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/… – phdstudent 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. – phdstudent 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? – Quantuple Mar 28 '16 at 19:13
• @Quantuple Is it though? - a particular case? – Henrik 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 $(A)$ – Quantuple 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 – Henrik 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). – Quantuple Mar 28 '17 at 9:50