Phase 14 - Lesson 14.4

The Black–Scholes PDE and Formula

Delta-hedging a continuously-traded option leads to a PDE whose solution is the closed-form price.

⏱ 60 min● Advanced🔗 Prereqs: 14.2, Ito calculus (Phase 9)
↖ Phase 14 hub
Builds on: 14.2’s dynamic replication becomes continuous here; the risk-neutral measure comes from Phase 9 (Girsanov).
Leads to: Phase 15 relaxes constant volatility (local/stochastic vol, Gatheral).

Learning Objectives

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Key Vocabulary

Black–Scholes PDE
The PDE \(V_t+\tfrac12\sigma^2S^2V_{SS}+rSV_S-rV=0\) governing any European derivative under GBM.
Delta-hedged portfolio
Long the option, short \(\Delta=V_S\) shares; instantaneously risk-free, so it must earn \(r\).
d1, d2
The standardized moneyness terms \(d_{1,2}=\tfrac{\ln(S/K)+(r\pm\tfrac12\sigma^2)T}{\sigma\sqrt T}\).
Vega
Sensitivity of price to volatility, \(\partial V/\partial\sigma\); positive for vanilla calls and puts.
Implied volatility
The \(\sigma\) that makes the Black–Scholes price match a quoted market price; the market’s language for option prices.
Model assumptions
Constant \(\sigma\) and \(r\), GBM dynamics, no jumps, continuous frictionless trading, no dividends - idealizations, each violable.

Intuition & Motivation

Intuition
Black and Scholes made continuous the dynamic hedge of 14.2. Hold one option and continuously short \(\Delta=V_S\) shares; by Ito’s lemma the random \(dW\) terms cancel, so over an instant the portfolio is risk-free. No-arbitrage then forces it to earn the risk-free rate \(r\) - and that single equation, written out, is the Black–Scholes PDE. Because the hedge annihilates the \(dW\), the stock’s real drift \(\mu\) disappears; only \(r\) and \(\sigma\) survive. Solving the PDE with the call’s payoff boundary condition gives the famous closed form. It is elegant and enormously useful - and it rests on assumptions (constant volatility, no jumps, costless trading) that markets routinely violate.

Deriving the PDE by delta-hedging

Assume \(dS=\mu S\,dt+\sigma S\,dW\) and a derivative price \(V(t,S)\). By Ito’s lemma,

\[dV=\Big(V_t+\mu S V_S+\tfrac12\sigma^2S^2V_{SS}\Big)dt+\sigma S V_S\,dW.\] (14.7)

Form the portfolio \(\Pi=V-\Delta S\) with \(\Delta=V_S\). The \(dW\) terms cancel:

\[d\Pi=dV-V_S\,dS=\Big(V_t+\tfrac12\sigma^2S^2V_{SS}\Big)dt.\] (14.8)

Now \(d\Pi\) is deterministic over \(dt\), so by no-arbitrage it must equal the risk-free growth of the money invested, \(d\Pi=r\Pi\,dt=r(V-SV_S)\,dt\). Equating the two expressions and cancelling \(dt\) gives the Black–Scholes PDE:

\[\boxed{\,V_t+\tfrac12\sigma^2S^2V_{SS}+rSV_S-rV=0\,},\qquad V(T,S)=(S-K)^+\ \text{for a call}.\] (14.9)
Key Idea
The physical drift \(\mu\) has vanished: hedging removes it, and only the risk-free rate \(r\) and the volatility \(\sigma\) enter. This is the PDE face of risk-neutral valuation - by Feynman–Kac (Phase 9), the solution is \(V_0=e^{-rT}\E_{\mathbb{Q}}[(S_T-K)^+]\).

The Black–Scholes formula

Solving (14.9) with the call payoff (or evaluating the risk-neutral integral over the lognormal \(S_T\)) gives

\[C=S_0N(d_1)-Ke^{-rT}N(d_2),\qquad P=Ke^{-rT}N(-d_2)-S_0N(-d_1),\] (14.10)
\[d_1=\frac{\ln(S_0/K)+(r+\tfrac12\sigma^2)T}{\sigma\sqrt T},\qquad d_2=d_1-\sigma\sqrt T,\] (14.11)

with \(N\) the standard normal CDF. Interpret \(N(d_2)=\mathbb{Q}(S_T\gt K)\) (risk-neutral exercise probability) and \(S_0N(d_1)\) as the discounted expected stock received conditional on exercise. Puts follow from put–call parity \(C-P=S_0-Ke^{-rT}\).

The Greeks

Differentiating (14.10) gives the sensitivities used for hedging. With \(\varphi\) the normal density:

GreekSymbolCall formulaSign (call)
Delta\(\partial_S V\)\(N(d_1)\)+
Gamma\(\partial_{SS}V\)\(\varphi(d_1)/(S_0\sigma\sqrt T)\)+
Vega\(\partial_\sigma V\)\(S_0\varphi(d_1)\sqrt T\)+
Theta\(\partial_t V\)\(-\tfrac{S_0\varphi(d_1)\sigma}{2\sqrt T}-rKe^{-rT}N(d_2)\)− (usually)
Rho\(\partial_r V\)\(KTe^{-rT}N(d_2)\)+
Worked Example - Price and Greeks at the benchmark
1
Data: \(S_0=100,\ K=100,\ r=0.03,\ \sigma=0.2,\ T=1\).
2
Moneyness: \(d_1=\tfrac{0+(0.03+0.02)\cdot1}{0.2}=0.25\), \(d_2=0.25-0.2=0.05\).
3
Normals: \(N(0.25)=0.5987,\ N(0.05)=0.5199\).
4
Price: \(C=100(0.5987)-100e^{-0.03}(0.5199)=59.87-50.46=9.4134\).
5
Greeks: \(\Delta=0.5987,\ \Gamma=0.0193,\ \text{vega}=38.67,\ \Theta=-5.38,\ \rho=50.46\).

Interactive: the Black–Scholes surface and Monte Carlo check

Common Mistakes to Avoid
  • Putting the real-world drift \(\mu\) in the PDE or formula; hedging removes it, leaving only \(r\) and \(\sigma\).
  • Reading \(N(d_2)\) as the real probability of finishing in the money; it is the risk-neutral probability \(\mathbb{Q}(S_T\gt K)\).
  • Forgetting the discount factor \(e^{-rT}\) on the strike term, or mis-signing \(d_2=d_1-\sigma\sqrt T\).
  • Trusting a single Black–Scholes \(\sigma\) across strikes - real option prices imply a volatility smile, evidence the constant-vol assumption fails.
Quant Practitioner Tips
  • Memorize \(d_1,d_2\) and the parity \(C-P=S_0-Ke^{-rT}\); together they reconstruct most vanilla results fast.
  • Vega and gamma are largest near the money and (for gamma) near expiry - where hedging is hardest and most valuable.
  • Quote and think in implied volatility, not price; it is the market’s common currency and exposes the smile the model ignores.
  • Treat Black–Scholes as a coordinate system, not truth: its assumptions (constant \(\sigma\), no jumps, frictionless continuous trading) are approximations to be stress-tested, and it makes no promise of trading profit.

Knowledge Check

Q1 Medium
The Black–Scholes PDE is derived by:
Assuming investors are risk-neutral
Constructing a delta-hedged portfolio that is instantaneously risk-free and must earn \(r\)
Maximizing expected utility
Averaging historical returns
Q2 Medium
In the call formula \(C=S_0N(d_1)-Ke^{-rT}N(d_2)\), the term \(N(d_2)\) is:
The real-world probability of exercise
The risk-neutral probability \(\mathbb{Q}(S_T\gt K)\)
The option’s delta
Always 0.5
Q3 Medium
Which statement about the model’s assumptions is correct?
Volatility is observed to be constant across strikes
A volatility smile shows the constant-\(\sigma\) assumption fails
The formula accounts for jumps
Trading is assumed costly

Practical Exercise

For \(S_0=100,K=100,r=0.03,\sigma=0.2,T=1\) you are given the call value \(9.4134\). (a) Use put–call parity to find the put. (b) State the call’s delta and explain what hedge it prescribes. (c) Vega is \(38.67\); estimate the price change if implied volatility rises from 20% to 21%.

▶ Show full solution

(a) Parity: \(P=C-S_0+Ke^{-rT}=9.4134-100+100e^{-0.03}=9.4134-100+97.0446=6.458\).

(b) Delta \(=N(d_1)=0.5987\): to hedge one long call, short \(0.5987\) shares; the position is then locally insensitive to small moves in \(S\).

(c) \(\Delta C\approx\text{vega}\times\Delta\sigma=38.67\times0.01=0.387\), so the call rises by about \(\$0.39\) to roughly \(9.80\) (a first-order estimate; gamma-of-vega and higher terms adjust it slightly).

After the reveal, answer for yourself: Why does the delta hedge need continuous rebalancing, and what quantity measures how fast delta itself moves?

Lesson Summary

Delta-hedging a European option under GBM makes a portfolio instantaneously risk-free; no-arbitrage then yields the Black–Scholes PDE \(V_t+\tfrac12\sigma^2S^2V_{SS}+rSV_S-rV=0\), whose call solution is \(C=S_0N(d_1)-Ke^{-rT}N(d_2)\) (9.4134 at the benchmark). The physical drift \(\mu\) drops out, leaving \(r\) and \(\sigma\); the Greeks give the hedge sensitivities. The model’s constant-volatility, no-jump, frictionless assumptions are idealizations - the volatility smile is the market’s reminder, and no version of it guarantees a trading profit.

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers
Q: Write the Black–Scholes PDE and the call formula.
A: \(V_t+\tfrac12\sigma^2S^2V_{SS}+rSV_S-rV=0\); \(C=S_0N(d_1)-Ke^{-rT}N(d_2)\) with \(d_1=\tfrac{\ln(S_0/K)+(r+\tfrac12\sigma^2)T}{\sigma\sqrt T}\), \(d_2=d_1-\sigma\sqrt T\).
Q: Where did the stock’s real drift \(\mu\) go, and what does \(N(d_2)\) mean?
A: The delta hedge cancels the \(dW\) term and eliminates \(\mu\); only \(r\) and \(\sigma\) remain. \(N(d_2)=\mathbb{Q}(S_T\gt K)\), the risk-neutral exercise probability.

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