Constant Drift Changes for Brownian Motion

From Gaussian Tilting to Constant Drift

Changing measure can turn a standard Brownian motion into one with a constant drift. The change is encoded by an exponential density: paths with larger terminal Brownian value receive more weight when the desired drift is positive, and less weight when it is negative. The key check is that this weighting changes every increment over an interval of length \(h\) from a centered normal variable to one with mean \(\theta h\), without changing its variance \(h\).

Fix a finite horizon \(T>0\). Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{0\le t\le T},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) with respect to this filtration. In particular, \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\) for \(0\le s<t\le T\). The constant process \(H_s=\theta\) is adapted and progressively measurable, and belongs to \(\mathcal{L}^2\), since \(\mathbb{E}\int_0^T H_s^2\,ds=\theta^2T<\infty\). Its Ito integral is \(\int_0^t\theta\,dB_s=\theta B_t\).

The density is strictly positive. Its expectation is one, so \(\mathbb{Q}\) is a probability measure equivalent to \(\mathbb{P}\) on \(\mathcal{F}_T\). Here equivalence means that the two measures have the same null events. These assertions, and the Brownian property under \(\mathbb{Q}\), are established below rather than inferred just from the exponential form.

The Density and the Brownian Motion Under the New Measure

Proof. The normal moment-generating formula gives \(\mathbb{E}[\exp(\theta(B_t-B_s))]=\exp(\tfrac12\theta^2(t-s))\). Since the increment is independent of \(\mathcal{F}_s\),

The same normal formula gives \(\mathbb{E}[Z_t]=1\), and \(Z_t\) is integrable because it is nonnegative with finite expectation. Thus this is a true martingale, not merely a local martingale. In particular \(Z_T\) is a probability density; since \(Z_T>0\) almost surely, \(\mathbb{Q}\) and \(\mathbb{P}\) are equivalent.

It remains to verify the Brownian increments under \(\mathbb{Q}\). For \(0\le s<t\le T\), write \(h=t-s\) and \(Y=B_t-B_s\). For any bounded measurable function \(g\), conditional expectation under the density gives

To see why no future part of \(Z_T\) remains in this expression, factor \(Z_T=Z_s e^{\theta Y-\frac12\theta^2h}e^{\theta(B_T-B_t)-\frac12\theta^2(T-t)}\). The last factor has conditional expectation one given \(\mathcal{F}_t\), by the same normal calculation. Also \(\mathbb{E}^{\mathbb{Q}}[g(Y)\mid\mathcal{F}_s]\) is obtained by dividing the conditional weighted expectation by \(\mathbb{E}^{\mathbb{P}}[Z_T\mid\mathcal{F}_s]=Z_s\). This yields the displayed formula.

Under \(\mathbb{P}\), \(Y\) has density \((2\pi h)^{-1/2}e^{-y^2/(2h)}\). Multiplying this density by \(e^{\theta y-\frac12\theta^2h}\) gives

because \(-y^2/(2h)+\theta y-\theta^2h/2=-(y-\theta h)^2/(2h)\). Thus the conditional law of \(Y\) under \(\mathbb{Q}\) is \(N(\theta h,h)\), and it does not depend on \(\mathcal{F}_s\). Therefore \(Y-\theta h=B_t^{\mathbb{Q}}-B_s^{\mathbb{Q}}\) is independent of \(\mathcal{F}_s\) under \(\mathbb{Q}\) and is \(N(0,h)\). The process \(B^{\mathbb{Q}}\) starts at zero and has continuous paths, so these increment properties make it a standard Brownian motion under \(\mathbb{Q}\). Equivalently, \(dB_t=\theta\,dt+dB_t^{\mathbb{Q}}\): the drift is \(\theta\), and the diffusion coefficient remains \(1\). \(\square\)

The change of measure does not alter the pathwise identity \(B_t=B_t^{\mathbb{Q}}+\theta t\); it changes which measure determines the distribution of the random path. The process \(B^{\mathbb{Q}}\) is a true \(\mathbb{Q}\)-martingale: it is integrable at each time, and its centered increments are independent of the past under \(\mathbb{Q}\), so \(\mathbb{E}^{\mathbb{Q}}[B_t^{\mathbb{Q}}\mid\mathcal{F}_s]=B_s^{\mathbb{Q}}\). In contrast, when \(\theta\ne0\), \(\mathbb{E}^{\mathbb{Q}}[B_t]=\theta t\), so \(B\) is not a \(\mathbb{Q}\)-martingale.

Gaussian Transition Densities

Proof. Set \(h=t-s\). Under \(\mathbb{Q}\), the increment \(B_t-B_s\) has density \(N(\theta h,h)\) by the theorem. Substituting \(y-x\) for the increment gives the first formula. Under \(\mathbb{P}\), the same increment has density \(N(0,h)\), so its density at \(y-x\) is \(p_{\mathbb{P}}(s,x;t,y)\). Multiplying by the exponential factor and completing the square gives the \(N(\theta h,h)\) density, as in the theorem’s proof. This proves both expressions and their equality. \(\square\)

Sign, Scope, and a Common Mistake

The sign in the exponential determines the sign of the drift. With \(Z_T=\exp(\theta B_T-\tfrac12\theta^2T)\), the density tilt shifts each increment’s mean by \(+\theta(t-s)\). A negative value of \(\theta\) shifts the mean down. The variance of an increment remains \(t-s\); only its mean changes.

Common mistake: reversing the sign. For \(\theta=1\), writing \(e^{-B_T-\frac12T}\) and claiming that it creates drift \(+1\) is incorrect. The exponent’s linear term is \(-B_T\), so completing the square gives \(-x^2/(2T)-x-T/2=-(x+T)^2/(2T)\). The resulting terminal distribution is \(N(-T,T)\), with drift \(-1\), not \(+1\). To create drift \(+1\), the correct density is \(e^{B_T-\frac12T}\), for which \(-x^2/(2T)+x-T/2=-(x-T)^2/(2T)\).

The construction here is for a fixed finite horizon and a constant drift. It gives an explicit density and verifies the changed law directly; it does not assert that every change of measure is valid on an infinite time horizon. If \(\theta=0\), then \(Z_t=1\), \(\mathbb{Q}=\mathbb{P}\), and the drift remains zero. In all cases the exponential is strictly positive, so the two measures on \(\mathcal{F}_T\) are equivalent.