The
equivalent martingale measure
was promised but never built. Girsanov's theorem is the construction. It answers a precise
question: if I reweight my probabilities by a clever density, what happens to a
Brownian motion? The astonishing answer is that
changing measure changes the drift, and nothing else. The randomness — the Brownian
wiggle — survives untouched; only the tilt of the path shifts.
Concretely, let W be a Brownian motion under
\mathbb{P}, fix an adapted process
\theta_t, and use as the
density
the
exponential martingale
Z_T = \frac{d\mathbb{Q}}{d\mathbb{P}} = \exp\!\left(-\int_0^T \theta_s\, dW_s - \tfrac{1}{2}\int_0^T \theta_s^2\, ds\right).
Then under the new measure \mathbb{Q} the shifted process
\tilde{W}_t = W_t + \int_0^t \theta_s\, ds
is itself a Brownian motion. The drift \int_0^t \theta_s\,ds that
\mathbb{Q} grafts on under \mathbb{P} is
exactly the drift it strips off going the other way: under
\mathbb{Q}, \tilde{W} is driftless and
clean.
The centrepiece: sending the stock risk-neutral, line by line
Watch Girsanov do the one job the whole theory was built for: turn the real-world stock
dynamics into the risk-neutral dynamics. Start with
geometric Brownian motion
under \mathbb{P}, where the stock drifts at its real rate
\mu:
dS_t = \mu S_t\, dt + \sigma S_t\, dW_t.
Step 1 — invert the Girsanov shift. Take
\theta constant. The relation
\tilde{W}_t = W_t + \theta t differentiates to
d\tilde{W}_t = dW_t + \theta\,dt, so the
\mathbb{P}-Brownian increment is
dW_t = d\tilde{W}_t - \theta\, dt.
Step 2 — substitute into the SDE. Replace every
dW_t with d\tilde{W}_t - \theta\,dt:
dS_t = \mu S_t\, dt + \sigma S_t\,\big(d\tilde{W}_t - \theta\, dt\big).
Step 3 — expand and collect the dt terms. Multiply
out the bracket and gather the two drift pieces:
dS_t = \mu S_t\, dt - \sigma\theta S_t\, dt + \sigma S_t\, d\tilde{W}_t = (\mu - \sigma\theta)\,S_t\, dt + \sigma S_t\, d\tilde{W}_t.
The diffusion coefficient \sigma S_t is unchanged — only
the drift moved, exactly as Girsanov advertised. The new drift is a knob we can turn by our
choice of \theta.
Step 4 — choose \theta to hit the riskless rate. We
want the stock to drift at r under
\mathbb{Q}, so demand
\mu - \sigma\theta = r and solve for
\theta:
\theta = \frac{\mu - r}{\sigma}.
This special \theta is the market price of risk —
the excess return \mu - r per unit of volatility
\sigma.
Step 5 — verify the drift collapses to r. Substitute
\theta = (\mu - r)/\sigma back into the drift coefficient and watch
the \sigma cancel:
\mu - \sigma\theta = \mu - \sigma\cdot\frac{\mu - r}{\sigma} = \mu - (\mu - r) = r.
Step 6 — read off the risk-neutral dynamics. The drift is now exactly the
riskless rate:
\boxed{\,dS_t = r S_t\, dt + \sigma S_t\, d\tilde{W}_t.\,}
Under \mathbb{Q} the stock earns r on
average — the real-world premium \mu has been reweighted clean away
— and \tilde{W} is the new Brownian motion. The discounted price
\tilde{S}_t = e^{-rt}S_t is then a
\mathbb{Q}-martingale, so \mathbb{Q} is
the equivalent martingale measure we were missing. Black–Scholes pricing is now a matter of
computing one expectation under \mathbb{Q}.
Let W be a \mathbb{P}-Brownian motion
and \theta_t an adapted process (with
\int_0^T \theta_s^2\,ds < \infty). Then:
-
the density
Z_T = \exp\!\big(-\int_0^T \theta_s\,dW_s - \tfrac12\int_0^T \theta_s^2\,ds\big)
is the exponential martingale; it is non-negative with
\mathbb{E}_{\mathbb{P}}[Z_T] = 1, so
d\mathbb{Q} = Z_T\, d\mathbb{P} defines a measure
\mathbb{Q} \sim \mathbb{P}.
-
under \mathbb{Q}, the shifted process
\tilde{W}_t = W_t + \int_0^t \theta_s\,ds is a Brownian motion.
-
the change of measure shifts the drift by
\sigma\theta and leaves the diffusion
\sigma untouched.
-
choosing the market price of risk
\theta = (\mu - r)/\sigma makes the discounted stock
e^{-rt}S_t a \mathbb{Q}-martingale.
The quantity \theta = (\mu - r)/\sigma is the market price
of risk (or Sharpe ratio): how much extra expected return the market pays for each
extra unit of volatility carried. Girsanov reweights probability until that price is exactly
zero — under \mathbb{Q} risk is not compensated, which is why
everything earns the same r.
One technical hazard: the density Z_T is only a
local martingale in general, and could secretly lose mass
(\mathbb{E}[Z_T] < 1), which would break the change of measure.
Novikov's condition rules this out — if
\mathbb{E}_{\mathbb{P}}\!\left[\exp\!\left(\tfrac12\int_0^T \theta_s^2\, ds\right)\right] < \infty,
then Z_T is a genuine martingale with
\mathbb{E}[Z_T] = 1 and Girsanov goes through. For constant
\theta the integral is just
\tfrac12\theta^2 T, finite always — so the textbook
Black–Scholes case is safe.
Finally, recognise the density itself. With
X_t = -\int_0^t \theta_s\,dW_s, its quadratic variation is
\langle X\rangle_t = \int_0^t \theta_s^2\,ds, and
Z_t = \exp\!\big(X_t - \tfrac12\langle X\rangle_t\big)
is precisely the stochastic exponential of
-\int\theta\,dW — the very
exponential martingale
\exp(\sigma W_t - \tfrac12\sigma^2 t) we built with
Itô's lemma, now with the integrand
-\theta in the role of \sigma. The
-\tfrac12\int\theta^2 ds in the exponent is the same drift
compensator, doing the same job: keeping the mean pinned at
1 so that Z_T is a legitimate density.