Question

*25. Explain why an American call option on a non-dividend-paying stock always has the same price as its European counterpart.

Answer 1

An American call option on a non-dividend-paying stock always has the same value as an european call option because it is never optimal to exercise an american option before maturity.

The reason is that if American calls are exercised early, Then We lose:-

- The benefit from the time value of money of paying strike in future and now, and

- The value of the right to exercise the option in the future.

Suppose, if you could sell an american call option before maturity (we can assume that S>K).

If you sell the call, you will receive the value of call, C

If you exerciese the call, you will receive the payoff of the call, S-K

This might not mean much, until we realize that C>S-K.

- You can plot the payoff diagram of a long call option against that of a portfolio containing long stock and short bond. you will see that the call option payoff is always higher.

- If C<S-K, an opportunity for arbitrage would occur. An intelligent person(Say, Ram) can long the call against shorting the stock and investing the proceeds.

At t=0, CF = -C+S-K > 0(since C<S-K)

At t=T, if it ends up that S>K, Ram will choose to exercise the call and CF = S-K+K-S = 0

At t=T, if it ends up that S<K, Ram will choose to not exercise the call and CF = K-S> 0 (Since S<K)

Hence, since C>S-K, then we will always choose selling the call over exercising it.

But what if we can’t sell the option? The thing is, we can always just keep the call, short the stock and invest the proceeds.

Say we exercise, then at t=T, we will have (S-K)e^rT (assuming we reinvest S-K at a continuously compounded interest rate till time T)

If alternatively, we indeed keep the call, short the stock and invest the proceeds, then at t=T, our payoff would be Se^rT - S + max(S-K,0):

If it ends up that S>K, payoff = Se^rT - K = (S-K)e^rT + K(e^rT - 1) > (S-K)e^rT
If it ends up that S<K, payoff = Se^rT - S > Se^rT - K (since S<K)

In both cases, the payoffs are higher than (S-K)e^rT.