I'm trying to show that the price of a European call option (payoff function is $(S_1-K)^+$) in a no-arbitrage market is a decreasing and convex function of K. That it shall be decreasing makes sense; as $K$ increases, $S_1-K$ decreases and we make less profit. But why shall it be convex?
Let the price of an option at strike $K$ be given by $V(K)$. To say that the price is convex in the strike means that
$$V(K-\delta) + V(K+\delta) > 2 V(K)$$
for all $K>0$ and $\delta>0$. Let's assume that the opposite is true, i.e. that there exist tradeable option contracts expiring on the same date such that
$$V(K-\delta) + V(K+\delta) \leq 2 V(K)$$
I therefore buy a contract at $K+\delta$ and one at $K-\delta$, and finance my purchase by selling two of the options at $K$ (which I can do, because the two options struck at $K$ are at least as expensive as the other two combined).
At expiry the price of the stock is $S$, and my total payout is
$$P = (S-(K-\delta))^+ + (S-(K+\delta))^+ - 2(S-K)^+$$
Now there are four regimes:
- $S<K-\delta$, which means $P=0$
- $K-\delta < S < K$, which means $P=S-(K-\delta) > 0$
- $K < S < K+\delta$, which means $P=S-K+\delta - 2(S-K)=K+\delta-S>0$
- $S>K+\delta$, which means $P = S-K+\delta + S-K-\delta - 2(S-K) = 0$
So I have the possibility of making a profit, but no possibility of making a loss - which is an arbitrage. Since no arbitrages exist, the option price must be convex in the strike price.
There is a much simpler solution to the question regarding the convexity of the call option function. Let $K$ be the strike price and $x$ the stock price. Since the function from $K$ to $\max(0, x-K)$ is convex and conditional expectations are linear, the function $K$ to the conditional expectation of $\max(0, x-K)$ (under the risk-neutral density function) is also convex. Therefore we have shown that the call option price function is convex.