Price of call/put is convex in $K$ (strike price)

Question

Let $\lambda\in(0,1)$. Then $$C(T, \lambda K_1 + (1 - \lambda)K_2, S, t) \leq \lambda C(T, K_1, S, t) + (1 - \lambda)C(T, K_2, S, t)$$

$T$ - the maturity

$K_1$,$K_2$ - Strike prices

$S$ - stock price

$t$ - current time

In other words, the price of the call/put option in convex in $K$. Show the same claim for the price of put options, American call options, and American put options.

I think you need to just apply the triangle inequality, but I am not sure. Any suggestions is greatly appreciated.

Answer 1

Using the answer from: Chris Taylor, on math stackexchange (link):

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.