Denote American call and put option values as $C$ and $P$ respectively. Similarly, denote European call and put options values as $c$ and $p$.

It is well known that time to maturity affects all $C,P,c,p$ given all else are fixed (such as stock price, strike price, risk-free interest rate, volatility, etc).

For American $P$ and $C,$ it is quite clear that longer time to maturity increases $P$ and $C.$ However, things are not so clear cut when it comes to European $p$ and $c.$

Question: How does longer time to maturity affect $p$ and $c?$


In simple terms, more time to expiry $T$ increases the value of an at-the-money (ATM) option as it gives more time for the stock to rise further (or fall further in the case of a put option). This means that the potential upside of the option is greater (the downside is not as it is floored at zero). So the option is worth more. This effect has nothing to do with being able to exercise the option early.

The situation is more complicated when the option is not ATM. For example, an in-the-money put option with no dividend may see its value decrease with increasing time to expiry due to its value being driven by its intrinsic value whose present-value declines, rather than increases with increasing $T$. Another exception would be an in-the-money call option on a currency with a high interest rate.

Unless the stock is paying a dividend, the value of an American and European call option are the same - there is no advantage to being able to exercise early. For a put there can be an advantage to exercising early even if there is no dividend. So the price of an American put will be slightly higher than that of the European put option. But this is not typically a significant difference in value.

To return to the special case of an ATM forward call option where $K=Se^{rT}$ on a zero-dividend stock, the price dependence of the option price $C$ with $T$ years to expiry and a stock price $S$ is approximately proportional to the square root of the time to expiry. The approximation is:

$C(S,T) \simeq 0.4 S \sigma \sqrt{T}$

where $\sigma$ is the volatility of the stock price.

This $T$ dependency does not change very much even if the option pays a dividend. For options that are not ATM forward the $T$ dependency is more complex.

  • $\begingroup$ From Falcon's Basic Option Pricing and Trading, he states that for non-dividend paying European put option, its value decreases whenever time to maturity increases. $\endgroup$
    – Idonknow
    Nov 28 '19 at 14:05
  • $\begingroup$ That's only true if the put option is in-the-money and there is no dividend. I will amend. $\endgroup$
    – Dom
    Nov 28 '19 at 14:20
  • $\begingroup$ Just adding that the relationship you've displayed holds for at-the-money options, namely when $S\approx K$. $\endgroup$ Nov 28 '19 at 14:22
  • $\begingroup$ Yes. I just made it clearer. $\endgroup$
    – Dom
    Nov 28 '19 at 14:41
  • $\begingroup$ @Dom 'To return to the special case of an ATM call option where $K=Se^{rT}$ on a zero-dividend stock,...' I do not understand how to obtain $K=Se^{rT}$ for ATM call option. From what I know, ATM call option refers to strike equals to current spot. So, how to derive $K=Se^{rT}$ from it? $\endgroup$
    – Idonknow
    Dec 4 '19 at 13:57

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