(T-t)}-S_t$ so if $r=0$ you have $p_t>K-S_t$. And by $p_t\geq0$ you get $p_t\geq(K-S_t)^+$. Hence the EU put can never be worth less than intrinsic in this case. (scotty.com.au/frm/options/optionbounds.html) $\endgroup$ – emcor Jun 10 '15 at 11:45
  • $\begingroup$ that was my idea as well, so there is no strict benefit from early exercising it then if $r = 0$ and in that case american and european prices shall be the same? $\endgroup$ – Ulysses Jun 10 '15 at 11:52
  • $\begingroup$ No and no: you would always early exercise the AM put if $S=0$ (and $r\geq0$) as described above. And because an AM put gives you the flexibility to early exercise as soon as $S=0$ which is the maximum payoff, the EU put does not have this flexibility and hence $p_t<P_t$ always ($\forall t<T$). $\endgroup$ – emcor Jun 10 '15 at 11:56
  • 1
    $\begingroup$ If the stock can hit zero and then go positive and also can never go negative, the model has an internal arbitrage. Buy when it hits zero for zero and you are laughing. $\endgroup$ – Mark Joshi Jun 11 '15 at 1:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.