Early exercise on American put

[chiyui]

I got stuck on this question, why is an American put on a non-income paying asset could be exercised early optimally?

I've searched for many sources, but all of them explain the reason as deeply in-the-money, rising interest rate, sth like that.

Is it possible to use mathematical arguments to answer this problem (e.g. to use American put lower/upper bounds, put-call parity for American option (an inequality), and so on)? I don't mind if numerical examples are used also.

Anyone who can help me? Thanks a lot......

 

[ShaktiRathore]

try this link:http://forum.bionicturtle.com/threa...ion-value-with-infinite-term.6160/#post-19506

thanks

 

[David Harper CFA FRM]

To complement Shakti's math, I got a lot from this thread http://forum.bionicturtle.com/threa...dividends-on-put-call-parity.4616/#post-16263

especially Aleks' post:

Early exercise of an American put becomes more attractive as the risk-free rate increases and volatility decreases, because:
The early exercise of an American put option is attractive when the interest earned on the strike price is greater than the insurance element lost. When interest rates increase, the value of the interest earned on the strike price increases making early exercise more attractive. When volatility decreases, the insurance element is less valuable. This makes early exercise more attractive.

 

[chiyui]

try this link:http://forum.bionicturtle.com/threa...ion-value-with-infinite-term.6160/#post-19506

thanks

What you said in the post is seeing what will happen when taking the limit T approaching infinity. How does it relate to the early exercise of American put?
In addition, is it possible to make the argument without using Black-Scholes?

To complement Shakti's math, I got a lot from this thread http://forum.bionicturtle.com/threa...dividends-on-put-call-parity.4616/#post-16263

especially Aleks' post:

What you're essentially saying (and most of the textbooks and websites also) is that, if Xe^(r(T-t)) is large enough, the interest benefit will outweigh the benefit of protection provided by the put.
But there's a point I don't understand.

The lower bound of an American put is P(t)≧X - S(t).
If we early exercise the put, we get the cash inflow of X - S(t), and we can invest it risk-free to capture the high interest rate as you mentioned about.
But why don't we just simply sell out the put? If we sell the put, we get P(t), which is larger than X - S(t).
Isn't it better to use this larger amount of money to generate more risk-free interest earnings than to use X - S(t)?

That's why I got stuck on the problem......

 

[chiyui]

I think I know why I got stuck on the question.
I know the lower bound of an American put is P(t)≧X - S(t).
If it is better to early exercise the American put, then of course P(t) = X - S(t).
But it is still consistent to the lower bound P(t)≧X - S(t).
And it is also consistent to the lower bound of European put p(t)≧Xe^(-r(T-t)) - S(t).
It can happen to be Xe^(-r(T-t)) - S(t) < p(t) < P(t) = X - S(t)
The American put could still be more expensive than the European put.
So actually there's no problem of early exercise.

For American call this does not work.
American call: C(t)≧S(t) - X
European call: c(t)≧S(t) - Xe^(-r(T-t))
If we early exercise the American call, we get C(t) = S(t) - X.
But this is not reasonable because S(t) - X < S(t) - Xe^(-r(T-t)).
That means C(t) < c(t) and this is a contradiction.
So that's why we never early exercise the American call.

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Putting it in another way, If we hold a deep in-the-money American call and early exercise it, of course we would get the profit S(t) - X.
But actually we have 3 choices:

1. Short a European call. (The obligation is met by exercising the American call at expiry.) The profit is at least S(t) - Xe^(-r(T-t))).
2. Early exercise the American call. The profit is S(t) - X.
3. Sell out the American call. The profit must be greater than the 1st choice because C(t) > c(t).

Which choice gives the smallest profit? Of course the 2nd one. So we never early exercise American call.
Which choice gives the largest profit? Of course the 3rd one. So we simply sell out the American call.


But it's another story for American put:

1. Short a European put (The obligation is met by exercising the American put at expiry.) The profit is at least Xe^r(T-t) - S(t).
2. Early exercise the American put. The profit is X - S(t).
3. Sell out the American put. The profit must be greater than the 1st choice because P(t) > p(t).

Which choice gives the smallest profit? The 1st one.
But which choice gives the largest profit? 2nd or 3rd? It depends!
If r is very large and the time to expiry is very long, the term e^(-r(T-t)) can drive down the Xe^(-r(T-t)) - S(t) a lot.
Even if the 3rd choice must give a profit greater than Xe^(-r(T-t)) - S(t), it can still be lower than X - S(t).
In that case, we should choose the 2nd choice - early exercise the American put getting X - S(t)!

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Is my thinking correct? May anyone help examining my idea please? Thanks a lot.​

 

[chiyui]

If the American put should be early exercised, what is its value? Of course X - S(t).
But does it violate the lower bound P(t)≧X - S(t)? No!
First, the lower bound is closed. It includes an equal sign.

Second, and more importantly, if you see that P(t) > X - S(t) in the option market, that just means the time does not yet come to early exercise the American put.
If you see that P(t) = X - S(t), that means the time comes. You can early exercise it.
If you see that Xe^(-r(T-t)) - S(t) < P(t) < X - S(t), that means you can earn arbitrage free money! You just buy the American put and the stock and exercise it at once!
In this case, the American put price will be drive up quickly by those arbitrageurs like you, until it equals X - S(t). Then no free money left at all.

Am I right?