Black-Scholes Assumptions and the American Call Delta

Worked Solution

How to Think About It: This is a foundational derivatives question that tests three things: do you know the modeling framework, can you reason about early exercise economically, and can you differentiate the Black-Scholes formula cleanly. Before diving into math, think about each part qualitatively. The assumptions are a checklist. The American call result follows from a simple "why would you ever exercise early?" argument. And the delta derivation is an exercise in careful differentiation where most of the messy terms cancel -- if you know that trick, the answer is clean.

Quick Sanity Checks: - Delta of a call should be between 0 and 1 (you benefit from spot going up, but the payoff is capped at linear growth). - Deep ITM: delta approaches 1 (the call behaves like the stock). Deep OTM: delta approaches 0 (the call is nearly worthless). - An American call on a non-dividend-paying stock should never be exercised early, so its value equals the European call. If there were dividends, this would break.

Derivation:

Part 1: Black-Scholes Assumptions

The standard assumptions are:

1. The stock price follows a geometric Brownian motion: $dS = \mu S \, dt + \sigma S \, dW$, where $\sigma$ is constant.
2. The risk-free interest rate $r$ is constant and the same for all maturities.
3. The stock pays no dividends.
4. Markets are frictionless: no transaction costs, no taxes, continuous trading is possible.
5. No arbitrage opportunities exist.
6. Short selling is permitted with full use of proceeds.
7. Securities are infinitely divisible.

These together imply that the stock price at time $T$ is log-normally distributed and that we can form a self-financing replicating portfolio.

Part 2: American Call Equals European Call (No Dividends)

We want to show that early exercise of an American call on a non-dividend-paying stock is never optimal, so $C_{\text{American}} = C_{\text{European}}$.

*Lower bound argument:* At any time $t < T$, the European call satisfies

$$C(S, t) \geq S - K e^{-r(T-t)}$$

This follows from put-call parity. Since the European call is worth at least $C \geq S - Ke^{-r(T-t)}$, and the stock pays no dividends, we have

$$C(S, t) \geq S - K e^{-r(T-t)} > S - K$$

where the last inequality holds because $e^{-r(T-t)} < 1$ for $r > 0$ and $t < T$. The quantity $S - K$ is exactly the payoff from early exercise.

So the value of holding the option always exceeds the value of exercising it. Therefore early exercise is never optimal, and the American call has the same value as the European call.

*Economic intuition:* By exercising early you give up two things -- the time value of money on the strike (you pay $K$ now instead of $Ke^{-r(T-t)}$ later) and the insurance value of the option (if the stock drops below $K$, you lose nothing by holding). With no dividends to capture, there is no offsetting benefit to early exercise.

Part 3: Deriving Delta

The Black-Scholes formula for a European call is

$$C = S N(d_1) - K e^{-r(T-t)} N(d_2)$$

where

$$d_1 = \frac{\ln(S/K) + (r + \sigma^2/2)(T-t)}{\sigma \sqrt{T-t}}, \quad d_2 = d_1 - \sigma \sqrt{T-t}$$

and $N(\cdot)$ is the standard normal CDF.

Differentiating with respect to $S$:

$$\Delta = \frac{\partial C}{\partial S} = N(d_1) + S N'(d_1) \frac{\partial d_1}{\partial S} - K e^{-r(T-t)} N'(d_2) \frac{\partial d_2}{\partial S}$$

Since $d_2 = d_1 - \sigma\sqrt{T-t}$, we have $\partial d_2 / \partial S = \partial d_1 / \partial S$. Also,

$$\frac{\partial d_1}{\partial S} = \frac{1}{S \sigma \sqrt{T-t}}$$

So the remaining terms become

$$S N'(d_1) \frac{1}{S \sigma \sqrt{T-t}} - K e^{-r(T-t)} N'(d_2) \frac{1}{S \sigma \sqrt{T-t}}$$

$$= \frac{1}{S \sigma \sqrt{T-t}} \left[ S N'(d_1) - K e^{-r(T-t)} N'(d_2) \right]$$

The key identity is that $S N'(d_1) = K e^{-r(T-t)} N'(d_2)$. To see this, note $N'(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$ and

$$\frac{d_1^2 - d_2^2}{2} = \frac{(d_1 - d_2)(d_1 + d_2)}{2} = \frac{\sigma\sqrt{T-t} \cdot (2d_1 - \sigma\sqrt{T-t})}{2}$$

Expanding and simplifying with the definition of $d_1$, one finds $N'(d_2)/N'(d_1) = S / (Ke^{-r(T-t)})$, confirming the identity. The bracketed term vanishes, and we are left with

$$\boxed{\Delta = N(d_1)}$$

where $d_1 = \dfrac{\ln(S/K) + (r + \sigma^2/2)(T-t)}{\sigma\sqrt{T-t}}$.

Practical Interpretation: The delta $N(d_1)$ tells a trader the hedge ratio -- to delta-hedge a short call, buy $N(d_1)$ shares of stock per option. Since $N(d_1) \in [0, 1]$, you never need more than one share per call. As the option moves deeper ITM, $d_1 \to +\infty$ and $\Delta \to 1$ (the option behaves like stock). As it moves OTM, $d_1 \to -\infty$ and $\Delta \to 0$ (the option is nearly worthless). Near the money and close to expiry, delta swings rapidly -- this is where gamma risk is highest and hedging becomes expensive.

Answer: The Black-Scholes assumptions define a frictionless, continuous-time market with constant volatility and rates, no dividends, and log-normal stock dynamics. Under these assumptions, an American call is never optimally exercised early (its alive value $S - Ke^{-r(T-t)}$ always exceeds the exercise value $S - K$), so it equals the European call. The European call delta is $\Delta = N(d_1)$ where $d_1 = \frac{\ln(S/K) + (r + \sigma^2/2)(T-t)}{\sigma\sqrt{T-t}}$.