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Define \begin{align} c & \text{ the price of a European call option at time } t = 0, \\ S & \text{ the spot price of a share at time } t=0, \\ X & \text{ the strike price for the option}, \\ T & \text{ the expiry time of the option}, \\ r & \text{ the } T \text{-year spot interest rate}, \\ S_T & \text{ the spot price of the share at time } t=T. \end{align} My course notes contain a proposition,

Assume that the stock does not pay dividends. Then $c>S-Xe^{-rT}$.

To prove this consider:

Portfolio A: one European call option plus an amount of cash equal to $Xe^{−rT}$ deposited for $T$ years at an interest rate of $r$.

Portfolio B: one share.

After $T$ years portfolio A will yield an amount of cash equal to $X$. If, after $T$ years, the stock price $S_T$ is above $X$, the call option in portfolio A will be exercised, the share sold and the portfolio will be worth $S_T$ . Otherwise, after $T$ years, $S_T ≤ X$, the option is not exercised and the portfolio will be worth $X$. So after $T$ years portfolio A is worth max$(S_T, X) \geq S_T$, and since portfolio B is always worth $S_T$ after $T$ years, the initial value of portfolio A must be no less that the initial value of portfolio B, which is just $S$. But since sometimes portfolio A is worth more than portfolio B we have a strict inequality $c+Xe^{−rT}>S$.

In cases where $S_T>X$ we have $V(A)=V(B)$ at time $t=T$, which means in those cases we have $V(A)=V(B) \; \forall \; t \in[0,T]$; so I assume "sometimes portfolio A is worth more than portfolio B" doesn't mean "in every case there is a time at which portfolio A is worth more than portfolio B". The other interpretation I can think of is "there are cases in which at time $t=T$ portfolio A is worth more than portfolio B". If that's the intended meaning, why is the inequality made strict without specifying that restriction?

Jose Avilez
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mjc
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1 Answers1

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The correct way of interpreting the last statement is by adding probabilities (to an otherwise model-free argument). Let $V_T^A$ and $V_T^B$ be the terminal values of portfolios $A$ and $B$, respectively. Then, the conclusion of your argument can be translated to: $V_T^A \geq V_T^B$ and $P(V_T^A > V_T^B) > 0$.

Thus, the initial value of portfolio $A$ must be strictly larger.

Jose Avilez
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  • Would you agree that $P(V_T^A>V_T^B)>0$ is an assumption? For instance, if the share is effectively a risk-free bond then $S=Xe^{-rT}$ and $P(V_T^A>V_T^B)=0$ and $c=0=S-Xe^{-rT}$. – mjc Oct 23 '23 at 06:34
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    @mjc Sure, but one usually assumes the stock is not the riskless asset :) – Jose Avilez Oct 23 '23 at 06:39