# Proof European call price is always less than stock price. (proof verification)

Let $$C_K(t,T)$$ be the value of a European call with strike $$K$$ and maturity $$T$$ on a stock with value $$S_t$$ at time $$t$$. Then for all $$t\leq T$$ we have $$C_K(t,T)\leq S_t.$$

$$\textbf{Proof}$$: We derive an arbitrage opportunity by contradiction.

Assume $$C_K(t,T)>S_t$$ for some $$t\leq T$$. Consider the following:

At this time $$t$$: write a call option with strike $$K$$ and maturity $$T$$ on the stock and buy said stock. Since we cash the premium $$C_K(t,T)$$ of the call we have $$C_K(t,T)-S_t>0$$ left.

At time $$t=T$$, we do the following:

1. If $$S_T, the call is worthless and nothing happens.
2. If $$S_T \geq K$$, the call is made and we sell the holder the stock (which we still own).

In both cases, no money is lost and we end up with money. This is an arbitrage opportunity.

Can someone check if this proof is decent and correct?

$$\textbf{Edit:}$$ I did not think this was necessary, but since a comment was made I feel obliged to add the following.

This is a proof by contradiction. Since the principle of arbitrage is an axiom to financial theory, all true statements (in the theory) are implied by this principle. In other words: it is enough to show that the negation ($$C_K(t,T)>S_t$$) is in contradiction with the principle of arbitrage.

• why is $C_K(t,T)-S_t>0$ ? This implies that $C_K(t,T) > S_t$, which is exactly the opposite of the original relation you're tring to prove...
– will
Jan 18, 2020 at 18:45
• @will It's a proof by contradiction. Proving $P\implies Q$ is equivalent to showing $P\land \neg Q$ is not true. Jan 18, 2020 at 18:55
• @will in this case $P$ is the principle of arbitrage and $Q$ is the proposition in the yellow box. Jan 18, 2020 at 18:56
• The proof is fine; however, it does rely on some implicit assumptions- e.g., limited liability associated with stock ownership, meaning stock price can’t go negative. Jan 19, 2020 at 17:00
– will
Jan 19, 2020 at 20:55

What you need to note is the following: \begin{align*} S_T - \max(S_T-K, \,0) &= S_T + \min(K-S_T, \,0)\\ &=\min(K, \, S_T) >0. \end{align*}
• Yes, that can more or less be made into a formal proof. It is indeed a corollary of the law of one price that if $K_1<K_2$, then for all $t\leq T: C_{K_2}(t,T)<C_{K_1}(t,T)$. Jan 18, 2020 at 19:14