Am Call = Euro Call if r is non-negative and Am Put = Euro Put if r is negative

Question

It can be proven that under non-negative interest rates, it is never optimal to exercise an American call option, such that:

We know, if R >= 0, the current price C of a Europen (and American) call option, with strike price K and time to expriry T, on a non- divided paying stock with current price S satisfies:

C >= max {S-exp(-rT)K, 0}

then, we also know that C >= 0, otherwise buying the call would give a riskless profit now and no obligations later.

To prove that under non-negative interest rates, it is never optimal to exercise an american option we asssume that:

C < S-exp(-rT)K

The we get an arbitrage table like:

we have a non-negative return in all possible states of the world at expiry which has a positive current cash flow. This is clearly an arbitrage opportunity and hence the assumption is wrong.

Suppose now that the American call is exercised at some time t strictly less than expiry T , i.e. t < T . The financial agent thereby realises a cash-flow St − K. From the above proposition we know that the value of the call must be greater or equal to St − exp(−r(T − t))K, which is greater than St − K, if r ≥ 0. Hence selling the call would have realised a higher cash-flow and the early exercise of the call was suboptimal. In conclusion the price of an American call equals the price of an European call: AC = EC

I would like to do an analogous proof to show that it is never optimal to exercise an american put option on a non-dividend pying stock with r =< 0 : EP = AP

I am stuck with the arbitrage table.

1. What does the portfolio consist of for an put call option ?
2. Is there an easier way how to prove this?

Answer 1

It is symetrical, you would buy the share and borrow from the bank account.

In any case, regarding AP and EP, here is an easier way to look at this. Under your assumptions, you know the following always holds:

P >= K.df - S

Where df is your discount factor df = exp{-rT}

Now your AP can be exercised at any time to yield the following payoff

K - S

As a result it is optimal to exercise AP when EP is worth somewhere between the two:

K - S > EP >= K.df - S

Now if r <= 0 then df >= 1 so that:

EP >= K.df - S >= K - S if r <= 0

There is no region where exercising AP for K - S dominates the value of EP.

Accordingly, AP = EP.