<p>Proposition. If <span class="math-container">$f:A\rightarrow\mathbb{R}$</span> and <span class="math-container">$g:A\rightarrow\mathbb{R}$</span> are convex functions defined on convex subset <span class="math-container">$A$</span> of <span class="math-container">$\mathbb{R}^n$</span>, then <span class="math-container">$\max(f,g):A\rightarrow\mathbb{R}$</span>, defined as <span class="math-container">$\max(f,g)(x) = \max(f(x),g(x))$</span> for all <span class="math-container">$x\in A$</span>, is also a convex function.</p>
<p>Proof. Consider arbitrary <span class="math-container">$x',x''\in A$</span>, and arbitrary <span class="math-container">$\lambda\in [0,1]$</span>,</p>
<p><span class="math-container">\begin{eqnarray*} &amp;&amp; \max(f,g)(\lambda x'+(1-\lambda)x'') \\ &amp; = &amp; \max(f(\lambda x'+(1-\lambda)x''),g(\lambda x'+(1-\lambda)x'')) \\ &amp; \leq  &amp; \max(\lambda f( x')+(1-\lambda)f(x''),\lambda g(x')+(1-\lambda)g(x'')) \\ &amp; \leq  &amp; \max(\lambda \max(f,g)( x')+(1-\lambda)\max(f,g)(x''),\lambda \max(f,g)(x')+(1-\lambda)\max(f,g)(x'')) \\ &amp;=&amp; \lambda\max(f,g)(x')+(1-\lambda)\max(f,g)(x'') \end{eqnarray*}</span></p>

CC BY-SA 4.0
2023-09-03T12:28:17.423Z
Could you pls specify the one to one correspondence between your f,g and the variables in the OP?
dm63
CC BY-SA 4.0
2023-09-03T15:12:46.253Z
$\max(f,g)$ is $\max(S_T-K,0)$. Please correct me if I am wrong. Thanks.
Amit
CC BY-SA 4.0
2023-09-03T20:53:33.353Z
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.
dm63
CC BY-SA 4.0
2023-09-02T05:59:32.130Z
2023-09-02T05:59:32.130Z
Amit
<p>I know nothing about finance so please do not be too harsh.</p>
<p>Let's sell 1 call option with underlying price <span class="math-container">$\alpha S_a + (1-\alpha) S_b$</span> and strike price <span class="math-container">$K$</span>. This means that I need to pay
<span class="math-container">$$
\max\{\alpha S_a + (1-\alpha) S_b- K, 0\}
$$</span></p>
<p>Let's use this money to buy <span class="math-container">$\alpha$</span> call options with price <span class="math-container">$S_a$</span> and <span class="math-container">$(1-\alpha)$</span> call options with price <span class="math-container">$S_b$</span> and strike price <span class="math-container">$K$</span>. So we will get
<span class="math-container">$$
\alpha \max\{S_a - K,0\} + (1-\alpha) \max\{S_b - K,0\}.
$$</span>
Given that:
<span class="math-container">$$
C^E(\alpha S_a + (1-\alpha) S_b) &gt; \alpha C^E(S_a) + (1-\alpha) C^E(S_b),
$$</span>
we have some money left, which we can invest in a risk free asset.</p>
<p>Let us show that we can always pay out the call option that we sold.</p>
<ol>
<li><p>If <span class="math-container">$K \ge S_a, S_b$</span>, then also <span class="math-container">$K \ge \alpha S_a + (1-\alpha) S_b$</span>, so we get nothing but we also need to pay nothing.</p>
</li>
<li><p>If <span class="math-container">$S_a, S_b \ge K$</span>, then also <span class="math-container">$\alpha S_a + (1-\alpha) S_b \ge K$</span> So we get <span class="math-container">$\alpha (S_a - K) + (1-\alpha) (S_b - K)$</span> and we need to pay <span class="math-container">$\alpha S_a + (1-\alpha) S_b - K$</span> which is the same.</p>
</li>
<li><p>If <span class="math-container">$S_a &gt; K &gt; S_b$</span> and <span class="math-container">$K &gt; \alpha S_a + (1-\alpha) S_b$</span> then we receive <span class="math-container">$\alpha(S_a - K)$</span> and need to pay nothing. I'm super happy. The same reasoning holds if <span class="math-container">$S_b &gt; K &gt; S_a$</span> and <span class="math-container">$K &gt; \alpha S_a + (1-\alpha) S_b$</span>.</p>
</li>
<li><p>If <span class="math-container">$S_a &gt; K &gt; S_b$</span> and <span class="math-container">$\alpha S_a + (1-\alpha) S_b &gt; K$</span> then we receive <span class="math-container">$\alpha (S_a - K)$</span> and need to pay <span class="math-container">$\alpha S_a + (1-\alpha) S_b - K$</span>. Note that:
<span class="math-container">$$
\begin{align*}
&amp;\alpha (S_a - K) &gt; \alpha S_a + (1-\alpha) S_b - K\\
\leftrightarrow &amp; (1-\alpha) K &gt; (1-\alpha) S_b\\
\leftrightarrow &amp; K &gt; S_b
\end{align*}
$$</span>
which is indeed the case. Again I'm happy. If <span class="math-container">$S_b &gt; K &gt; S_a$</span> and <span class="math-container">$\alpha S_a + (1-\alpha) S_b &gt; K$</span>, we obtain a similar result.</p>
</li>
</ol>
<p>If there is no arbitrage opportunity, then the convexity condition should hold.</p>

CC BY-SA 4.0
2023-09-01T09:08:48.827Z
2023-09-01T09:08:48.827Z
tdm
<p>I know the payoff of a European Call at expiration is <span class="math-container">$\max\{S_T - K,0\}$</span>, where <span class="math-container">$S_T$</span> is the underlying price at expiration and <span class="math-container">$K$</span> is the strike price.</p>
<p>I want to prove convexity of the value of a European Call <span class="math-container">$C^E$</span> in the underlying price <span class="math-container">$S$</span> , from first principles, i.e.</p>
<p><span class="math-container">$C^E (\alpha S_a + (1-\alpha) S_b) \leq \alpha C^E (S_a) + (1-\alpha) C^E (S_b)$</span></p>
<p>Note I don’t write the other variables as they’re held constant in this statement.</p>
<p>I’m taking a course on Financial Derivatives, in which many equations are proved by constructing an arbitrage supposing the equality doesn’t hold.</p>
<p>My course just covered Forwards and Futures and now we‘re starting to study Options.</p>
<p>The professor told us an arbitrage argument also works for this statement.</p>
<p>So I’d start by supposing the opposite statement, i.e. there exists <span class="math-container">$\alpha \in (0,1), S_a, S_b &gt; 0$</span> such that</p>
<p><span class="math-container">$C^E (\alpha S_a + (1-\alpha) S_b) &gt; \alpha C^E (S_a) + (1-\alpha) C^E (S_b)$</span></p>
<p>Usually the way to go in these proofs (kind of) is</p>
<p>At <span class="math-container">$t=0$</span>:</p>
<ul>
<li>Buy the cheaper side</li>
<li>Short the expensive side</li>
<li>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 <span class="math-container">$0$</span>.</li>
</ul>
<p>Then check that at <span class="math-container">$t=T$</span> all your positions cancel out and you’re left with a positive profit no matter what.</p>
<p>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 <span class="math-container">$S_a, S_b$</span> and <span class="math-container">$\alpha S_a + (1-\alpha) S_b$</span>.</p>
<p>I could only guarantee one by defining the current underlying price as either of them.</p>
<p>I know how to prove convexity in the strike price:</p>
<ul>
<li>Buy <span class="math-container">$\alpha$</span> calls with strike price <span class="math-container">$K_a$</span></li>
<li>Buy <span class="math-container">$1-\alpha$</span> calls with strike price <span class="math-container">$K_b$</span></li>
<li>Short <span class="math-container">$1$</span> call with strike price <span class="math-container">$\alpha K_a + (1-\alpha) K_b$</span></li>
<li>Use the excess money to buy T-bills.</li>
</ul>
<p>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.</p>
<p>I know about Put-Call parity but don’t see how that would be useful here, since everything is in terms of calls.</p>
<p>I would appreciate any insights on this proof.</p>

<finance><financial-markets><corporate-finance><financial><financial-derivatives>
How to prove the value of a European Call Option is convex on the underlying’s price?
Nicolas Torres
# How to prove the value of a European Call Option is convex on the underlying’s price?

(Asked by: username_0 on 2023-08-30)

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.

**Tags:** finance, financial-markets, corporate-finance, financial, financial-derivatives

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## 2 answers

### Answer by username_3 on 2023-09-01. Score: 3 | Accepted answer

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.


---

### Answer by username_1 on 2023-09-02. Score: 3

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\*}

> **Comments:**
> **username_2**: Could you pls specify the one to one correspondence between your f,g and the variables in the OP?  
> **username_1**: $\max(f,g)$ is $\max(S_T-K,0)$. Please correct me if I am wrong. Thanks.  
> **username_2**: 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.  

---

