Why European Call Prices Are Convex Across Strike Prices
Summary
The document presents an arbitrage argument for convexity of European call option prices as a function of strike. If the price at a middle strike were at least as high as the average prices of two equally spaced neighboring strikes, a trader could buy the lower- and higher-strike calls and sell two middle-strike calls for a nonnegative initial cost under the assumed price inequality.
At expiration, the resulting payoff is zero when the stock finishes outside the two outer strikes and positive within them. That payoff pattern would create an arbitrage, so the assumed pricing inequality is ruled out in an arbitrage-free market. The proof is framed for calls with the same expiration. Although the question also asks about puts and American options, the supplied answer does not extend the argument to those contracts or discuss conditions such as early exercise.
Key ideas
- For calls with a common expiration, the payoff combination of two outer strikes against two middle-strike calls is nonnegative at expiration.
- The payoff is positive when the terminal stock price lies between the outer strikes and zero outside that interval.
- An initial pricing inequality that funds this combination would imply an arbitrage.
- The argument provided addresses European calls and does not prove the requested claims for puts or American options.
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Full text
# Price of call/put is convex in $K$ (strike price) # Price of call/put is convex in $K$ (strike price) 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 by phdstudent (score 5, accepted) https://quant.stackexchange.com/a/22770 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.
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