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.

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.