Deriving The Black-Scholes PDE

You can write down an option's payoff in one line. Working out what it is worth today is another matter entirely, and the obvious approach fails immediately: average the payoff over where you think the stock might end up, and your answer depends on how bullish you are. Two traders with different views would quote different prices for the same contract, and there would be no such thing as a market. Black and Scholes found the way out, and the surprise is that the answer does not depend on your view at all.

Noise-cancelling headphones

Noise-cancelling headphones listen to the sound around you and play the exact opposite into your ear. Two loud signals arrive, one the mirror of the other, and you hear silence. An option and its stock are two loud signals driven by the same noise: when the stock jumps the option jumps too, just by a different amount. So hold the option and short the right number of shares, and the jumps cancel. What is left in your hands has no randomness in it at all.

A thing with no randomness has an obvious price. It must grow at the bank rate, because if it grew faster you would borrow to buy it and collect the difference for free, and if it grew slower you would sell it and do the same. "Riskless things earn the risk-free rate" is the only economics in the whole derivation. Everything else is bookkeeping.

Step 1: how the stock moves

$$dS_t \;=\; \mu\, S_t\, dt \;+\; \sigma\, S_t\, dW_t .$$

In plain English: over a tiny slice of time, the stock drifts up a little and gets shoved randomly, both in proportion to its current price. The pieces: $S_t$ is the price now, $\mu$ ("mu") is the expected growth rate per year, $\sigma$ ("sigma") is the volatility, and $dW_t$ is the random shove, a coin flip whose typical size is the square root of the time slice. Note where $\mu$ sits, because in three steps it will be gone.

Step 2: how the option moves

Call the option's value $V(S, t)$, a function of the stock price and the date. Because $S$ is jittery rather than smooth, ordinary calculus is not enough; you need Itô's lemma, which gives

$$dV = \left( \frac{\partial V}{\partial t} + \mu S \frac{\partial V}{\partial S} + \tfrac{1}{2}\sigma^{2} S^{2} \frac{\partial^{2} V}{\partial S^{2}} \right) dt \;+\; \sigma S \frac{\partial V}{\partial S}\, dW .$$

In plain English: the option's value changes for four reasons. It loses value simply because time passed ($\partial V/\partial t$, called theta). It moves because the stock drifted. It gains from the bend in its own value curve ($\partial^{2}V/\partial S^{2}$, called gamma) being shaken back and forth. And it gets the same random shove the stock did, scaled by the slope $\partial V/\partial S$, called delta.

That third term is the one that is not in ordinary calculus. It exists because the option's value curve is bent, so a shake up and an equal shake down do not cancel: the up gains more than the down loses. It is the entire reason volatility is worth money.

Step 3: cancel the noise

Build the portfolio: own one option, short delta shares.

$$\Pi \;=\; V \;-\; \frac{\partial V}{\partial S}\, S .$$

Over the next instant, the shares lose exactly what the option's random term gains. Both $dW$ pieces vanish, and so, silently, does every $\mu$:

$$d\Pi \;=\; \left( \frac{\partial V}{\partial t} + \tfrac{1}{2}\sigma^{2} S^{2} \frac{\partial^{2} V}{\partial S^{2}} \right) dt .$$

In plain English: the hedged portfolio's change over the next instant is a known number. No coin flip, and no trace of how fast anyone expected the stock to grow.

Step 4: charge it the bank rate

A riskless portfolio must earn $r$, otherwise there is free money: $d\Pi = r\,\Pi\, dt$. Substitute the two expressions for $d\Pi$, cancel $dt$, and rearrange:

$$\frac{\partial V}{\partial t} \;+\; \tfrac{1}{2}\sigma^{2} S^{2} \frac{\partial^{2} V}{\partial S^{2}} \;+\; r S \frac{\partial V}{\partial S} \;-\; r V \;=\; 0 .$$

In plain English, reading left to right: time decay, plus the value earned from shaking a bent position, plus the financing carry on the shares you hold, must equal the interest on the money tied up in the option. That is the Black-Scholes partial differential equation.

The version worth carrying in your head drops the interest terms, which are small for short-dated options:

$$\Theta \;\approx\; -\tfrac{1}{2}\sigma^{2} S^{2} \Gamma .$$

The equation does not know whether you hold a call, a put or something exotic. What makes it a call is the boundary condition $V(S, T) = \max(S - K, 0)$, imposed at expiry and solved backwards.

Worked example 1: checking the theta-gamma balance

Take a 30-day at-the-money call: stock at $100, strike $100, volatility 20 percent, and set $r = 0$ so the arithmetic stays clean. Here $T = 30/365 = 0.0822$ years, so $\sqrt{T} = 0.2867$ and $\sigma\sqrt{T} = 0.0573$.

1. Gamma. $\Gamma = \phi(d_1)/(S\sigma\sqrt{T})$, where $\phi$ is the bell-curve height, here $0.3988$. So $\Gamma = 0.3988 / (100 \times 0.0573) = 0.0695$.
2. The gamma term. $\tfrac{1}{2}\sigma^{2}S^{2}\Gamma = 0.5 \times 0.04 \times 10{,}000 \times 0.0695 = 13.91$ per year.
3. So theta must be $-13.91$ per year, which is $13.91 / 365 = 0.0381$ per calendar day.
4. Check it independently. The textbook theta formula gives $-S\phi(d_1)\sigma / (2\sqrt{T}) = -(100)(0.3988)(0.2) / (2 \times 0.2867) = -13.91$ per year. Identical, as the PDE promised.

For context, that option is worth about $2.29, and it loses 3.8 cents a day to time. Nearly 2 percent of its value, every day, for standing still.

Worked example 2: what the rent buys you

You own that call and hedge it. The stock moves $1.50 tomorrow, up or down, it does not matter which.

• Gamma pays you $\tfrac{1}{2}\Gamma(\Delta S)^{2} = 0.5 \times 0.0695 \times 1.50^{2} = 0.0782$.
• Theta charges you $0.0381$.
• Net: $0.040 on the day, from a position with no directional view.

So what size move makes you break even? Solve $\tfrac{1}{2}\Gamma x^{2} = 0.0381$: $x^{2} = 0.0762 / 0.0695 = 1.096$, so $x = 1.047$, a move of about 1.05 percent. Annualise it by multiplying by the square root of 365: $1.047 \times 19.10 = 20.0$ percent.

That is the volatility you paid for, recovered exactly. The PDE is telling you something completely concrete: a hedged option position breaks even if and only if the stock realises the volatility priced into it, makes money if the stock is wilder, and loses if it is calmer. This is the trade behind Gamma Scalping.

Where the drift went

Drag the sliders below. Move the drift and the whole bundle of paths tilts, yet nothing about the option's fair value changes. Move the volatility and the paths thrash, and the value changes a great deal. The PDE contains $\sigma$ and not $\mu$; this is the picture of why.

What this means in practice

Every finite-difference pricer on a derivatives desk is this equation on a grid, marched backwards from the payoff, and most model extensions are the same equation with one coefficient changed: dividends turn $rS$ into $(r - q)S$, and Local Volatility and Dupire's Formula replaces the constant $\sigma$ with $\sigma(S, t)$.

More usefully, each assumption in the derivation names a way real hedging loses money. Prices gap rather than glide, so the cancellation in step three is imperfect. Rebalancing is neither continuous nor free. And volatility is not a constant, which is why one $\sigma$ cannot fit every strike and the smile exists at all (Black-Scholes Assumptions And Failure Modes).

Key terms

• Delta ($\partial V/\partial S$) — the slope; how many shares to short to cancel the noise.
• Gamma ($\partial^{2}V/\partial S^{2}$) — the bend; what pays you when the stock shakes.
• Theta ($\partial V/\partial t$) — time decay; the rent charged for holding gamma.
• Hedged portfolio — one option less delta shares, momentarily riskless.
• Boundary condition — the payoff at expiry, the only thing that says which option this is.