Proof Sketch and Key Ideas¶

This section develops the main ideas behind the proof of Girsanov's theorem systematically.

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.

Method: Apply Itô's lemma to \(Z_t = e^{X_t}\) where \(X_t = -\int_0^t \theta_s\,dW_s - \frac{1}{2}\int_0^t \theta_s^2\,ds\).

Computing the differential:

\[ dX_t = -\theta_t\,dW_t - \frac{1}{2}\theta_t^2\,dt \]

Apply Itô to \(f(X) = e^X\):

\[ dZ_t = e^{X_t}\,dX_t + \frac{1}{2}e^{X_t}(dX_t)^2 \]

Compute \((dX_t)^2\):

\[ (dX_t)^2 = \left(-\theta_t\,dW_t - \frac{1}{2}\theta_t^2\,dt\right)^2 = \theta_t^2\,(dW_t)^2 = \theta_t^2\,dt \]

Therefore:

\[ dZ_t = Z_t\left(-\theta_t\,dW_t - \frac{1}{2}\theta_t^2\,dt\right) + \frac{1}{2}Z_t\,\theta_t^2\,dt \]

\[ = -Z_t\theta_t\,dW_t - \frac{1}{2}Z_t\theta_t^2\,dt + \frac{1}{2}Z_t\theta_t^2\,dt \]

\[ \boxed{dZ_t = -Z_t\theta_t\,dW_t} \]

Key observation: There is no \(dt\) term, only the \(dW_t\) term. This shows \(Z_t\) is a local martingale; the Novikov condition then ensures it is a true martingale.

Since \(Z_t\) is a true martingale (by Novikov):

\[ \mathbb{E}^{\mathbb{P}}[Z_t] = \mathbb{E}^{\mathbb{P}}[Z_0] = 1 \]

Step 2: Unit Expectation (Justification)¶

When \(\theta_t\) is deterministic (or constant), \(X_t = -\int_0^t \theta_s\,dW_s - \frac{1}{2}\int_0^t \theta_s^2\,ds\) is Gaussian (since the Ito integral of a deterministic integrand is Gaussian):

\[ X_t \sim \mathcal{N}\left(-\frac{1}{2}\int_0^t \theta_s^2\,ds, \int_0^t \theta_s^2\,ds\right) \]

For any Gaussian random variable \(Y \sim \mathcal{N}(\mu, \sigma^2)\):

\[ \mathbb{E}[e^Y] = \exp\left(\mu + \frac{\sigma^2}{2}\right) \]

Applying this with \(\mu = -\frac{1}{2}\int_0^t \theta_s^2\,ds\) and \(\sigma^2 = \int_0^t \theta_s^2\,ds\):

\[ \mathbb{E}^{\mathbb{P}}[Z_t] = \exp\left(-\frac{1}{2}\int_0^t \theta_s^2\,ds + \frac{1}{2}\int_0^t \theta_s^2\,ds\right) = 1 \quad \checkmark \]

Step 3: The Shifted Process is a Q-Martingale¶

Goal: Show that \(\widetilde{W}_t = W_t + \int_0^t \theta_s\,ds\) is a martingale under \(\mathbb{Q}\).

Method: Use the change-of-measure formula. For any \(\mathcal{F}_t\)-adapted process \(X_t\):

\[ \mathbb{E}^{\mathbb{Q}}[X_t | \mathcal{F}_s] = \frac{\mathbb{E}^{\mathbb{P}}[X_t Z_T | \mathcal{F}_s]}{\mathbb{E}^{\mathbb{P}}[Z_T | \mathcal{F}_s]} \]

For \(s \leq t\), by the martingale property of \(Z\):

\[ \mathbb{E}^{\mathbb{P}}[Z_T | \mathcal{F}_s] = Z_s \]

Therefore, for the shifted Brownian motion:

\[ \mathbb{E}^{\mathbb{Q}}\left[\widetilde{W}_t | \mathcal{F}_s\right] = \frac{\mathbb{E}^{\mathbb{P}}\left[\left(W_t + \int_0^t \theta_u\,du\right) Z_T | \mathcal{F}_s\right]}{Z_s} \]

By the martingale property and adapted property of \(\theta\):

\[ = \frac{\mathbb{E}^{\mathbb{P}}\left[\left(W_t + \int_0^t \theta_u\,du\right) Z_T | \mathcal{F}_s\right]}{Z_s} \]

The key step is to compute \(d(\widetilde{W}_t Z_t)\) via the Ito product rule and verify that no \(dt\) term survives. The explicit calculation (carried out in the cancellation mechanism section below) proceeds as:

Compute \(d(\widetilde{W}_t Z_t) = Z_t\,d\widetilde{W}_t + \widetilde{W}_t\,dZ_t + d\langle \widetilde{W}, Z\rangle_t\)
The \(dt\) contributions \(+Z_t\theta_t\,dt\) (from the drift in \(\widetilde{W}_t\)) and \(-Z_t\theta_t\,dt\) (from the cross-variation) cancel exactly
Therefore \(\widetilde{W}_t Z_t\) is a \(\mathbb{P}\)-martingale. Dividing by \(Z_s\) in the conditional expectation (using Bayes' rule for conditional expectations under measure change) yields that \(\widetilde{W}_t\) is a \(\mathbb{Q}\)-martingale

Result: \(\widetilde{W}_t\) is a \(\mathbb{Q}\)-martingale.

Step 4: Quadratic Variation is Measure-Invariant¶

Key principle: Quadratic variation is a pathwise property—it depends only on the paths of the process, not on the probability measure used.

Since:

\[ \widetilde{W}_t = W_t + \int_0^t \theta_s\,ds \]

and the integral term has finite variation (in particular, absolutely continuous):

\[ d\langle\widetilde{W}\rangle_t = d\langle W\rangle_t = dt \]

Conclusion: The quadratic variation of \(\widetilde{W}_t\) equals \(t\), the same as for standard Brownian motion.

Step 5: 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 \(\mathbb{Q}\)-martingale (Step 3)
\(\langle\widetilde{W}\rangle_t = t\) (Step 4)

By Lévy's theorem:

\[ \boxed{\widetilde{W}_t \text{ is a standard Brownian motion under } \mathbb{Q}} \]

Step 6: Alternative Derivation Using Change of Variables¶

Another approach uses direct computation. This derivation is most transparent when \(\theta\) is constant; for general adapted \(\theta_t\), the characteristic-function argument requires additional care. Under \(\mathbb{Q}\), we want to show \(\widetilde{W}_t\) is Brownian.

Key formula: For a \(\mathbb{Q}\)-martingale \(M_t\) with \(\langle M\rangle_t = t\), the process must be Brownian.

We can verify this by computing the characteristic function (shown here for constant \(\theta\)):

\[ \mathbb{E}^{\mathbb{Q}}[e^{i\lambda\widetilde{W}_t}] = e^{-\lambda^2 t/2} \]

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.

Summary of the Proof Strategy¶

Step	What we prove	Why it's needed
1	\(Z_t\) is a \(\mathbb{P}\)-martingale	To define the measure \(\mathbb{Q}\) properly
2	\(\mathbb{E}^{\mathbb{P}}[Z_t] = 1\)	Ensures \(\mathbb{Q}\) is a probability measure
3	\(\widetilde{W}_t\) is a \(\mathbb{Q}\)-martingale	Necessary property of Brownian motion
4	\(d\langle\widetilde{W}\rangle_t = dt\)	Identifies the volatility as unit constant
5	Apply Lévy's theorem	Concludes \(\widetilde{W}_t\) is Brownian motion

Key Insight: The Cancellation Mechanism¶

The proof's elegance lies in a concrete \(dt\)-cancellation that occurs inside the Itô product rule for \(\widetilde{W}_t Z_t\).

Recall the two SDEs in play:

\[ d\widetilde{W}_t = dW_t + \theta_t\,dt, \qquad dZ_t = -Z_t\theta_t\,dW_t \]

Itô's product rule gives:

\[ d(\widetilde{W}_t Z_t) = Z_t\,d\widetilde{W}_t + \widetilde{W}_t\,dZ_t + d\langle \widetilde{W}, Z \rangle_t \]

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:

\[ d(\widetilde{W}_t Z_t) = \bigl(Z_t - \widetilde{W}_t Z_t \theta_t\bigr)\,dW_t \]

No \(dt\) term survives, so \(\widetilde{W}_t Z_t\) is a \(\mathbb{P}\)-martingale, and \(\widetilde{W}_t\) is a \(\mathbb{Q}\)-martingale. This cancellation is not coincidental — \(Z_t\) is defined by \(dZ_t = -Z_t\theta_t\,dW_t\) precisely so that its cross-variation with \(\widetilde{W}_t\) kills the drift \(\theta_t\,dt\) term.

Technical Note: Novikov's Condition¶

The Novikov condition \(\mathbb{E}^{\mathbb{P}}[\exp(\frac{1}{2}\int_0^T \theta_s^2\,ds)] < \infty\) is sufficient to guarantee:

\(Z_t\) is a true martingale (not just a local martingale)
\(\mathbb{E}^{\mathbb{P}}[Z_T] = 1\)
The measure change is valid

Without this condition, \(Z_t\) may only be a local martingale, and the measure change may not define a valid probability measure.

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 \(Z_t = e^{f(W_t, t)}\):

\[ dZ_t = \frac{\partial Z_t}{\partial t}\,dt + \frac{\partial Z_t}{\partial W_t}\,dW_t + \frac{1}{2}\frac{\partial^2 Z_t}{\partial W_t^2}\,(dW_t)^2 \]

Computing the partial derivatives:

\[ \frac{\partial Z_t}{\partial t} = -\frac{1}{2}\theta^2 Z_t, \qquad \frac{\partial Z_t}{\partial W_t} = -\theta Z_t, \qquad \frac{\partial^2 Z_t}{\partial W_t^2} = \theta^2 Z_t \]

Substituting and using \((dW_t)^2 = dt\):

\[ dZ_t = -\frac{1}{2}\theta^2 Z_t\,dt + (-\theta Z_t)\,dW_t + \frac{1}{2}\theta^2 Z_t\,dt \]

\[ = -\theta Z_t\,dW_t + \left(-\frac{1}{2}\theta^2 + \frac{1}{2}\theta^2\right)Z_t\,dt \]

\[ = -\theta Z_t\,dW_t \]

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. For constant \(\theta\), the exponent \(X_t = -\theta W_t - \frac{1}{2}\theta^2 t\) is Gaussian with mean \(-\frac{1}{2}\theta^2 t\) and variance \(\theta^2 t\). Using the moment generating function of a Gaussian random variable, verify that \(\mathbb{E}^{\mathbb{P}}[Z_t] = 1\) for all \(t \geq 0\).

Solution to Exercise 2

For constant \(\theta\), the exponent is \(X_t = -\theta W_t - \frac{1}{2}\theta^2 t\). Since \(W_t \sim \mathcal{N}(0, t)\) under \(\mathbb{P}\):

\[ -\theta W_t \sim \mathcal{N}(0, \theta^2 t) \]

Therefore:

\[ X_t = -\theta W_t - \frac{1}{2}\theta^2 t \sim \mathcal{N}\!\left(-\frac{1}{2}\theta^2 t,\; \theta^2 t\right) \]

with mean \(\mu_X = -\frac{1}{2}\theta^2 t\) and variance \(\sigma_X^2 = \theta^2 t\).

For a Gaussian random variable \(Y \sim \mathcal{N}(\mu, \sigma^2)\), the moment generating function gives:

\[ \mathbb{E}[e^Y] = \exp\!\left(\mu + \frac{\sigma^2}{2}\right) \]

Applying this to \(Z_t = e^{X_t}\):

\[ \mathbb{E}^{\mathbb{P}}[Z_t] = \exp\!\left(-\frac{1}{2}\theta^2 t + \frac{\theta^2 t}{2}\right) = \exp(0) = 1 \]

This holds for all \(t \geq 0\), confirming the unit expectation property of the exponential martingale.

Exercise 3. In Step 3 of the proof, the change-of-measure formula states

\[ \mathbb{E}^{\mathbb{Q}}[X_t | \mathcal{F}_s] = \frac{\mathbb{E}^{\mathbb{P}}[X_t Z_T | \mathcal{F}_s]}{Z_s} \]

Explain why the denominator is \(Z_s\) rather than \(Z_T\). What property of \(Z_t\) makes this valid?

Solution to Exercise 3

The change-of-measure formula states:

\[ \mathbb{E}^{\mathbb{Q}}[X_t | \mathcal{F}_s] = \frac{\mathbb{E}^{\mathbb{P}}[X_t Z_T | \mathcal{F}_s]}{Z_s} \]

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 \(A \in \mathcal{F}_s\):

\[ \mathbb{E}^{\mathbb{Q}}[\mathbf{1}_A X_t] = \mathbb{E}^{\mathbb{P}}[\mathbf{1}_A X_t Z_T] \]

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:

\[ \mathbb{E}^{\mathbb{Q}}[X_t | \mathcal{F}_s] = \frac{\mathbb{E}^{\mathbb{P}}[X_t Z_T | \mathcal{F}_s]}{\mathbb{E}^{\mathbb{P}}[Z_T | \mathcal{F}_s]} = \frac{\mathbb{E}^{\mathbb{P}}[X_t Z_T | \mathcal{F}_s]}{Z_s} \]

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 4. 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 4

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:

\[ \langle M \rangle_t = \sigma^2 \langle B \rangle_t = \sigma^2 t \neq t \]

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 5. 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 5

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 \(0 = t_0 < t_1 < \cdots < t_n = T\):

\[ \sum_{k=0}^{n-1} |A_{t_{k+1}} - A_{t_k}| = \sum_{k=0}^{n-1} \left|\int_{t_k}^{t_{k+1}} \theta_s\,ds\right| \leq \int_0^T |\theta_s|\,ds < \infty \]

A fundamental property of quadratic variation is that any finite-variation process has zero quadratic variation. This is because:

\[ \sum_{k=0}^{n-1} (A_{t_{k+1}} - A_{t_k})^2 \leq \max_k |A_{t_{k+1}} - A_{t_k}| \cdot \sum_{k=0}^{n-1} |A_{t_{k+1}} - A_{t_k}| \]

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 \(\widetilde{W}_t\):

\[ \langle \widetilde{W} \rangle_t = \langle W + A \rangle_t = \langle W \rangle_t + 2\langle W, A \rangle_t + \langle A \rangle_t \]

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:

\[ \langle \widetilde{W} \rangle_t = \langle W \rangle_t = t \]

Exercise 6. 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 6

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:

\[ \mathbb{E}^{\mathbb{P}}[Z_T | \mathcal{F}_s] \leq Z_s \]

In particular, \(\mathbb{E}^{\mathbb{P}}[Z_T] \leq Z_0 = 1\). Without the Novikov condition, strict inequality can occur:

\[ \mathbb{E}^{\mathbb{P}}[Z_T] < 1 \]

This means the "measure" \(\mathbb{Q}\) defined by \(d\mathbb{Q}/d\mathbb{P} = Z_T\) 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 7. Verify the characteristic function computation in Step 6: show that \(\mathbb{E}^{\mathbb{Q}}[e^{i\lambda \widetilde{W}_t}] = e^{-\lambda^2 t / 2}\) for constant \(\theta\). Start from the definition \(\mathbb{E}^{\mathbb{Q}}[e^{i\lambda \widetilde{W}_t}] = \mathbb{E}^{\mathbb{P}}[e^{i\lambda \widetilde{W}_t} Z_T]\) and use the fact that \(\widetilde{W}_t = W_t + \theta t\) and \(Z_T = \exp(-\theta W_T - \frac{1}{2}\theta^2 T)\).

Solution to Exercise 7

We want to compute \(\mathbb{E}^{\mathbb{Q}}[e^{i\lambda \widetilde{W}_t}]\) for constant \(\theta\). By definition of the \(\mathbb{Q}\)-expectation:

\[ \mathbb{E}^{\mathbb{Q}}[e^{i\lambda \widetilde{W}_t}] = \mathbb{E}^{\mathbb{P}}[e^{i\lambda \widetilde{W}_t} Z_T] \]

Substituting \(\widetilde{W}_t = W_t + \theta t\) and \(Z_T = \exp(-\theta W_T - \frac{1}{2}\theta^2 T)\):

\[ = \mathbb{E}^{\mathbb{P}}\!\left[\exp\!\left(i\lambda(W_t + \theta t) - \theta W_T - \frac{1}{2}\theta^2 T\right)\right] \]

\[ = e^{i\lambda\theta t - \frac{1}{2}\theta^2 T} \cdot \mathbb{E}^{\mathbb{P}}\!\left[\exp\!\left(i\lambda W_t - \theta W_T\right)\right] \]

Since \(W_T = W_t + (W_T - W_t)\) and \(W_t\) is independent of \(W_T - W_t\) under \(\mathbb{P}\):

\[ = e^{i\lambda\theta t - \frac{1}{2}\theta^2 T} \cdot \mathbb{E}^{\mathbb{P}}\!\left[e^{(i\lambda - \theta)W_t}\right] \cdot \mathbb{E}^{\mathbb{P}}\!\left[e^{-\theta(W_T - W_t)}\right] \]

Using the moment generating function of Gaussian variables, \(W_t \sim \mathcal{N}(0, t)\) and \(W_T - W_t \sim \mathcal{N}(0, T-t)\):

\[ \mathbb{E}^{\mathbb{P}}\!\left[e^{(i\lambda - \theta)W_t}\right] = \exp\!\left(\frac{(i\lambda - \theta)^2 t}{2}\right) = \exp\!\left(\frac{(-\lambda^2 - 2i\lambda\theta + \theta^2)t}{2}\right) \]

\[ \mathbb{E}^{\mathbb{P}}\!\left[e^{-\theta(W_T - W_t)}\right] = \exp\!\left(\frac{\theta^2(T-t)}{2}\right) \]

Combining all factors:

\[ \mathbb{E}^{\mathbb{Q}}[e^{i\lambda\widetilde{W}_t}] = \exp\!\left(i\lambda\theta t - \frac{\theta^2 T}{2} + \frac{(-\lambda^2 - 2i\lambda\theta + \theta^2)t}{2} + \frac{\theta^2(T-t)}{2}\right) \]

Collecting terms in the exponent:

\[ = i\lambda\theta t - \frac{\theta^2 T}{2} - \frac{\lambda^2 t}{2} - i\lambda\theta t + \frac{\theta^2 t}{2} + \frac{\theta^2 T}{2} - \frac{\theta^2 t}{2} \]

The terms \(i\lambda\theta t\) cancel, \(-\frac{\theta^2 T}{2} + \frac{\theta^2 T}{2}\) cancel, and \(\frac{\theta^2 t}{2} - \frac{\theta^2 t}{2}\) cancel, leaving:

\[ \mathbb{E}^{\mathbb{Q}}[e^{i\lambda\widetilde{W}_t}] = e^{-\lambda^2 t/2} \]

This is the characteristic function of \(\mathcal{N}(0, t)\), confirming that \(\widetilde{W}_t\) is a standard Brownian motion under \(\mathbb{Q}\).