The Black–Scholes PDE
gave us a price in which the stock's real-world drift \mu had
mysteriously vanished. There is a second, probabilistic road to the same price that makes the
disappearance of \mu not a surprise but the whole point. It is called
risk-neutral pricing, and it says: the value of an option is the discounted
expected payoff — computed under a cleverly tilted probability measure in which every
asset drifts at the risk-free rate r.
Under the real-world measure \mathbb{P} the stock obeys
dS = \mu S\,dt + \sigma S\,dW. By
Girsanov's theorem
we can switch to an equivalent risk-neutral measure
\mathbb{Q} under which the Brownian motion
\tilde W absorbs the drift change, and the stock becomes a geometric
Brownian motion with drift r:
dS_t = r S_t\,dt + \sigma S_t\,d\tilde W_t.
The volatility \sigma is untouched — Girsanov reweights
probabilities, it cannot change the size of the wiggle — and \mu is
gone for good.
The pricing formula, line by line
The engine is a single structural fact. Under \mathbb{Q} the
discounted stock price e^{-rt}S_t is a
martingale
— a fair game — because choosing the drift to be r exactly offsets
the discount factor's decay. The same holds for the discounted option value: it is built by
trading the stock and the (riskless) bank account, so it inherits the martingale property. We
ride that fact straight to the price.
Step 1 — the discounted value is a \mathbb{Q}
-martingale. Write V_t for the option's value at
time t. Then e^{-rt}V_t is a martingale
under \mathbb{Q}, so its best forecast equals its current value:
e^{-rt}V_t = \mathbb{E}_{\mathbb{Q}}\!\big[\,e^{-rT}V_T \,\big|\, \mathcal{F}_t\,\big].
Step 2 — solve for V_t.
Multiply both sides by e^{rt}, pulling the time-t
discount factor inside (it is \mathcal{F}_t-measurable, a constant
to the expectation):
V_t = e^{rt}\,\mathbb{E}_{\mathbb{Q}}\!\big[\,e^{-rT}V_T \,\big|\, \mathcal{F}_t\,\big] = e^{-r(T - t)}\,\mathbb{E}_{\mathbb{Q}}\!\big[\,V_T \,\big|\, \mathcal{F}_t\,\big].
Step 3 — substitute the terminal payoff. At expiry the option is its
payoff, V_T = \text{payoff}(S_T). For a European call this is
(S_T - K)^+:
V_t = e^{-r(T - t)}\,\mathbb{E}_{\mathbb{Q}}\!\big[\,(S_T - K)^+ \,\big|\, \mathcal{F}_t\,\big].
Step 4 — read it at t = 0.
With nothing yet observed, the conditional expectation is an ordinary one:
V_0 = e^{-rT}\,\mathbb{E}_{\mathbb{Q}}\!\big[\,(S_T - K)^+\,\big].
The recipe is now a slogan: price = discount × expected payoff, with the
expectation taken in the world where the stock drifts at r. No
\mu in sight — and no PDE either, though the two answers agree (the
Feynman–Kac formula
is the bridge).
Let \mathbb{Q} be the risk-neutral measure, under which
S is a geometric Brownian motion of drift
r,
dS_t = r S_t\,dt + \sigma S_t\,d\tilde W_t. Then for any
European derivative with payoff g(S_T):
-
its value today is the discounted risk-neutral expectation,
V_0 = e^{-rT}\,\mathbb{E}_{\mathbb{Q}}\!\big[\,g(S_T)\,\big];
-
the discounted price processes e^{-rt}S_t and
e^{-rt}V_t are both \mathbb{Q}-martingales;
-
the real-world drift \mu never appears — only
r and \sigma do.
The name is honest. A genuinely risk-neutral investor demands no extra reward for bearing
uncertainty, so in their world every asset — risky stock and safe bond alike — drifts at the
single rate r, and future cashflows are discounted at that same
r. We are not claiming investors are risk-neutral; we are
exploiting that the option's price, being perfectly hedgeable, is the same in that fictitious
world as in the real one.
That is why \mu may be replaced by r
for free. The replication argument behind the PDE and this change of measure are two faces of
one coin: hedging removes the directional bet, so the market's view of the stock's
direction (its \mu, its risk premium) is irrelevant to
the price. Only the spread \sigma matters. Reassuringly,
this expectation gives the same number the PDE does.
Pricing by simulation
The formula V_0 = e^{-rT}\mathbb{E}_{\mathbb{Q}}[(S_T - K)^+] is
also a recipe: simulate many risk-neutral paths of the stock, collect each path's
discounted payoff, and average. Below are several sample paths of
dS = rS\,dt + \sigma S\,d\tilde W from
S_0 = 100; the dashed line is the strike
K, and the figure prints the Monte-Carlo estimate
\hat V_0 from this batch. Refresh for a fresh draw — each run prints
the \sigma it used.
Paths finishing above K pay
S_T - K; those below pay nothing. The average of the discounted
payoffs is exactly the Black–Scholes price — the same number the PDE produces, which the
Feynman–Kac formula
guarantees.