From Forwards to Options¶

Key Idea

Forwards are easy because the payoff is linear. Options are hard because the payoff is nonlinear. This single distinction---linearity versus nonlinearity---explains why forward pricing requires only algebra and static replication, while option pricing demands the full machinery of stochastic calculus.

A Two-Scenario Hedge That Fails¶

Before any general argument, watch a single static hedge break. Suppose $K = 100$ and the stock can end at either $S_T = 90$ or $S_T = 120$ — only two scenarios. A long call $(S_T - K)^+$ pays $0$ in the first scenario and $20$ in the second. Try to match these payoffs with a static portfolio of $\phi$ shares and a constant cash amount $c$:

Scenario down ($S_T = 90$): $\;\; 90 \phi + c = 0$
Scenario up ($S_T = 120$): $\;\; 120 \phi + c = 20$

Solving gives $\phi = \tfrac{2}{3}$ and $c = -60$. The static hedge works — in exactly two states. Now add a third scenario, $S_T = 110$. The call pays $10$, but the same portfolio pays $\tfrac{2}{3}(110) - 60 = 13.33$. The "hedge" is now wrong by $\$3.33$, with neither $\phi$ nor $c$ free to fix it.

Repeat the same exercise with a long forward $S_T - K$ instead. The system is

Down: $\;\; 90 \phi + c = -10$
Up: $\;\; 120 \phi + c = 20$

giving $\phi = 1$, $c = -100$. Plug in $S_T = 110$: the portfolio pays $110 - 100 = 10$ — exactly the forward payoff. Add a fourth scenario, a fifth, a continuum: the same $(\phi, c) = (1, -100)$ keeps working. The static hedge survives every state of the world precisely because the forward payoff lies on the single straight line $\phi S + c$.

The call's payoff lies on two straight pieces joined at $S_T = K$. No single line, and hence no fixed $(\phi, c)$, can sit on both. The hedger must change $\phi$ as $S$ moves — the dynamic replication that powers Black–Scholes — and that change is what makes the option price depend on volatility.

The previous sections developed forward and futures pricing from first principles: no-arbitrage, cost of carry, and the replication argument. Every result rested on the fact that the forward payoff $S_T - K$ is a linear function of the terminal stock price. This linearity made replication simple and pricing elegant. We now examine what happens when linearity breaks down, and why that breakdown leads directly to the Black-Scholes theory.

Linear Payoff and Static Replication¶

Recall (see § Payoff of Forwards and Futures and § No-Arbitrage Pricing of Forwards): the long forward payoff $S_T - K$ is linear in $S_T$, and a one-time portfolio — buy one share at $S_0$, borrow $K e^{-rT}$ — replicates it exactly, pinning the forward price at

\[ F_0 = S_0 e^{rT} \]

The crucial observation is that this derivation required no assumptions about the stock's volatility $\sigma$, no model for the stock price dynamics, and no stochastic calculus. Linearity of the payoff made the problem purely algebraic — the hedge is constructed once at $t = 0$ and held until $T$ without adjustment.

Nonlinear Payoff and the Need for Dynamic Replication¶

Now consider a European call option with the same strike $K$ and maturity $T$. Its payoff is

\[ (S_T - K)^+ = \max(S_T - K, \, 0) \]

This payoff is nonlinear in $S_T$. The $\max$ operator introduces a kink at $S_T = K$: the payoff rises one-for-one with $S_T$ above the strike but is flat (zero) below it. Unlike the forward, a static portfolio of stock and bonds cannot replicate this kinked payoff for all possible values of $S_T$.

Why not? A portfolio of $\phi$ shares and $\psi$ bonds has terminal value $\phi S_T + \psi e^{rT}$, which is always linear in $S_T$. No fixed choice of $(\phi, \psi)$ can match a function that is linear on one piece and flat on another. The replication strategy must therefore change over time as the stock price moves---it must be dynamic.

In a dynamic replication strategy, the portfolio weights $\phi_t$ and $\psi_t$ are adjusted continuously (or at least frequently) in response to changes in $S_T$. At each instant, the portfolio is rebalanced to maintain a local match with the option's payoff profile. This continuous adjustment is called delta hedging, and it is the mechanism through which option pricing becomes tied to the volatility $\sigma$ of the underlying asset.

Put-Call Parity: The Bridge¶

The connection between the linear world of forwards and the nonlinear world of options is made precise by put-call parity. For European options on a non-dividend-paying stock:

\[ C - P = S_0 - Ke^{-rT} \]

where $C$ is the call price, $P$ is the put price, $S_0$ is the current stock price, and $K$ is the common strike. The right-hand side is exactly the value of a forward contract with delivery price $K$:

\[ C - P = e^{-rT}(F_0 - K) \]

Put-call parity says that a long call and short put with the same strike and maturity replicates a forward. This is not surprising once we observe the payoff identity:

\[ (S_T - K)^+ - (K - S_T)^+ = S_T - K \]

The nonlinear pieces---the call and the put---combine to produce a linear payoff. Individually, each option is nonlinear and difficult to price. Together, their nonlinearities cancel perfectly. Put-call parity is therefore the algebraic bridge between forward pricing and option pricing: it shows that the difference between a call and a put is a forward, and the individual prices carry the additional complexity of nonlinearity.

Convexity, Gamma, and Continuous Hedging¶

The kink in the call payoff $(S_T - K)^+$ at $S_T = K$ is the source of all the additional mathematical complexity in option pricing. To understand why, consider the option's value $C(S, t)$ as a function of the current stock price $S$.

Far above the strike, the call behaves almost like a forward: $C \approx S - Ke^{-r(T-t)}$, and the hedge ratio (delta) is close to 1. Far below the strike, the call is nearly worthless and delta is close to 0. Near the strike, delta transitions rapidly from 0 to 1. The rate of this transition is measured by gamma:

\[ \Gamma = \frac{\partial^2 C}{\partial S^2} \]

Gamma is largest near the strike and near expiration. It quantifies the convexity (curvature) of the option price with respect to the stock price. A forward contract, by contrast, has $\Gamma = 0$ everywhere because its value is linear in $S$.

Nonzero gamma means that delta changes as the stock price moves, which is precisely why the hedge must be continuously adjusted. Each small move $dS$ in the stock price changes the option's delta by approximately $\Gamma \, dS$, requiring a rebalancing trade of $\Gamma \, dS$ shares. The cost of maintaining this hedge over the option's life depends on how much the stock price fluctuates---that is, on the volatility $\sigma$.

This is the fundamental reason that forward prices depend on $r$ alone while option prices depend on both $r$ and $\sigma$:

Forwards: $\Gamma = 0$, static hedge, price depends on $r$ only.
Options: $\Gamma \neq 0$, dynamic hedge, price depends on $r$ and $\sigma$.

What Comes Next¶

Big Picture

Forward pricing is the prototype for all derivative pricing. The principle is always the same: replicate the payoff, and no-arbitrage determines the price. For forwards, replication is static and the mathematics is elementary. Options add nonlinearity, which demands dynamic replication---and with it, the full machinery of stochastic calculus, Ito's lemma, and the Black-Scholes PDE.

In practice, many of the same underlying assets are traded as both futures and options. For example, S&P 500 E-mini futures (ES) — a linear instrument covered in this section — trade alongside SPX index options on the same underlying. SPX options are European-style and cash-settled with a $100 multiplier; their nonlinear payoff requires the Black-Scholes framework rather than the algebraic no-arbitrage arguments that suffice for ES. The transition from pricing ES futures to pricing SPX options is a concrete instance of the shift from linear to nonlinear payoffs described next.

The table below summarizes the conceptual transition from forwards to options:

	Forward	Option
Payoff	$S_T - K$ (linear)	$(S_T - K)^+$ (nonlinear)
Replication	Static	Dynamic (delta hedging)
Key parameters	$r, T, S_0$	$r, T, S_0, \sigma$
Gamma	$0$	$> 0$ (near the strike)
Pricing method	Algebraic no-arbitrage	Black-Scholes PDE / risk-neutral expectation

Section Summary (Futures and Forwards)

Forward pricing shows that derivative values can be determined entirely by replication under no-arbitrage. The payoff is linear, the hedge is static, and the mathematics is algebraic. This simplicity is not a limitation but a foundation: it reveals that prices are determined by replication, not by forecasting future outcomes. Options extend this principle to nonlinear payoffs, where replication becomes dynamic and pricing depends on volatility. The transition from static to dynamic replication is the central step toward the Black-Scholes model.

Exercises¶

Exercise 1. A European call and a European put on a non-dividend-paying stock share strike $K = 50$ and maturity $T = 1$ year. The current stock price is $S_0 = 52$ and the continuously compounded risk-free rate is $r = 0.04$. If the call is priced at $C = 5.50$, use put-call parity to determine the put price $P$.

Solution to Exercise 1

Put-call parity states

\[ C - P = S_0 - Ke^{-rT} \]

Substituting the given values:

\[ 5.50 - P = 52 - 50e^{-0.04} = 52 - 50 \times 0.9608 = 52 - 48.04 = 3.96 \]

Therefore

\[ P = 5.50 - 3.96 = 1.54 \]

Exercise 2. Explain why a static portfolio consisting of $\phi$ shares of stock and $\psi$ units of a zero-coupon bond cannot replicate the payoff $(S_T - K)^+$ for all possible values of $S_T$. Your argument should be precise: state what the portfolio pays at maturity and why it cannot match the option payoff on both regions $S_T > K$ and $S_T \leq K$ simultaneously.

Solution to Exercise 2

A static portfolio of $\phi$ shares and $\psi$ bonds has terminal value

\[ V_T = \phi S_T + \psi e^{rT} \]

which is a linear (affine) function of $S_T$. Suppose this portfolio replicates $(S_T - K)^+$ for all $S_T \geq 0$.

For $S_T > K$, the payoff is $S_T - K$, so we need $\phi S_T + \psi e^{rT} = S_T - K$, giving $\phi = 1$ and $\psi = -Ke^{-rT}$.

For $S_T \leq K$, the payoff is $0$, so we need $\phi S_T + \psi e^{rT} = 0$ for all $S_T \in [0, K]$. This requires $\phi = 0$ and $\psi = 0$.

These two systems are contradictory: we cannot have $\phi = 1$ and $\phi = 0$ simultaneously. Therefore no static portfolio of stock and bonds can replicate the call payoff. The kink at $S_T = K$ makes the payoff piecewise linear with different slopes on the two regions, which a single affine function cannot match. $\square$

Exercise 3. Show that the payoff identity $(S_T - K)^+ - (K - S_T)^+ = S_T - K$ holds for all $S_T \geq 0$ by considering the two cases $S_T \geq K$ and $S_T < K$ separately.

Solution to Exercise 3

Case 1: $S_T \geq K$.

In this case $(S_T - K)^+ = S_T - K$ and $(K - S_T)^+ = 0$, so

\[ (S_T - K)^+ - (K - S_T)^+ = (S_T - K) - 0 = S_T - K \]

Case 2: $S_T < K$.

In this case $(S_T - K)^+ = 0$ and $(K - S_T)^+ = K - S_T$, so

\[ (S_T - K)^+ - (K - S_T)^+ = 0 - (K - S_T) = S_T - K \]

In both cases the result is $S_T - K$, establishing the identity for all $S_T \geq 0$. $\square$

Exercise 4. A forward contract has $\Gamma = 0$, while a European call option has $\Gamma > 0$ near the strike. Explain in plain language why $\Gamma = 0$ implies that a forward can be hedged with a static portfolio, and why $\Gamma > 0$ implies that an option requires continuous rebalancing. Relate your answer to the dependence (or independence) of the price on volatility $\sigma$.

Solution to Exercise 4

Gamma measures how much delta (the hedge ratio) changes when the stock price moves. For a forward, $\Gamma = 0$ means delta is constant: it is always exactly 1. A hedge of one share of stock perfectly offsets the forward's exposure to stock price movements at all times, regardless of where the stock trades. Since the hedge never needs adjustment, it is static, and the cost of maintaining it does not depend on how much the stock price fluctuates. This is why the forward price is independent of $\sigma$.

For a call option, $\Gamma > 0$ means delta changes as the stock moves. When the stock rises, delta increases (the option becomes more sensitive to the stock), and the hedger must buy additional shares. When the stock falls, delta decreases, and the hedger must sell shares. The frequency and magnitude of these adjustments depend directly on how much the stock price moves---that is, on volatility $\sigma$. Higher volatility means more frequent and larger rebalancing trades, which in turn means higher hedging costs. These costs are reflected in the option price. This is why the call price depends on $\sigma$: volatility determines the cost of the dynamic hedge required to replicate the nonlinear payoff. $\square$

Exercise 5. Consider a portfolio consisting of one long call and one short put on a non-dividend-paying stock, both with strike $K$ and maturity $T$. Using put-call parity, show that this portfolio has the same value today as a forward contract with delivery price $K$. Conclude that, at inception of an at-the-money forward, $C = P$.

Solution to Exercise 5

Put-call parity gives

\[ C - P = S_0 - K e^{-rT} = e^{-rT}(F_0 - K) \]

The right-hand side is the present value of the long forward payoff $S_T - K$ — i.e., the value of a forward contract with delivery price $K$.

Recall (see § Payoff of Forwards and Futures): when $K = F_0$, the forward has zero value at inception. Then $C - P = e^{-rT}(F_0 - F_0) = 0$, so $C = P$. An at-the-money-forward call and put have equal price. $\square$

Exercise 6. Far in-the-money calls have delta $\Delta \approx 1$ and $\Gamma \approx 0$. Explain why such options price very nearly like a forward, and what the residual price difference represents.

Solution to Exercise 6

When $\Delta \approx 1$ and $\Gamma \approx 0$, the call's value behaves locally like a linear function of $S$ with slope $1$: $C(S, t) \approx S - K e^{-r(T-t)}$. This is exactly the present value of the forward payoff $S_T - K$. Intuitively, the option is so deep in the money that exercise at maturity is virtually certain, and the kink in $(S_T - K)^+$ is irrelevant.

The residual price difference is the out-of-the-money insurance: a small probability that $S_T$ ends below $K$, in which case the option pays $0$ rather than the negative forward value $S_T - K$. This residual is exactly the value of the embedded put — i.e., put-call parity reads $C = (S_0 - K e^{-rT}) + P$, with $P$ small but positive. As $S_0 \to \infty$, $P \to 0$ and $C \to S_0 - K e^{-rT}$, the forward value.

	Forward	Option
Payoff	\(S_T - K\) (linear)	\((S_T - K)^+\) (nonlinear)
Replication	Static	Dynamic (delta hedging)
Key parameters	\(r, T, S_0\)	\(r, T, S_0, \sigma\)
Gamma	\(0\)	\(> 0\) (near the strike)
Pricing method	Algebraic no-arbitrage	Black-Scholes PDE / risk-neutral expectation