Girsanov Change of Measure in the Black-Scholes Model

From the Physical Drift to the Risk-Neutral Drift

In the Black–Scholes model, the stock’s drift under the physical probability measure describes its expected growth in that model. It need not equal the interest rate. A change of probability measure can alter the drift assigned to the Brownian driver while leaving the stock’s volatility unchanged. The key is to choose a positive density process whose weighting of Brownian paths is consistent over time.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\), and fix a finite horizon \(T>0\). Assume the Black–Scholes stock satisfies \(dS_t=\mu S_t\,dt+\sigma S_t\,dB_t\), with \(S_0>0\), constant \(\mu,r\in\mathbb{R}\), and constant volatility \(\sigma>0\). Here \(\mu\) is the drift under \(\mathbb{P}\), \(\sigma\) is the diffusion coefficient, and \(r\) is the constant interest rate. The strict assumption \(\sigma>0\) makes the market price of risk below well-defined.

The negative sign in the exponent fixes the convention used here: the Brownian motion under \(\mathbb{Q}\) will be \(B_t^{\mathbb{Q}}=B_t+\theta t\). All stochastic integrals below are Itô integrals. Their integrands are adapted and progressively measurable, with the stated square-integrability condition whenever an \(L^2\) integral is used.

Why the Exponential Is a Valid Density

For a density process, expectation one at the terminal time is essential; the stronger conditional property \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s\) for \(s\le t\) ensures that the densities at different times are consistent. Here both facts can be checked directly using independent Brownian increments. For a constant integrand, the integral \(\int_0^t\theta\,dB_s\) is defined with an adapted, progressively measurable integrand satisfying \(\mathbb{E}\int_0^t\theta^2\,ds=\theta^2t<\infty\), and equals \(\theta B_t\).

Proof. Fix \(0\le s\le t\le T\), and put \(h=t-s\). Brownian increments are independent of \(\mathcal{F}_s\), with \(B_t-B_s\sim N(0,h)\). The density factors as \(Z_t=Z_s\exp(-\theta(B_t-B_s)-\tfrac12\theta^2h)\). The normal moment-generating function gives \(\mathbb{E}[\exp(-\theta(B_t-B_s))]=\exp(\tfrac12\theta^2h)\). Therefore \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s\). In particular, taking \(s=0\) gives \(\mathbb{E}[Z_t]=Z_0=1\). Since \(Z_t\) is finite and strictly positive, \(Z_T\) integrates to one and is positive almost surely; hence the measure it defines is a probability measure equivalent to \(\mathbb{P}\). Integrability follows from positivity and \(\mathbb{E}[Z_t]=1\), so this is a true martingale, not merely a local martingale. \(\square\)

For comparison with a common sufficient criterion, Novikov’s condition for an integrand \(\theta_s\) on \([0,T]\) is \(\mathbb{E}[\exp(\tfrac12\int_0^T\theta_s^2\,ds)]<\infty\). It is a standard sufficient condition for the associated stochastic exponential to be a true martingale; its general proof involves technical localization and is not given here. In this constant case the condition is verified explicitly: \(\mathbb{E}[\exp(\tfrac12\int_0^T\theta^2\,ds)]=\exp(\tfrac12\theta^2T)<\infty\). The direct conditional-expectation argument above independently verifies the martingale property.

The Brownian Motion Under the New Measure

Define \(B_t^{\mathbb{Q}}=B_t+\theta t\) for \(0\le t\le T\). The measure change does not make \(B_t\) itself a Brownian motion under \(\mathbb{Q}\); it is the shifted process \(B_t^{\mathbb{Q}}\) that is Brownian. The next argument verifies this directly for constant \(\theta\), rather than relying on the general Girsanov theorem.

Proof. Its paths are continuous and start at zero. To identify its increments, fix \(0\le s<t\le T\), write \(h=t-s\) and \(\Delta B=B_t-B_s\), and let \(g\) be any bounded measurable function. For an event \(A\in\mathcal{F}_s\), the density martingale property gives \(\mathbb{E}_{\mathbb{P}}[Z_T\mid\mathcal{F}_t]=Z_t\). Consequently, \(\mathbb{E}_{\mathbb{Q}}[\mathbf{1}_A g(\Delta B)] =\mathbb{E}_{\mathbb{P}}[\mathbf{1}_A Z_t g(\Delta B)]\). Factor \(Z_t=Z_s\exp(-\theta\Delta B-\tfrac12\theta^2h)\), then condition on \(\mathcal{F}_s\). Under \(\mathbb{P}\), \(\Delta B\) is independent of \(\mathcal{F}_s\) and has density \(\phi_h(x)=(2\pi h)^{-1/2}\exp(-x^2/(2h))\). Weighting this density by the increment factor gives \[ \phi_h(x)e^{-\theta x-\theta^2h/2} =(2\pi h)^{-1/2}\exp\left(-\frac{(x+\theta h)^2}{2h}\right). \] This is the normal density with mean \(-\theta h\) and variance \(h\), and integrates to one. The calculation with \(A\) shows that the conditional law of \(\Delta B\) given \(\mathcal{F}_s\) under \(\mathbb{Q}\) is this same law, independent of \(\mathcal{F}_s\). Therefore \(\Delta B+\theta h=B_t^{\mathbb{Q}}-B_s^{\mathbb{Q}}\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,h)\). Together with continuity, these increment properties establish that \(B^{\mathbb{Q}}\) is Brownian under \(\mathbb{Q}\). \(\square\)

Transforming the Stock Dynamics

By definition, \(dB_t^{\mathbb{Q}}=dB_t+\theta\,dt\), so \(dB_t=dB_t^{\mathbb{Q}}-\theta\,dt\). Substituting into the stock equation gives \(dS_t=(\mu-\sigma\theta)S_t\,dt+\sigma S_t\,dB_t^{\mathbb{Q}}\). The choice \(\theta=(\mu-r)/\sigma\) makes \(\mu-\sigma\theta=r\). Thus the stock drift under \(\mathbb{Q}\) is \(r\), while its diffusion coefficient remains \(\sigma S_t\).

Proof. The drift identity follows from substitution and \(\mu-\sigma\theta=\mu-(\mu-r)=r\). To verify the discounted-martingale statement, the explicit stock solution under \(\mathbb{Q}\) is \(S_t=S_0\exp((r-\tfrac12\sigma^2)t+\sigma B_t^{\mathbb{Q}})\). This formula is also obtained by applying Itô’s formula to the exponential function: for \(f(t,x)=S_0\exp((r-\tfrac12\sigma^2)t+\sigma x)\), \(f_t=(r-\tfrac12\sigma^2)f\), \(f_x=\sigma f\), and \(f_{xx}=\sigma^2f\). The Itô correction is \(\tfrac12 f_{xx}\,dt=\tfrac12\sigma^2f\,dt\); hence the total \(dt\) coefficient is \(rf\), giving \(dS_t=rS_t\,dt+\sigma S_t\,dB_t^{\mathbb{Q}}\).

The stochastic-integral integrand \(\sigma S_t\) is adapted and progressively measurable, and it is square-integrable under \(\mathbb{Q}\) on every finite horizon because \(\mathbb{E}^{\mathbb{Q}}[S_t^2]=S_0^2e^{(2r+\sigma^2)t}\), whose integral over \([0,T]\) is finite. For \(0\le s\le t\le T\), the explicit solution and the independent \(N(0,t-s)\) increment of \(B^{\mathbb{Q}}\) give \(\mathbb{E}^{\mathbb{Q}}[S_t\mid\mathcal{F}_s] =S_s e^{(r-\sigma^2/2)(t-s)}e^{\sigma^2(t-s)/2} =S_s e^{r(t-s)}\). The conditional expectation is finite, so \(\mathbb{E}^{\mathbb{Q}}[e^{-rt}S_t\mid\mathcal{F}_s]=e^{-rs}S_s\). This proves the true-martingale claim, including its integrability; no step relies only on a local-martingale property. \(\square\)

Common Pitfalls and Scope

Omitting the quadratic correction destroys the density. A tempting but incorrect choice is \(\widetilde Z_t=e^{-\theta B_t}\). For \(\theta\ne0\), its expectation is \(\mathbb{E}[\widetilde Z_t]=e^{\theta^2t/2}>1\) when \(t>0\), so it cannot be a probability density. The term \(-\tfrac12\theta^2t\) makes the expectation equal to one.

The sign determines the drift shift. With the convention \(B_t^{\mathbb{Q}}=B_t+\theta t\), substitution gives \(\mu-\sigma\theta\), not \(\mu+\sigma\theta\). Using the opposite sign in the density requires changing the definition of the shifted Brownian motion as well. Under the convention used here, the chosen \(\theta\) makes the drift exactly \(r\).

The measure \(\mathbb{P}\) describes the model before the change of measure; \(\mathbb{Q}\) is the equivalent risk-neutral measure on the specified finite horizon. The construction assumes constant coefficients and \(\sigma>0\). If \(\sigma=0\), the ratio defining \(\theta\) is unavailable: the stock has no Brownian risk to tilt, and a deterministic drift different from \(r\) cannot be changed by weighting Brownian paths. This tutorial establishes the measure change and the resulting stock dynamics; it does not derive prices of particular claims.