-1

I'm trying to prove that the price of an European call option, denoted by $C(T,K),$ where $T$ is the maturity date and $K$ is the strike, is an increasing function of $T.$

I can't see why the map $T\rightarrow C(T,K)$ is an increasing function.I was using the definition of a Call option but I don't find clear the reason of the proposition.

My little Squid
  • 37

2 Answers2

1

Suppose that $T_1 < T_2$ but $C(T_1, K) > C(T_2,K)$ for a contradiction. Call the option with maturity time $T_i$ option $i$ and write $S_t$ for the price of the underlying asset at time $t$ (e.g. the stock price). I proceed by a no arbitrage argument.

We begin by buying the cheap option and writing the expensive one. That is we buy option $2$ and we sell option $1$ giving a profit of $C(T_1,K) - C(T_2,K) > 0$ which we place in a risk-free asset.

At time $T_1$, we may have $S_{T_1} < K$ in which case the holder of option $1$ does nothing and we can then also do nothing to guarantee the profit from our risk-free asset (though this may be suboptimal).

Otherwise $S_{T_1} \geq K$ and the holder of option $1$ exercises the option so we must short-sell them the stock for the strike price $K$. We are left to close out the short sell, which we do by exercising option $2$ at time $T_2$ to use the $K$ profit from the short-sell to buy the underlying stock leaving us again with the profit from the risk-free asset.

In total this gives a risk-free profit of $[C(T_1,K) - C(T_2,K)] e^{rT_2}$ where $r$ is the risk-free rate so this strategy gives an arbitrage. Therefore by the no-arbitrage assumption we must in fact have $C(T_1,K) \leq C(T_2,K)$.

Rhys Steele
  • 19,671
  • 1
  • 19
  • 50
  • Thanks for answering @nobody. Your proof is very clear. Could you give me some advices about how to create strategies to prove this kind of propositions? – My little Squid Aug 22 '17 at 07:38
  • My initial strategy for this kind of no arbitrage argument is to suppose that whatever I want to be true isn't true and then buy whatever ends up being cheap and sell what ends up being (relatively) expensive. Often then all that's left is to work through the cases and check that you never make a loss and you can make a positive profit in at least one of the cases. – Rhys Steele Aug 22 '17 at 08:59
0

You can look at this in a simplistic way. The holder of the option as time increases $T$ has a higher probability of the stock fluctuation get higher or lower. The holder basically is paying for more time to watch the stock fluctuate. As the stock movement is more unpredictable and therefore riskier as the time passes by, the return $S-K$(S-underlying asset price) can be higher once the volatility measured by the slander-deviation gets higher. If it gets much lower that does not affect the holder of the option once he\she can "opt out"(not exercise the option).

Pedro Gomes
  • 3,891