# 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.

• This has been answered on math stackexchange. Please check: math.stackexchange.com/questions/112063/… Jan 15 '16 at 18:16
• I will still post it as an answer, so that this does not go unanswered. Tried to vote for close, but the duplicate must be on quant.stackexchange. Jan 15 '16 at 18:29

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.

• This answer starts from the identity $V (K+\delta)+V (K-\delta) > 2V (K)$ which is only a *particular case* of the full convexity definition stated in the OP's question (corresponds to taking $K_1=K-\delta$, $K_2=K+\delta$ and $\lambda=1/2$). Wouldn't it be more general to show that convexity means $\partial_K^2 C (K) = C^{\prime\prime}(K) \geq 0$ (if this derivative exists of course) and then re-derive the Breeden-Litzenberger identity to show that this is indeed true, because $\text {sign}[C^{\prime\prime}(K)] = \text {sign}[q(S_0;T,S_T=K)] \geq 0$ by definition of a pdf? Mar 28 '16 at 19:13
• @Quantuple Is it though? - a particular case?
– htd
Mar 27 '17 at 12:39
• What do you mean @Henrik? Don't you agree that starting from $$C(T, \lambda K_1 + (1 - \lambda)K_2, S, t) \leq \lambda C(T, K_1, S, t) + (1 - \lambda)C(T, K_2, S, t) \tag{A}$$ letting $K_1 = K-\delta$, $K_2=K+\delta$ and $\lambda=1/2$ one gets: $$C(T, \frac{1}{2} (K + \delta + K - \delta), S, t) \leq \frac{1}{2} \left( C(T, K-\delta, S, t) + C(T, K+\delta, S, t)\right)$$ which is equivalent to writing: $$2 C(T,K, S, t) \leq C(T, K-\delta, S, t) + C(T, K+\delta, S, t) \tag{B}$$ hence inequality $(B)$ is a particular instance of inequality $(A)$ Mar 27 '17 at 13:06
• @Quantuple $K\mapsto C(K)$ needs to be really messed up for the definitions to not be equivalent. See e.g. math.stackexchange.com/q/83383/28157 and math.stackexchange.com/q/83383/28157
– htd
Mar 28 '17 at 9:26
• Thank you @Henrik for the reference. I guess what I was saying then amounts to put up a "caveat" stating that it's sufficient to prove midpoint-convexity assuming the mapping is continuous (midpoint convexity -> rational convexity -> convexity). Mar 28 '17 at 9:50