The hedging argument, line by line
Form a portfolio that is long one option and short
\Delta units of the stock:
\Pi = V - \Delta\, S.
We will write down how \Pi moves over an instant, choose
\Delta to kill the random part, then impose that a riskless
portfolio earns r. Throughout we hold \Delta
fixed across the instant (the self-financing assumption examined in
replication).
Step 1 — Itô's lemma on V(t, S).
Since V is a function of time and of the Itô process
S,
Itô's lemma
gives dV = V_t\,dt + V_S\,dS + \tfrac12 V_{SS}\,(dS)^2. Substituting
dS = \mu S\,dt + \sigma S\,dW and the box-algebra value
(dS)^2 = \sigma^2 S^2\,dt and collecting the
dt and dW pieces:
dV = \Big(V_t + \mu S\,V_S + \tfrac12\sigma^2 S^2\,V_{SS}\Big)\,dt + \sigma S\,V_S\,dW.
Step 2 — differentiate the portfolio. With
\Delta held fixed over the instant,
d\Pi = dV - \Delta\,dS. Substitute the
dV just found and
dS = \mu S\,dt + \sigma S\,dW:
d\Pi = \Big(V_t + \mu S\,V_S + \tfrac12\sigma^2 S^2\,V_{SS}\Big)\,dt + \sigma S\,V_S\,dW - \Delta\big(\mu S\,dt + \sigma S\,dW\big).
Step 3 — group the dt and
dW terms. Collect the deterministic drift and the
random kick separately:
d\Pi = \Big(V_t + \tfrac12\sigma^2 S^2\,V_{SS} + \mu S\,(V_S - \Delta)\Big)\,dt + \sigma S\,(V_S - \Delta)\,dW.
Step 4 — choose \Delta = V_S to cancel the
randomness. The dW coefficient is
\sigma S\,(V_S - \Delta); setting
\Delta = V_S makes it zero. Notice the same choice annihilates the
\mu S\,(V_S - \Delta) drift term — so the entire
\mu contribution disappears at once:
d\Pi = \Big(V_t + \tfrac12\sigma^2 S^2\,V_{SS}\Big)\,dt.
This is the magic stroke: holding exactly V_S shares against the
option leaves a portfolio with no dW — it is
instantaneously riskless.
Step 5 — a riskless portfolio must earn r.
If \Pi carried no risk yet grew faster or slower than the bank rate,
a no-arbitrage
trade would mint risk-free profit. So over the instant
d\Pi = r\,\Pi\,dt, and with
\Pi = V - \Delta S = V - S\,V_S:
d\Pi = r\Pi\,dt = r\big(V - S\,V_S\big)\,dt.
Step 6 — equate the two expressions for
d\Pi. Steps 4 and 5 describe the same drift, so their
dt coefficients match:
V_t + \tfrac12\sigma^2 S^2\,V_{SS} = r\big(V - S\,V_S\big).
Step 7 — rearrange into the standard form. Expand the right-hand side and
move every term to the left:
V_t + \tfrac12\sigma^2 S^2\,V_{SS} + rS\,V_S - rV = 0.
There it is — the Black–Scholes partial differential equation. Every option
on this stock, whatever its payoff, satisfies the very same equation; only the boundary and
terminal conditions distinguish a call from a put from anything else.
Under the Black–Scholes model, the value V(t, S) of any
derivative on a non-dividend stock obeys
\frac{\partial V}{\partial t} + \tfrac12\sigma^2 S^2\,\frac{\partial^2 V}{\partial S^2} + rS\,\frac{\partial V}{\partial S} - rV = 0,
a backward, second-order parabolic PDE, together with:
-
the terminal condition
V(T, S) = \text{payoff}(S) — for a call,
(S - K)^+;
-
the hedge ratio \Delta = V_S, the shares to hold per option
(these are the Greeks);
-
no dependence on the real-world drift \mu — it cancelled in
the hedging argument.
The most startling feature of the derivation is what is missing. The stock's
expected return \mu entered in Step 1, survived into Step 3, then
vanished completely the instant we set \Delta = V_S in Step 4.
The final PDE contains r and \sigma but
not \mu.
The interpretation is profound: an optimist who thinks the stock will soar and a pessimist
who thinks it will languish must agree on the option's price, because both can hedge away the
directional bet. Only the volatility \sigma — the size of
the wiggle, not its direction — survives. This is precisely the seed of
risk-neutral pricing:
we may as well pretend the stock drifts at the risk-free rate r,
and the answer is unchanged.
The PDE alone is solved by infinitely many functions; the conditions pick out the
one that is the call. For a European call C(t, S) with strike
K and expiry T:
-
Terminal: at expiry the option is worth its payoff,
C(T, S) = (S - K)^+ — the hockey stick.
-
At S = 0: a worthless stock keeps a worthless
call, C(t, 0) = 0.
-
As S \to \infty: exercise is near-certain, so
the call behaves like the stock minus the discounted strike,
C(t, S) \sim S - K e^{-r(T - t)}.
Because the terminal condition is fixed at T and the PDE is solved
backward in time, the value today is the smooth hockey stick relaxing into a curved
surface as time to expiry grows — exactly the picture below.