The Black-Scholes Model

The Black-Scholes-Merton model is the founding result of quantitative finance: a self-financing hedge in the stock and cash can replicate an option, so the option's price is pinned by no-arbitrage, independent of anyone's view on where the stock is going. The output is a single, closed-form formula. Every later model (local vol, Heston, SABR) is best understood as a repair of a specific Black-Scholes assumption, so this derivation is the one to know cold.

Assumptions

1. The stock follows geometric Brownian motion with constant volatility: $dS = \mu S\,dt + \sigma S\,dW$.
2. Constant risk-free rate $r$; no dividends (extendable).
3. Continuous, frictionless trading, no transaction costs, infinitely divisible shares, unlimited short-selling.
4. No arbitrage.

Derivation 1: delta-hedging (the PDE)

Let $V(S,t)$ be the option value. Form a portfolio long one option and short $\Delta$ shares: $\Pi = V - \Delta S$. Over $dt$, using Itô's lemma on $V$,

$dV = \Big(V_t + \tfrac12\sigma^2 S^2 V_{SS}\Big)dt + V_S\,dS,$

so the portfolio changes by

$d\Pi = dV - \Delta\,dS = \Big(V_t + \tfrac12\sigma^2 S^2 V_{SS}\Big)dt + (V_S - \Delta)\,dS.$

Choose $\Delta = V_S$, the delta hedge. The random $dS$ term vanishes and the portfolio is instantaneously riskless:

$d\Pi = \Big(V_t + \tfrac12\sigma^2 S^2 V_{SS}\Big)dt.$

A riskless portfolio must, by no-arbitrage, earn the risk-free rate: $d\Pi = r\Pi\,dt = r(V - V_S S)\,dt$. Equating the two expressions for $d\Pi$ and cancelling $dt$ gives the Black-Scholes PDE:

$\boxed{\,V_t + \tfrac12\sigma^2 S^2 V_{SS} + rS\,V_S - rV = 0.\,}$

Notice $\mu$ has disappeared, the drift never entered because the hedge removed all exposure to $dS$. The PDE holds for any European claim; the payoff enters only through the terminal condition, e.g. $V(S,T) = (S-K)^+$ for a call.

Derivation 2: risk-neutral expectation

By Feynman-Kac, the solution of that PDE is a discounted expectation under the risk-neutral measure $\mathbb{Q}$, where $dS = rS\,dt + \sigma S\,dW^{\mathbb{Q}}$:

$V_0 = e^{-rT}\,\mathbb{E}^{\mathbb{Q}}\big[(S_T - K)^+\big], \qquad S_T = S_0\exp\!\Big(\big(r - \tfrac12\sigma^2\big)T + \sigma\sqrt{T}\,Z\Big),\ Z\sim\mathcal{N}(0,1).$

The call pays off when $S_T > K$, i.e. when $Z > -d_2$ where

$d_2 = \frac{\ln(S_0/K) + (r - \tfrac12\sigma^2)T}{\sigma\sqrt T}.$

Split the expectation:

$V_0 = e^{-rT}\Big(\underbrace{\mathbb{E}^{\mathbb{Q}}[S_T\mathbf{1}_{S_T>K}]}_{\text{(I)}} - K\,\underbrace{\mathbb{Q}(S_T>K)}_{\text{(II)}}\Big).$

Term (II) is $\mathbb{Q}(Z > -d_2) = N(d_2)$. Term (I) requires completing the square in the lognormal integral: $\mathbb{E}^{\mathbb{Q}}[S_T \mathbf 1_{S_T>K}] = S_0 e^{rT} N(d_1)$ with $d_1 = d_2 + \sigma\sqrt T$. Substituting,

$\boxed{\,C = S_0\,N(d_1) - K e^{-rT}\,N(d_2),\qquad d_{1,2} = \frac{\ln(S_0/K) + (r \pm \tfrac12\sigma^2)T}{\sigma\sqrt T}.\,}$

Reading the formula

The two terms are the split of the payoff $(S_T - K)^+ = S_T\mathbf 1 - K\mathbf 1$:

• $N(d_2) = \mathbb{Q}(S_T > K)$ is the risk-neutral probability the option finishes in the money. So $Ke^{-rT}N(d_2)$ is the present value of paying the strike, conditional on exercise.
• $N(d_1)$ is the option's delta, $\partial C/\partial S$; the term $S_0 N(d_1)$ is the present value of receiving the stock upon exercise (formally, the exercise probability under the stock numeraire).

The gap between them, $d_1 - d_2 = \sigma\sqrt T$, is the total volatility to expiry, the same $\tfrac12\sigma^2$ Itô correction that appears in geometric Brownian motion.

The formula's economic content is easiest to read off a picture: the call price is a smooth, convex curve that lies strictly above its own expiry payoff, and the vertical gap between the two is the option's time value, which decays to zero as expiry approaches.

Worked example

$S_0 = 100$, $K = 100$ (at the money), $r = 0$, $\sigma = 20\%$, $T = 1$. Then $d_1 = (0 + \tfrac12(0.2)^2)/0.2 = 0.1$, $d_2 = -0.1$. So $C = 100\,N(0.1) - 100\,N(-0.1) = 100(0.5398 - 0.4602) = 7.97$. A one-year ATM call on a 20-vol stock costs about 8% of spot. The rule-of-thumb $C_{\text{ATM}} \approx 0.4\,S\sigma\sqrt T = 0.4\times100\times0.2 = 8.0$ nails it, a shortcut worth memorizing.

What breaks in practice

• Volatility is not constant. The single biggest failure. Real option prices imply different $\sigma$ for different strikes and maturities, the smile, which is a direct contradiction of the constant-$\sigma$ assumption. This spawned local vol and stochastic vol.
• Continuous, costless hedging is impossible. You rebalance discretely and pay spreads; the replication is approximate, and the hedging error is governed by gamma times realized-minus-implied variance.
• Lognormal tails are too thin. Crashes are far more likely than GBM allows; the model underprices deep OTM puts, which is why the equity skew exists. Jumps and fat tails need extensions (Merton jump-diffusion, Lévy models).
• Constant rates, no dividends. Both are easily patched (Black-76 for forwards, $q$ for dividend yield: replace $S_0$ with $S_0 e^{-qT}$), but the naive formula ignores them.

Despite all this, Black-Scholes survives as the lingua franca: traders quote and risk-manage in its language via implied volatility, using it as a nonlinear translator between price and vol rather than a literal model of returns.

In interviews

You should be able to derive the PDE via delta-hedging and explain the risk-neutral-expectation route, and to write the call formula with the correct $d_1, d_2$. Be ready for: "what is $N(d_2)$?" (risk-neutral prob of finishing ITM), "what is $N(d_1)$?" (the delta), "where did $\mu$ go?" (removed by the hedge, the option price is preference-free), and "what's the ATM call worth?" ($\approx 0.4 S\sigma\sqrt T$). The deepest follow-up is "which assumption does the market violate most, and how do you know?", answer: constant volatility, and the evidence is that inverting the formula strike-by-strike gives a non-flat implied-vol curve.