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I know the payoff of a European Call at expiration is $\max\{S_T - K,0\}$, where $S_T$ is the underlying price at expiration and $K$ is the strike price.

I want to prove convexity of the value of a European Call $C^E$ in the underlying price $S$ , from first principles, i.e.

$C^E (\alpha S_a + (1-\alpha) S_b) \leq \alpha C^E (S_a) + (1-\alpha) C^E (S_b)$

Note I don’t write the other variables as they’re held constant in this statement.

I’m taking a course on Financial Derivatives, in which many equations are proved by constructing an arbitrage supposing the equality doesn’t hold.

My course just covered Forwards and Futures and now we‘re starting to study Options.

The professor told us an arbitrage argument also works for this statement.

So I’d start by supposing the opposite statement, i.e. there exists $\alpha \in (0,1), S_a, S_b > 0$ such that

$C^E (\alpha S_a + (1-\alpha) S_b) > \alpha C^E (S_a) + (1-\alpha) C^E (S_b)$

Usually the way to go in these proofs (kind of) is

At $t=0$:

  • Buy the cheaper side
  • Short the expensive side
  • If you have leftover money, but T-bills. If the opposite happens, then short T-bills. This is done in a way such that the initial value of the portfolio is $0$.

Then check that at $t=T$ all your positions cancel out and you’re left with a positive profit no matter what.

However, I don’t know how to proceed in this case, since only one underlying price exists at a given time and the underlying isn’t guaranteed to hit all of the three price points $S_a, S_b$ and $\alpha S_a + (1-\alpha) S_b$.

I could only guarantee one by defining the current underlying price as either of them.

I know how to prove convexity in the strike price:

  • Buy $\alpha$ calls with strike price $K_a$
  • Buy $1-\alpha$ calls with strike price $K_b$
  • Short $1$ call with strike price $\alpha K_a + (1-\alpha) K_b$
  • Use the excess money to buy T-bills.

This works as at any time there exists contracts for any strike. As I said before, this wouldn’t work when varying the underlying price.

I know about Put-Call parity but don’t see how that would be useful here, since everything is in terms of calls.

I would appreciate any insights on this proof.

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I know nothing about finance so please do not be too harsh.

Let's sell 1 call option with underlying price $\alpha S_a + (1-\alpha) S_b$ and strike price $K$. This means that I need to pay $$ \max\{\alpha S_a + (1-\alpha) S_b- K, 0\} $$

Let's use this money to buy $\alpha$ call options with price $S_a$ and $(1-\alpha)$ call options with price $S_b$ and strike price $K$. So we will get $$ \alpha \max\{S_a - K,0\} + (1-\alpha) \max\{S_b - K,0\}. $$ Given that: $$ C^E(\alpha S_a + (1-\alpha) S_b) > \alpha C^E(S_a) + (1-\alpha) C^E(S_b), $$ we have some money left, which we can invest in a risk free asset.

Let us show that we can always pay out the call option that we sold.

  1. If $K \ge S_a, S_b$, then also $K \ge \alpha S_a + (1-\alpha) S_b$, so we get nothing but we also need to pay nothing.

  2. If $S_a, S_b \ge K$, then also $\alpha S_a + (1-\alpha) S_b \ge K$ So we get $\alpha (S_a - K) + (1-\alpha) (S_b - K)$ and we need to pay $\alpha S_a + (1-\alpha) S_b - K$ which is the same.

  3. If $S_a > K > S_b$ and $K > \alpha S_a + (1-\alpha) S_b$ then we receive $\alpha(S_a - K)$ and need to pay nothing. I'm super happy. The same reasoning holds if $S_b > K > S_a$ and $K > \alpha S_a + (1-\alpha) S_b$.

  4. If $S_a > K > S_b$ and $\alpha S_a + (1-\alpha) S_b > K$ then we receive $\alpha (S_a - K)$ and need to pay $\alpha S_a + (1-\alpha) S_b - K$. Note that: $$ \begin{align*} &\alpha (S_a - K) > \alpha S_a + (1-\alpha) S_b - K\\ \leftrightarrow & (1-\alpha) K > (1-\alpha) S_b\\ \leftrightarrow & K > S_b \end{align*} $$ which is indeed the case. Again I'm happy. If $S_b > K > S_a$ and $\alpha S_a + (1-\alpha) S_b > K$, we obtain a similar result.

If there is no arbitrage opportunity, then the convexity condition should hold.

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Proposition. If $f:A\rightarrow\mathbb{R}$ and $g:A\rightarrow\mathbb{R}$ are convex functions defined on convex subset $A$ of $\mathbb{R}^n$, then $\max(f,g):A\rightarrow\mathbb{R}$, defined as $\max(f,g)(x) = \max(f(x),g(x))$ for all $x\in A$, is also a convex function.

Proof. Consider arbitrary $x',x''\in A$, and arbitrary $\lambda\in [0,1]$,

\begin{eqnarray*} && \max(f,g)(\lambda x'+(1-\lambda)x'') \\ & = & \max(f(\lambda x'+(1-\lambda)x''),g(\lambda x'+(1-\lambda)x'')) \\ & \leq & \max(\lambda f( x')+(1-\lambda)f(x''),\lambda g(x')+(1-\lambda)g(x'')) \\ & \leq & \max(\lambda \max(f,g)( x')+(1-\lambda)\max(f,g)(x''),\lambda \max(f,g)(x')+(1-\lambda)\max(f,g)(x'')) \\ &=& \lambda\max(f,g)(x')+(1-\lambda)\max(f,g)(x'') \end{eqnarray*}

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  • $\begingroup$ Could you pls specify the one to one correspondence between your f,g and the variables in the OP? $\endgroup$
    – dm63
    Sep 3, 2023 at 12:28
  • $\begingroup$ $\max(f,g)$ is $\max(S_T-K,0)$. Please correct me if I am wrong. Thanks. $\endgroup$
    – Amit
    Sep 3, 2023 at 15:12
  • $\begingroup$ Well then f and g are linear functions of $S_T$, so you have proved that the payoff of a call is convex. But the OP is to prove the current value of the call is convex as a function of today’s stock price. $\endgroup$
    – dm63
    Sep 3, 2023 at 20:53

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