Drift Adjustment and Financial Meaning

Girsanov’s theorem is used in finance to remove risk premia from asset prices, making arbitrage-free pricing possible.

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Asset Price Dynamics

Under the physical measure \(\mathbb{P}\), assume the stock price follows geometric Brownian motion:

\[ dS_t = \mu S_t\,dt + \sigma S_t\,dW_t^{\mathbb{P}} \]

where \(\mu\) is the real-world expected return and \(\sigma > 0\) is the volatility.

For derivative pricing we need the discounted stock price \(\widetilde{S}_t := e^{-rt}S_t\) to be a martingale. By Itô's lemma:

\[ d\widetilde{S}_t = d(e^{-rt}S_t) = -r e^{-rt}S_t\,dt + e^{-rt}dS_t = e^{-rt}S_t\bigl[(\mu - r)\,dt + \sigma\,dW_t^{\mathbb{P}}\bigr] \]

The \(dt\) coefficient is \((\mu - r)S_te^{-rt}\), which is non-zero whenever \(\mu \neq r\). Discounting alone does not eliminate the drift — we need a change of measure to remove it.

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Market Price of Risk

To eliminate the residual drift \((\mu - r)\) we need to shift the Brownian motion by exactly the right amount. Define the market price of risk:

\[ \theta := \frac{\mu - r}{\sigma} \]

This has a concrete financial interpretation: it is the excess return per unit of volatility, i.e. how many units of risk premium the investor earns for each unit of volatility exposure. In the scalar GBM setting it coincides with the Sharpe ratio of the asset, though "market price of risk" is the safer general terminology.

The sign of \(\theta\) encodes the direction of the adjustment:

• If \(\mu > r\): the asset earns a positive risk premium, so \(\theta > 0\). The measure change will shift probability mass toward lower paths (removing the upward drift).
• If \(\mu < r\): \(\theta < 0\), and the adjustment goes the other way.
• If \(\mu = r\): \(\theta = 0\), no measure change is needed, and \(\widetilde{S}_t\) is already a martingale.

The factor of \(\sigma\) in the denominator appears because the Brownian motion enters the dynamics with coefficient \(\sigma\) — a unit shift in \(W_t\) produces a \(\sigma\)-sized move in \(S_t\), so we need to shift \(W_t\) by \(\theta = (\mu-r)/\sigma\) to cancel a drift of size \((\mu-r)\).

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Measure Change

Define the Radon-Nikodym density process:

\[ Z_t = \exp\!\left(-\theta W_t^{\mathbb{P}} - \frac{1}{2}\theta^2 t\right), \qquad \frac{d\mathbb{Q}}{d\mathbb{P}}\bigg|_{\mathcal{F}_T} = Z_T \]

By Girsanov's theorem, under \(\mathbb{Q}\) the process

\[ W_t^{\mathbb{Q}} := W_t^{\mathbb{P}} + \theta t \]

is a standard Brownian motion. Equivalently, \(W_t^{\mathbb{P}} = W_t^{\mathbb{Q}} - \theta t\).

Why this specific \(Z_t\)? The exponential form arises from Itô's lemma: applying it to \(e^{X_t}\) with \(X_t = -\theta W_t^{\mathbb{P}} - \frac{1}{2}\theta^2 t\) gives \(dZ_t = -\theta Z_t\,dW_t^{\mathbb{P}}\), which has no \(dt\) term, guaranteeing \(Z_t\) is a \(\mathbb{P}\)-martingale with \(\mathbb{E}^{\mathbb{P}}[Z_T] = 1\) so that \(\mathbb{Q}\) is a valid probability measure.

How the reweighting works: Paths where \(W_T^{\mathbb{P}}\) is large and positive make \(Z_T = e^{-\theta W_T^{\mathbb{P}} - \frac{1}{2}\theta^2 T}\) small (for \(\theta > 0\)), downweighting the high-return paths. Paths where \(W_T^{\mathbb{P}}\) is negative get upweighted. The net effect is that the average drift is shifted from \(\mu\) down to \(r\).

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Risk-Neutral Dynamics

We now substitute \(W_t^{\mathbb{P}} = W_t^{\mathbb{Q}} - \theta t\) into the original SDE:

\[ dS_t = \mu S_t\,dt + \sigma S_t\,dW_t^{\mathbb{P}} = \mu S_t\,dt + \sigma S_t\left(dW_t^{\mathbb{Q}} - \theta\,dt\right) \]

\[ = \mu S_t\,dt - \sigma\theta S_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}} = (\mu - \sigma\theta)S_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}} \]

Substituting \(\theta = (\mu - r)/\sigma\), so \(\sigma\theta = \mu - r\):

\[ \boxed{dS_t = r S_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}}} \]

The drift has shifted from \(\mu\) to \(r\). To confirm that \(\widetilde{S}_t = e^{-rt}S_t\) is now a \(\mathbb{Q}\)-martingale, apply Itô's lemma:

\[ d\widetilde{S}_t = e^{-rt}S_t\bigl[(r - r)\,dt + \sigma\,dW_t^{\mathbb{Q}}\bigr] = \sigma\widetilde{S}_t\,dW_t^{\mathbb{Q}} \]

There is no \(dt\) term, confirming that \(\widetilde{S}_t\) is a \(\mathbb{Q}\)-martingale. The option pricing formula follows immediately:

\[ V_t = e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}\bigl[\Phi(S_T)\,\big|\,\mathcal{F}_t\bigr] \]

where \(\Phi\) is the payoff function. Assuming absence of arbitrage and market completeness, this is the risk-neutral pricing formula, and it holds for any \(\mathcal{F}_T\)-measurable payoff \(\Phi(S_T)\) integrable under \(\mathbb{Q}\). This construction corresponds to the existence of an equivalent martingale measure, as guaranteed by the Fundamental Theorem of Asset Pricing.

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Interpretation

Expected returns are adjusted to the risk-free rate. Under \(\mathbb{P}\), a risky asset earns \(\mu > r\) to compensate investors for bearing risk. Under \(\mathbb{Q}\), every asset grows at \(r\) regardless of its risk. This does not mean we believe the stock actually grows at \(r\) — it means we have reweighted the probability space so that the pricing formula \(V = e^{-rT}\mathbb{E}^{\mathbb{Q}}[\Phi]\) is internally consistent and arbitrage-free. The real-world drift \(\mu\) drops out entirely, which is why option prices do not depend on investors' return expectations.

Volatility is unchanged. The Girsanov shift only moves the drift; the diffusion coefficient \(\sigma\) multiplies \(dW_t\) in both the \(\mathbb{P}\) and \(\mathbb{Q}\) dynamics. This is because quadratic variation — which determines \(\sigma\) — is a pathwise property of the sample paths, independent of which measure is used to weight them. The paths themselves are identical under \(\mathbb{P}\) and \(\mathbb{Q}\); only their probabilities differ.

Sample paths are identical; probabilities are reweighted. Both \(\mathbb{P}\) and \(\mathbb{Q}\) are defined on the same probability space and generate the same set of possible paths. What changes is how much weight each path receives. Specifically, for \(\theta > 0\) (i.e. \(\mu > r\)), paths that drifted strongly upward under \(\mathbb{P}\) (large positive \(W_T^{\mathbb{P}}\)) are downweighted by \(Z_T = e^{-\theta W_T^{\mathbb{P}} - \frac{1}{2}\theta^2 T}\), while paths that drifted downward are upweighted. The net effect is that the average drift seen under \(\mathbb{Q}\) equals \(r\) rather than \(\mu\).

This drift adjustment is the mathematical foundation of risk-neutral pricing: it converts the no-arbitrage condition (the discounted price is a martingale) into a concrete computational tool.

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Exercises

Exercise 1. A stock follows \(dS_t = 0.08\,S_t\,dt + 0.30\,S_t\,dW_t^{\mathbb{P}}\) with risk-free rate \(r = 0.02\). Compute the market price of risk \(\theta\), write the density process \(Z_t\), and derive the risk-neutral dynamics of \(S_t\).

Solution to Exercise 1

Given \(dS_t = 0.08\,S_t\,dt + 0.30\,S_t\,dW_t^{\mathbb{P}}\) with \(r = 0.02\), the parameters are \(\mu = 0.08\), \(\sigma = 0.30\).

Market price of risk:

\[ \theta = \frac{\mu - r}{\sigma} = \frac{0.08 - 0.02}{0.30} = \frac{0.06}{0.30} = 0.2 \]

Density process:

\[ Z_t = \exp\!\left(-0.2\,W_t^{\mathbb{P}} - \frac{1}{2}(0.2)^2 t\right) = \exp\!\left(-0.2\,W_t^{\mathbb{P}} - 0.02\,t\right) \]

Risk-neutral dynamics: Define \(W_t^{\mathbb{Q}} = W_t^{\mathbb{P}} + 0.2\,t\), so \(W_t^{\mathbb{P}} = W_t^{\mathbb{Q}} - 0.2\,t\). Substituting:

\[ dS_t = 0.08\,S_t\,dt + 0.30\,S_t\left(dW_t^{\mathbb{Q}} - 0.2\,dt\right) \]

\[ = 0.08\,S_t\,dt + 0.30\,S_t\,dW_t^{\mathbb{Q}} - 0.06\,S_t\,dt \]

\[ = 0.02\,S_t\,dt + 0.30\,S_t\,dW_t^{\mathbb{Q}} \]

Under \(\mathbb{Q}\): \(dS_t = rS_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}} = 0.02\,S_t\,dt + 0.30\,S_t\,dW_t^{\mathbb{Q}}\). The drift has changed from \(\mu = 0.08\) to \(r = 0.02\) while volatility remains \(\sigma = 0.30\).

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Exercise 2. Starting from the physical dynamics \(dS_t = \mu S_t\,dt + \sigma S_t\,dW_t^{\mathbb{P}}\) and the definition \(W_t^{\mathbb{Q}} = W_t^{\mathbb{P}} + \theta t\), show step by step that the discounted price process \(e^{-rt}S_t\) is a \(\mathbb{Q}\)-martingale. In particular, verify that the \(dt\) term vanishes after substitution.

Solution to Exercise 2

Start with \(dS_t = \mu S_t\,dt + \sigma S_t\,dW_t^{\mathbb{P}}\) and \(W_t^{\mathbb{Q}} = W_t^{\mathbb{P}} + \theta t\) where \(\theta = (\mu - r)/\sigma\).

Step 1: Substitute \(dW_t^{\mathbb{P}} = dW_t^{\mathbb{Q}} - \theta\,dt\) into the SDE:

\[ dS_t = \mu S_t\,dt + \sigma S_t(dW_t^{\mathbb{Q}} - \theta\,dt) = (\mu - \sigma\theta)S_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}} \]

Step 2: Since \(\sigma\theta = \sigma \cdot \frac{\mu - r}{\sigma} = \mu - r\):

\[ dS_t = (\mu - (\mu - r))S_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}} = rS_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}} \]

Step 3: Apply Itô's lemma to \(\widetilde{S}_t = e^{-rt}S_t\):

\[ d\widetilde{S}_t = -re^{-rt}S_t\,dt + e^{-rt}\,dS_t \]

\[ = -re^{-rt}S_t\,dt + e^{-rt}\bigl(rS_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}}\bigr) \]

\[ = e^{-rt}S_t(-r + r)\,dt + \sigma e^{-rt}S_t\,dW_t^{\mathbb{Q}} \]

\[ = \sigma\widetilde{S}_t\,dW_t^{\mathbb{Q}} \]

Step 4: The \(dt\) term is zero. Since \(d\widetilde{S}_t = \sigma\widetilde{S}_t\,dW_t^{\mathbb{Q}}\) contains only a \(dW_t^{\mathbb{Q}}\) term and no drift, \(\widetilde{S}_t = e^{-rt}S_t\) is a \(\mathbb{Q}\)-martingale.

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Exercise 3. Assume a complete market driven by a single Brownian motion. An asset has physical drift \(\mu = 0.05\) and volatility \(\sigma = 0.40\) with \(r = 0.03\). A second asset has \(\mu' = 0.10\) and \(\sigma' = 0.40\), driven by the same Brownian motion. Compute \(\theta\) for each asset. Are the two values consistent? What does this tell you about arbitrage in this market?

Solution to Exercise 3

For the first asset: \(\mu = 0.05\), \(\sigma = 0.40\), \(r = 0.03\):

\[ \theta_1 = \frac{\mu - r}{\sigma} = \frac{0.05 - 0.03}{0.40} = \frac{0.02}{0.40} = 0.05 \]

For the second asset: \(\mu' = 0.10\), \(\sigma' = 0.40\), \(r = 0.03\):

\[ \theta_2 = \frac{\mu' - r}{\sigma'} = \frac{0.10 - 0.03}{0.40} = \frac{0.07}{0.40} = 0.175 \]

The two values are not consistent: \(\theta_1 = 0.05 \neq 0.175 = \theta_2\).

Since both assets are driven by the same Brownian motion \(W_t\), a single Girsanov change of measure can only shift \(W_t\) by one value of \(\theta\). If we use \(\theta_1 = 0.05\), then the first asset's discounted price is a martingale under \(\mathbb{Q}\), but the second asset's drift under \(\mathbb{Q}\) would be \(\mu' - \sigma'\theta_1 = 0.10 - 0.40 \times 0.05 = 0.08 \neq r = 0.03\). The second asset's discounted price would not be a martingale.

This means there is no single equivalent martingale measure under which both discounted prices are martingales. By the fundamental theorem of asset pricing, this implies the market admits arbitrage. Intuitively, one could exploit the inconsistency by trading the two assets against each other, since they share the same risk factor but offer different compensation per unit of risk. (The precise construction of the arbitrage strategy requires care and is not immediate from the \(\theta\)-mismatch alone.)

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Exercise 4. Explain why the volatility \(\sigma\) is unchanged under the Girsanov measure change, while the drift changes from \(\mu\) to \(r\). Relate your answer to the fact that quadratic variation is a pathwise quantity.

Solution to Exercise 4

The drift changes from \(\mu\) to \(r\) because it depends on how probabilities are assigned to paths. The Girsanov measure change reweights the probability of each path via the Radon-Nikodym derivative \(Z_T\), which shifts the expected value of \(dW_t\) from \(0\) (under \(\mathbb{P}\)) to \(-\theta\,dt\) (effectively, by making \(W_t^{\mathbb{Q}} = W_t^{\mathbb{P}} + \theta t\) a \(\mathbb{Q}\)-Brownian motion). This reweighting changes the first moment (mean/drift) of the process.

The volatility \(\sigma\), however, is determined by the quadratic variation, which is a pathwise quantity. Quadratic variation is computed as:

\[ \langle S \rangle_t = \lim_{n \to \infty} \sum_{k} (S_{t_{k+1}} - S_{t_k})^2 \]

This limit depends only on the individual sample paths, not on how those paths are weighted by the probability measure. Since \(\mathbb{P}\) and \(\mathbb{Q}\) are defined on the same sample space with the same set of paths (they are equivalent measures), the quadratic variation is identical under both measures.

Formally, in the SDE \(dS_t = \mu S_t\,dt + \sigma S_t\,dW_t^{\mathbb{P}}\), the Girsanov substitution replaces \(dW_t^{\mathbb{P}} = dW_t^{\mathbb{Q}} - \theta\,dt\), which only modifies the \(dt\) coefficient (the drift). The coefficient of \(dW_t^{\mathbb{Q}}\) remains \(\sigma S_t\), leaving volatility unchanged.

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Exercise 5. Under \(\mathbb{P}\), a zero-coupon bond price satisfies \(dP(t,T) = \mu_P P(t,T)\,dt + \sigma_P P(t,T)\,dW_t^{\mathbb{P}}\). Apply the Girsanov drift adjustment to derive the dynamics under \(\mathbb{Q}\) and show that the discounted bond price \(e^{-\int_0^t r_s\,ds}P(t,T)\) is a \(\mathbb{Q}\)-martingale.

Solution to Exercise 5

Under \(\mathbb{P}\), the zero-coupon bond dynamics are:

\[ dP(t,T) = \mu_P P(t,T)\,dt + \sigma_P P(t,T)\,dW_t^{\mathbb{P}} \]

The market price of risk is \(\lambda_t = (\mu_P - r_t)/\sigma_P\), where \(r_t\) is the short rate. Note that in interest rate models the market price of risk \(\lambda_t\) may depend on state variables such as \(r_t\), making \(\lambda_t\) adapted rather than constant. Define \(W_t^{\mathbb{Q}} = W_t^{\mathbb{P}} + \int_0^t \lambda_s\,ds\). Substituting \(dW_t^{\mathbb{P}} = dW_t^{\mathbb{Q}} - \lambda_t\,dt\):

\[ dP(t,T) = \mu_P P\,dt + \sigma_P P\bigl(dW_t^{\mathbb{Q}} - \lambda_t\,dt\bigr) \]

\[ = (\mu_P - \sigma_P\lambda_t)P\,dt + \sigma_P P\,dW_t^{\mathbb{Q}} \]

Since \(\sigma_P\lambda_t = \sigma_P \cdot \frac{\mu_P - r_t}{\sigma_P} = \mu_P - r_t\):

\[ dP(t,T) = r_t P(t,T)\,dt + \sigma_P P(t,T)\,dW_t^{\mathbb{Q}} \]

Now consider the discounted bond price \(\widetilde{P}(t,T) = e^{-\int_0^t r_s\,ds}P(t,T)\). By the product rule:

\[ d\widetilde{P} = -r_t e^{-\int_0^t r_s\,ds}P\,dt + e^{-\int_0^t r_s\,ds}\,dP \]

\[ = -r_t\widetilde{P}\,dt + e^{-\int_0^t r_s\,ds}\bigl(r_t P\,dt + \sigma_P P\,dW_t^{\mathbb{Q}}\bigr) \]

\[ = -r_t\widetilde{P}\,dt + r_t\widetilde{P}\,dt + \sigma_P\widetilde{P}\,dW_t^{\mathbb{Q}} \]

\[ = \sigma_P\widetilde{P}\,dW_t^{\mathbb{Q}} \]

The \(dt\) term vanishes, confirming that \(e^{-\int_0^t r_s\,ds}P(t,T)\) is a \(\mathbb{Q}\)-martingale.

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Exercise 6. Suppose \(\theta\) is not constant but depends on the current stock price: \(\theta_t = (\mu(S_t) - r) / \sigma(S_t)\). Write the Radon-Nikodym derivative \(Z_T\) in integral form and state the Novikov condition that must hold. Explain why this condition may fail for certain choices of \(\mu(\cdot)\) and \(\sigma(\cdot)\).

Solution to Exercise 6

When \(\theta_t = (\mu(S_t) - r)/\sigma(S_t)\) depends on the stock price, the Radon-Nikodym derivative in integral form is:

\[ Z_T = \exp\!\left(-\int_0^T \theta_t\,dW_t^{\mathbb{P}} - \frac{1}{2}\int_0^T \theta_t^2\,dt\right) \]

\[ = \exp\!\left(-\int_0^T \frac{\mu(S_t) - r}{\sigma(S_t)}\,dW_t^{\mathbb{P}} - \frac{1}{2}\int_0^T \left(\frac{\mu(S_t) - r}{\sigma(S_t)}\right)^2 dt\right) \]

The Novikov condition requires:

\[ \mathbb{E}^{\mathbb{P}}\!\left[\exp\!\left(\frac{1}{2}\int_0^T \theta_t^2\,dt\right)\right] = \mathbb{E}^{\mathbb{P}}\!\left[\exp\!\left(\frac{1}{2}\int_0^T \left(\frac{\mu(S_t) - r}{\sigma(S_t)}\right)^2 dt\right)\right] < \infty \]

This condition may fail in several scenarios:

• If \(\sigma(S_t) \to 0\) as \(S_t\) approaches some value (e.g., in the CEV model \(\sigma(S) = \sigma_0 S^{\beta-1}\) with \(\beta < 1\), the volatility vanishes as \(S \to 0\)), then \(\theta_t = (\mu(S_t) - r)/\sigma(S_t) \to \infty\), making the integral \(\int_0^T \theta_t^2\,dt\) potentially infinite.
• If \(\mu(S_t)\) grows without bound (e.g., superlinear drift), \(\theta_t\) can become unbounded.
• In stochastic volatility models where \(\sigma(S_t)\) can become very small with positive probability, the exponential moment may diverge.

When Novikov fails, \(Z_t\) is only a local martingale, and the measure change may not define a valid probability measure.

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Exercise 7. For the density process \(Z_t = \exp(-\theta W_t^{\mathbb{P}} - \frac{1}{2}\theta^2 t)\) with constant \(\theta = 0.3\) and \(T = 2\), compute \(\mathbb{E}^{\mathbb{P}}[Z_T]\) and verify it equals 1. Then compute \(Z_T\) for a specific path where \(W_T^{\mathbb{P}} = 1.5\) and interpret the result: is this path upweighted or downweighted under \(\mathbb{Q}\)?

Solution to Exercise 7

With \(\theta = 0.3\) and \(T = 2\):

\[ Z_T = \exp\!\left(-0.3\,W_T^{\mathbb{P}} - \frac{1}{2}(0.3)^2 \cdot 2\right) = \exp\!\left(-0.3\,W_T^{\mathbb{P}} - 0.09\right) \]

Computing \(\mathbb{E}^{\mathbb{P}}[Z_T]\): Since \(W_T^{\mathbb{P}} \sim \mathcal{N}(0, T) = \mathcal{N}(0, 2)\) under \(\mathbb{P}\), the exponent \(-0.3\,W_T^{\mathbb{P}} - 0.09\) is Gaussian with mean \(-0.09\) and variance \((0.3)^2 \cdot 2 = 0.18\). Using the lognormal moment formula:

\[ \mathbb{E}^{\mathbb{P}}[Z_T] = \exp\!\left(-0.09 + \frac{0.18}{2}\right) = \exp(-0.09 + 0.09) = e^0 = 1 \quad \checkmark \]

Specific path with \(W_T^{\mathbb{P}} = 1.5\):

\[ Z_T = \exp(-0.3 \times 1.5 - 0.09) = \exp(-0.45 - 0.09) = \exp(-0.54) \approx 0.583 \]

Since \(Z_T \approx 0.583 < 1\), this path is downweighted under \(\mathbb{Q}\).

Interpretation: The path where \(W_T^{\mathbb{P}} = 1.5\) is a path where the Brownian motion drifted upward (1.5 standard deviations above zero at \(T = 2\), since \(\sqrt{T} = \sqrt{2} \approx 1.41\)). Since \(\theta > 0\) (meaning \(\mu > r\)), the Girsanov measure change needs to reduce the apparent upward drift. It does so by assigning lower probability to upward-trending paths and higher probability to downward-trending paths. A path with positive \(W_T^{\mathbb{P}}\) gets a weight \(Z_T < 1\), meaning it is less likely under \(\mathbb{Q}\) than under \(\mathbb{P}\). Conversely, a path with \(W_T^{\mathbb{P}} = -1.5\) would give \(Z_T = \exp(0.45 - 0.09) = e^{0.36} \approx 1.433 > 1\), meaning it would be upweighted.