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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?

Marie. P.
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2 Answers2

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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.

Chris Taylor
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    Why this definition of convexity is equivalent to the std definition : $\forall K_1, K_2$, $t \in(0,1)$, $V(tK_1 + (1-t) K_2) \leq tV(K_1) + (1-t)K_2$...?? – noob-mathematician Oct 12 '16 at 16:26
  • Choose $K_1=K-\delta$, $K_2=K+\delta$ and $t=1/2$ to see the equivalence. – Chris Taylor Dec 15 '17 at 08:01
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    @ChrisTaylor, but by taking t=1/2, aren't we considering a particular case only? – User Aug 11 '20 at 09:33
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    @User you can show that the case $t =1/2$ is equivalent to standard convexity when the function is continuous. In this case, the black scholes price is continuous in $K$/ – rubikscube09 Oct 29 '20 at 23:06
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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.

Selos
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  • Your answer concerns the expected pay-out instead of the price. The latter includes a risk premium. – LinAlg Jan 27 '17 at 13:12
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    The expected pay-out (under the risk-neutral measure) is the call option price. – Selos Jan 27 '17 at 15:55
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    By the Fundamental Theorem of Asset Pricing it's known that the market is arbitrage-free if there exists at least one risk-neutral probability measure. (The market is even complete if there exists only one risk-neutral probability measure). Therefore, if there is no risk-neutral measure, the market is not arbitrage-free. So we can "assume" that there exists at least one risk-neutral probability measure, since we want the market to be arbitrage free for our model/calculations. – Selos Mar 28 '17 at 12:57
  • The call price is the expected payout under the (unique) Martingale risk-neutral measure. This can be written as an iterated expectation first conditioning on the stock price at expiry, then the expectation over those stock prices is a linear combination of those simple convex functions Quant Finance uses. Linear combos of convex fns are convex fns. – Don Slowik Nov 16 '18 at 16:53