Proof Sketch and Key Ideas¶
This section justifies the theorem statement through a sequence of steps.
Toy mechanism: one cancellation does all the work
The whole proof has a single load-bearing identity. The shifted process \(\widetilde W_t = W_t + \int_0^t\theta\,ds\) carries an unwanted drift \(+\theta_t\,dt\); the density \(Z_t\) is defined with the term \(-\theta_t\,dW_t\) precisely so that the Itô product \(d(\widetilde W_t Z_t)\) generates a cross-variation \(-Z_t\theta_t\,dt\) that cancels the drift exactly. After that one cancellation, \(\widetilde W_t Z_t\) is a \(\mathbb{P}\)-martingale, \(\widetilde W_t\) is a continuous \(\mathbb{Q}\)-local martingale with \(\langle\widetilde W\rangle_t = t\), and Lévy's theorem closes the argument. The four steps below verify this in turn; the cancellation table at the bottom makes the bookkeeping explicit.
Step 1: Verify the Exponential Martingale Property¶
Goal: Show that \(Z_t = \exp\left(-\int_0^t \theta_s\,dW_s - \frac{1}{2}\int_0^t \theta_s^2\,ds\right)\) is a \(\mathbb{P}\)-martingale.
Recall (see § The Stochastic Exponential): for a continuous local martingale \(M_t = -\int_0^t \theta_s\,dW_s\), the Doléans-Dade exponential \(\mathcal{E}(M)_t = \exp(M_t - \tfrac{1}{2}\langle M\rangle_t)\) satisfies the driftless SDE
so \(Z_t\) is automatically a \(\mathbb{P}\)-local martingale. Recall (see § Novikov and Kazamaki Conditions): the Novikov condition \(\mathbb{E}^{\mathbb{P}}[\exp(\tfrac{1}{2}\int_0^T \theta_s^2\,ds)] < \infty\) promotes this local martingale to a true martingale with \(\mathbb{E}^{\mathbb{P}}[Z_t] = 1\) for all \(t\).
Step 2: The Shifted Process is a Q-Martingale¶
Goal: Show that \(\widetilde{W}_t = W_t + \int_0^t \theta_s\,ds\) is a continuous local martingale under \(\mathbb{Q}\).
Method: The abstract Bayes rule relates \(\mathbb{Q}\)-conditional expectations to \(\mathbb{P}\)-conditional expectations weighted by the density:
where the denominator simplifies to \(Z_s\) by the martingale property of \(Z_t\). The key step is to show that \(\widetilde{W}_t Z_t\) is a \(\mathbb{P}\)-martingale (which implies \(\widetilde{W}_t\) is a \(\mathbb{Q}\)-martingale). This is verified by the Itô product rule in the cancellation mechanism below: the \(dt\) contributions from the drift in \(\widetilde{W}_t\) and from the cross-variation \(d\langle \widetilde{W}, Z\rangle_t\) cancel exactly.
Result: \(\widetilde{W}_t\) is a continuous \(\mathbb{Q}\)-local martingale.
Step 3: Quadratic Variation is Measure-Invariant¶
Since:
and the integral term has finite variation (in particular, absolutely continuous):
Therefore \(\langle\widetilde{W}\rangle_t = t\).
Step 4: Lévy Characterization Theorem¶
Theorem (Lévy): A continuous local martingale \(M_t\) with quadratic variation \(\langle M\rangle_t = t\) is a standard Brownian motion.
Application: We have shown:
- \(\widetilde{W}_t\) is continuous (sum of continuous processes)
- \(\widetilde{W}_t\) is a continuous \(\mathbb{Q}\)-local martingale (Step 2)
- \(\langle\widetilde{W}\rangle_t = t\) (Step 3)
By Lévy's theorem:
Step 5: Alternative Derivation Using Change of Variables (Optional)¶
Another approach uses direct computation. This is most transparent when \(\theta\) is constant; for general adapted \(\theta_t\), the characteristic-function argument does not generalize (use Steps 2–4 instead).
Key formula: For a continuous \(\mathbb{Q}\)-local martingale \(M_t\) with \(\langle M\rangle_t = t\), the process must be Brownian (Lévy's characterization).
We can verify this by computing the characteristic function (shown here for constant \(\theta\)):
This matches the characteristic function of \(\mathcal{N}(0, t)\), confirming that \(\widetilde{W}_t\) is a standard Brownian motion. For general adapted \(\theta_t\), Levy's characterization (Steps 3-5) is the standard route.
Constant θ only
The characteristic function argument above applies when \(\theta\) is constant. For general adapted \(\theta_t\), the proof requires the full machinery of stochastic exponentials and Lévy's characterization theorem — the direct computation does not generalize.
Summary of the Proof Strategy¶
| Step | What we prove | Why it's needed |
|---|---|---|
| 1 | \(Z_t\) is a \(\mathbb{P}\)-martingale with \(\mathbb{E}[Z_t]=1\) | Defines a valid probability measure \(\mathbb{Q}\) |
| 2 | \(\widetilde{W}_t\) is a continuous \(\mathbb{Q}\)-local martingale | Necessary property of Brownian motion |
| 3 | \(\langle\widetilde{W}\rangle_t = t\) | Identifies the volatility as unit constant |
| 4 | Apply Lévy's theorem | Concludes \(\widetilde{W}_t\) is Brownian motion |
Key Insight: The Cancellation Mechanism¶
Recall the two SDEs in play:
Itô's product rule gives:
Expanding each term:
| Term | Expands to | \(dt\) part | \(dW_t\) part |
|---|---|---|---|
| \(Z_t\,d\widetilde{W}_t\) | \(Z_t\,dW_t + Z_t\theta_t\,dt\) | \(+Z_t\theta_t\,dt\) | \(Z_t\,dW_t\) |
| \(\widetilde{W}_t\,dZ_t\) | \(-\widetilde{W}_t Z_t\theta_t\,dW_t\) | \(0\) | \(-\widetilde{W}_t Z_t\theta_t\,dW_t\) |
| \(d\langle \widetilde{W}, Z \rangle_t\) | \((dW_t)(-Z_t\theta_t\,dW_t) = -Z_t\theta_t\,dt\) | \(-Z_t\theta_t\,dt\) | \(0\) |
Summing the \(dt\) column: \(+Z_t\theta_t\,dt - Z_t\theta_t\,dt = 0\).
The two \(dt\) contributions are:
- \(+Z_t\theta_t\,dt\) comes from the drift \(\theta_t\,dt\) baked into \(\widetilde{W}_t = W_t + \int\theta\,ds\)
- \(-Z_t\theta_t\,dt\) comes from the cross-variation \(d\langle \widetilde{W}, Z\rangle_t\), which is non-zero precisely because \(Z_t\) was constructed with the \(-\theta_t\,dW_t\) term
These two cancel to give:
No \(dt\) term survives, so \(\widetilde{W}_t Z_t\) is a \(\mathbb{P}\)-martingale, and \(\widetilde{W}_t\) is a \(\mathbb{Q}\)-martingale.
Recall (see § Novikov and Kazamaki Conditions): without Novikov, \(Z_t\) is only a supermartingale with \(\mathbb{E}^{\mathbb{P}}[Z_T] \le 1\), and the measure change may fail.
Filtration is preserved
The measure change from \(\mathbb{P}\) to \(\mathbb{Q}\) does not alter the filtration \(\{\mathcal{F}_t\}\). Adaptedness is a pathwise property: a process \(X_t\) is \(\mathcal{F}_t\)-measurable if and only if it depends only on the history up to time \(t\). Since equivalent measure changes do not alter the sample space or the sigma-algebras, every adapted process under \(\mathbb{P}\) remains adapted under \(\mathbb{Q}\).
Exercises¶
Exercise 1. Let \(\theta\) be a constant. Apply Ito's lemma to \(Z_t = \exp(-\theta W_t - \frac{1}{2}\theta^2 t)\) and verify that \(dZ_t = -\theta Z_t\,dW_t\). Explain why the absence of a \(dt\) term guarantees that \(Z_t\) is a local martingale.
Solution to Exercise 1
Let \(\theta\) be constant and \(Z_t = \exp(-\theta W_t - \frac{1}{2}\theta^2 t)\). Write \(Z_t = e^{f(W_t, t)}\) where \(f(x, t) = -\theta x - \frac{1}{2}\theta^2 t\).
By Itô's lemma applied to \(Z_t = e^{f(W_t, t)}\):
Computing the partial derivatives:
Substituting and using \((dW_t)^2 = dt\):
The \(dt\) terms cancel exactly, leaving \(dZ_t = -\theta Z_t\,dW_t\).
The absence of a \(dt\) term means \(Z_t\) has no drift, which is precisely the defining property of a local martingale. A process whose stochastic differential contains only \(dW_t\) terms and no \(dt\) terms is a local martingale because its conditional expectation satisfies \(\mathbb{E}[Z_{t+h} - Z_t | \mathcal{F}_t] \approx 0\) (the Itô integral has zero expectation). The Novikov condition then promotes this local martingale to a true martingale.
Exercise 2. In Step 2 of the proof, the change-of-measure formula states
Explain why the denominator is \(Z_s\) rather than \(Z_T\). What property of \(Z_t\) makes this valid?
Solution to Exercise 2
The change-of-measure formula states:
The denominator is \(Z_s\) rather than \(Z_T\) because of the tower property combined with the martingale property of \(Z_t\).
Since \(Z_t\) is a \(\mathbb{P}\)-martingale, we have \(\mathbb{E}^{\mathbb{P}}[Z_T | \mathcal{F}_s] = Z_s\). This is the key property that allows us to replace \(Z_T\) by \(Z_s\) in the denominator. Formally, the derivation proceeds as follows. For any \(A \in \mathcal{F}_s\):
We want to express \(\mathbb{E}^{\mathbb{Q}}[X_t | \mathcal{F}_s]\) in terms of a \(\mathbb{P}\)-conditional expectation. The abstract Bayes formula for conditional expectations under a change of measure gives:
The final equality uses precisely the martingale property \(\mathbb{E}^{\mathbb{P}}[Z_T | \mathcal{F}_s] = Z_s\). Without this property, the formula would not simplify, and the denominator would remain as the conditional expectation of \(Z_T\), which would depend on the specific event and not factor out cleanly.
Exercise 3. Levy's characterization theorem states that a continuous local martingale with quadratic variation \(\langle M \rangle_t = t\) is a standard Brownian motion. Explain why both conditions (continuity and unit quadratic variation) are necessary. Give an example of a continuous martingale whose quadratic variation is not \(t\) and is therefore not a Brownian motion.
Solution to Exercise 3
Lévy's characterization requires both conditions:
- Continuity: The process must have continuous sample paths.
- Unit quadratic variation: \(\langle M \rangle_t = t\).
Both are necessary because:
- A continuous martingale can have quadratic variation different from \(t\). In that case, it is a time-changed Brownian motion but not a standard Brownian motion.
- A process with \(\langle M \rangle_t = t\) that is not continuous could be a jump process.
Example of a continuous martingale that is not Brownian: Let \(B_t\) be a standard Brownian motion and define \(M_t = \sigma B_t\) for some constant \(\sigma \neq 1\). Then \(M_t\) is a continuous martingale (since \(B_t\) is), but its quadratic variation is:
So \(M_t\) is not a Brownian motion. It is a scaled Brownian motion with the wrong volatility.
Another example: let \(M_t = B_{t^2}\) (Brownian motion evaluated at \(t^2\)). This is a continuous martingale (in a suitably defined filtration), but \(\langle M \rangle_t = t^2 \neq t\), so it is not a standard Brownian motion. It can, however, be represented as a time-changed Brownian motion.
Exercise 4. Consider the process \(\widetilde{W}_t = W_t + \int_0^t \theta_s\,ds\). Show that the integral \(\int_0^t \theta_s\,ds\) has zero quadratic variation, and conclude that \(\langle \widetilde{W} \rangle_t = \langle W \rangle_t = t\). Why does this step rely on the fact that the integral is a finite-variation process?
Solution to Exercise 4
The process \(\widetilde{W}_t = W_t + \int_0^t \theta_s\,ds\) is the sum of a Brownian motion and a finite-variation process.
Quadratic variation of the integral term: Let \(A_t = \int_0^t \theta_s\,ds\). This is a process of bounded (finite) variation on \([0, T]\) because, for any partition \(0 = t_0 < t_1 < \cdots < t_n = T\):
A fundamental property of quadratic variation is that any finite-variation process has zero quadratic variation. This is because:
As the partition mesh goes to zero, the maximum increment \(\max_k |A_{t_{k+1}} - A_{t_k}| \to 0\) (by continuity of \(A\)), while the total variation sum remains bounded. Therefore the quadratic variation sum converges to zero: \(\langle A \rangle_t = 0\).
For the quadratic variation of \(\widetilde{W}_t\):
The cross-variation \(\langle W, A \rangle_t = 0\) because \(A\) has finite variation (the cross-variation of a continuous semimartingale with a finite-variation process is always zero). Since \(\langle A \rangle_t = 0\) as well:
Exercise 5. Suppose the Novikov condition fails, meaning \(\mathbb{E}^{\mathbb{P}}[\exp(\frac{1}{2}\int_0^T \theta_s^2\,ds)] = \infty\). Explain why \(Z_t\) may still be a local martingale but could fail to be a true martingale. What goes wrong with the measure change in this case? State the consequence for \(\mathbb{E}^{\mathbb{P}}[Z_T]\).
Solution to Exercise 5
When the Novikov condition fails, the exponential process \(Z_t = \exp(-\int_0^t \theta_s\,dW_s - \frac{1}{2}\int_0^t \theta_s^2\,ds)\) is still a non-negative local martingale under \(\mathbb{P}\) (this follows from the Itô computation \(dZ_t = -Z_t\theta_t\,dW_t\), which has no drift term, regardless of whether the Novikov condition holds).
However, a non-negative local martingale is always a supermartingale (by Fatou's lemma), meaning:
In particular, \(\mathbb{E}^{\mathbb{P}}[Z_T] \leq Z_0 = 1\). Without the Novikov condition, strict inequality can occur:
This means the "measure" \(\mathbb{Q}\) defined by \(d\mathbb{Q}/d\mathbb{P}|_{\mathcal{F}_T} = Z_T\) satisfies \(\mathbb{Q}(\Omega) = \mathbb{E}^{\mathbb{P}}[Z_T] < 1\), so \(\mathbb{Q}\) is not a valid probability measure (it does not assign total mass 1 to the sample space). Some probability "leaks" to infinity.
Intuitively, when \(\theta_s\) can become very large, the exponential martingale \(Z_t\) can spend extended periods near zero, causing its expectation to drop below 1. The "lost mass" corresponds to paths along which \(Z_t\) collapses toward zero. The Novikov condition prevents this by ensuring \(\theta_s\) does not grow too fast, keeping \(Z_t\) well-behaved enough to be a true martingale with unit expectation.
Exercise 6. The Radon–Nikodym derivative is \(Z_t = \exp(-\theta W_t - \frac{1}{2}\theta^2 t)\). Explain why the correction term \(-\frac{1}{2}\theta^2 t\) is absolutely necessary. What goes wrong if you define \(\hat{Z}_t = \exp(-\theta W_t)\) instead?
Solution to Exercise 6
Apply Itô's lemma to \(\hat{Z}_t = \exp(-\theta W_t)\). Writing \(\hat{Z}_t = e^{f(W_t)}\) with \(f(x) = -\theta x\):
The \(dt\) term \(\frac{1}{2}\theta^2\hat{Z}_t\,dt\) does not vanish. This means \(\hat{Z}_t\) is not a local martingale — it has a positive drift. Consequently:
- \(\mathbb{E}^{\mathbb{P}}[\hat{Z}_T] = e^{\frac{1}{2}\theta^2 T} \neq 1\), so \(\hat{Z}_T\) cannot define a valid probability measure.
- The measure defined by normalizing \(\hat{Z}_T\) would not produce the correct Brownian shift — the resulting shifted process would not be a standard Brownian motion under the new measure.
The correction \(-\frac{1}{2}\theta^2 t\) is precisely the Itô correction that cancels this unwanted drift. With \(Z_t = \exp(-\theta W_t - \frac{1}{2}\theta^2 t)\):
No \(dt\) term survives, making \(Z_t\) a local martingale with \(\mathbb{E}^{\mathbb{P}}[Z_T] = 1\). This is the defining property of the stochastic exponential: the \(-\frac{1}{2}\theta^2 t\) term is the price of working with exponentials in Itô calculus, where the chain rule picks up an extra second-derivative (quadratic variation) term that must be compensated.