From a Pathwise Bound to a Change of Measure
A bounded drift integrand gives direct control of the quadratic energy that appears in Novikov’s condition. The key point is that the integrand may depend on the Brownian path and on the past: no deterministic-formula assumption is needed. Boundedness on a finite time interval alone provides the required exponential-integrability bound.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Fix a finite horizon \(T>0\). Throughout, stochastic integrals are Ito integrals: an integrand \(H\) is progressively measurable and belongs to \(\mathcal{L}^2\) on \([0,T]\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). In particular, the drift integrand \(\theta\) below will satisfy these requirements.
Here a true martingale means that each \(Z_t\) is integrable and \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s\) for \(s\le t\). The criterion is a standard result from the earlier material on Novikov’s condition. Once \(Z\) is a true martingale, \(\mathbb{E}[Z_T]=Z_0=1\), so it defines a probability measure \(\mathbb{Q}\) on \(\mathcal{F}_T\) by \(\mathbb{Q}(A)=\mathbb{E}[Z_T\mathbf{1}_A]\). The standard Girsanov theorem, established in the earlier material on change of measure, then says that \(B_t^{\mathbb{Q}}=B_t+\int_0^t\theta_s\,ds\) is Brownian motion under \(\mathbb{Q}\).
The Finite-Horizon Bound
Proof. The assumed bound gives \(\mathbb{E}\int_0^T\theta_s^2\,ds\le K^2T<\infty\), so \(\theta\in\mathcal{L}^2\). It also gives, almost surely, \(\frac12\int_0^T\theta_s^2\,ds\le\frac12K^2T\). Exponentiating and taking expectations yields \(\mathbb{E}\exp(\frac12\int_0^T\theta_s^2\,ds)\le\exp(\frac12K^2T)<\infty\). Thus Novikov’s condition holds, and the standard Novikov criterion implies that \(Z\) is a true martingale. In particular, \(Z_T\) is integrable and \(\mathbb{E}[Z_T]=1\). Its exponential formula shows \(Z_T>0\) almost surely; hence \(\mathbb{Q}\) and \(\mathbb{P}\) are equivalent on \(\mathcal{F}_T\). Applying the standard Girsanov theorem with density \(\mathcal{E}(-\int\theta\,dB)\) gives the stated Brownian motion with the plus sign. \(\square\)
The theorem controls the energy \(\int\theta^2\,dt\), not the size of the stochastic integral \(\int\theta\,dB\) path by path. Its conclusion is a martingale statement about the exponential, obtained from Novikov’s criterion; it does not assert that the stochastic integral itself is bounded.
Proof. Integrating the assumed inequality gives \(\int_0^T\theta_s^2\,ds\le\int_0^T k(s)^2\,ds\) almost surely. Therefore \(\mathbb{E}\exp(\frac12\int_0^T\theta_s^2\,ds)\le\exp(\frac12\int_0^T k(s)^2\,ds)<\infty\). Also \(\mathbb{E}\int_0^T\theta_s^2\,ds\le\int_0^T k(s)^2\,ds<\infty\), so the integral is defined in the stated \(\mathcal{L}^2\) class. Novikov’s criterion and Girsanov’s theorem give the conclusions as in the theorem. \(\square\)
Worked Adapted Examples
Worked Example: A Drift Chosen by the Brownian Sign
Take \(T=2\) and define \(\theta_t=0\) for \(0\le t\le1\), and \(\theta_t=\mathbf{1}_{\{B_1>0\}}\) for \(1<t\le2\). The value at the single time \(t=1\) does not affect either integral. This process is progressively measurable, and \(|\theta_t|\le1\), so it belongs to \(\mathcal{L}^2\) and satisfies Novikov’s condition with bound \(\mathbb{E}e^{\frac12\int_0^2\theta_s^2ds}\le e^{1/2}\).
Put \(I=\mathbf{1}_{\{B_1>0\}}\). The time integral of \(\theta^2\) is \(I\), and its stochastic integral is \(I(B_2-B_1)\). Thus the density is $$ Z_2=\exp\left(-I(B_2-B_1)-\frac12 I\right). $$
To check its expectation directly, condition on \(\mathcal{F}_1\). If \(I=0\), then \(Z_2=1\). If \(I=1\), then \(B_2-B_1\) is independent of \(\mathcal{F}_1\) and has distribution \(N(0,1)\). Its normal moment-generating function gives \(\mathbb{E}[e^{-(B_2-B_1)}\mid\mathcal{F}_1]=e^{1/2}\). Consequently \(\mathbb{E}[Z_2\mid\mathcal{F}_1]=e^{-1/2}e^{1/2}=1\) on \(\{I=1\}\), as well as on \(\{I=0\}\). Therefore \(\mathbb{E}[Z_2]=1\). The drift-adjusted Brownian motion is \(B_t^{\mathbb{Q}}=B_t+I(t-1)^+\), where \((t-1)^+=\max(t-1,0)\); it is Brownian under the measure defined by \(Z_2\). As a further check, \(Z_1=1\), so \(\mathbb{Q}(B_1>0)=\mathbb{E}[Z_1\mathbf{1}_{\{B_1>0\}}]=\mathbb{P}(B_1>0)=\frac12\).
Worked Example: Two Successive Adapted Drift Choices
On \([0,3]\), take \(\theta_t=\frac12\) for \(0\le t\le1\) and \(\theta_t=\mathbf{1}_{\{B_1\ge0\}}\) for \(1<t\le3\). The second choice depends only on information available at time \(1\), so the process is progressively measurable. It is bounded by \(1\), and therefore belongs to \(\mathcal{L}^2\). Write \(I=\mathbf{1}_{\{B_1\ge0\}}\) and \(\Delta=B_3-B_1\). Direct integration gives $$ Z_3=\exp\left(-\frac12B_1-\frac18-I\Delta-I\right). $$
Here \(\frac12\int_0^1(1/2)^2dt=\frac18\), and on \((1,3]\) the quadratic term is \(\frac12(2I)=I\). To check normalization, condition first on \(\mathcal{F}_1\). Since \(\Delta\sim N(0,2)\) independently of \(\mathcal{F}_1\), its moment-generating function gives \(\mathbb{E}[e^{-I\Delta-I}\mid\mathcal{F}_1]=e^{-I}e^{I^2}=1\): for \(I=0\) both factors are \(1\), and for \(I=1\) they are \(e^{-1}\) and \(e^1\). Next \(B_1\sim N(0,1)\), so \(\mathbb{E}[e^{-B_1/2-1/8}]=e^{-1/8}e^{(1/2)(1/2)^2}=1\). Thus \(\mathbb{E}[Z_3]=1\), agreeing with the martingale conclusion.
For \(0\le t\le3\), the adjusted process is \(B_t^{\mathbb{Q}}=B_t+\frac12(t\wedge1)+I(t-1)^+\). The first added term is the integral of the drift on the first interval; the second is its integral on the second interval. The sign is positive because the density uses \(-\int\theta\,dB\). Girsanov’s theorem says this process is Brownian under \(\mathbb{Q}\), not under the original measure \(\mathbb{P}\).
Worked Example: A Bounded State-Dependent Integrand
On \([0,1]\), let \(\theta_t=2\tanh(B_t)\). This is adapted and progressively measurable because \(B\) has continuous adapted paths and \(\tanh\) is continuous. Since \(|\tanh(x)|\le1\), we have \(|\theta_t|\le2\), \(\mathbb{E}\int_0^1\theta_s^2ds\le4\), and \(\mathbb{E}\exp(\frac12\int_0^1\theta_s^2ds)\le e^2\). Thus \(\theta\in\mathcal{L}^2\) and Novikov’s condition holds.
The resulting density and adjusted process are $$ Z_1=\exp\left(-2\int_0^1\tanh(B_s)\,dB_s-2\int_0^1\tanh^2(B_s)\,ds\right), \qquad B_t^{\mathbb{Q}}=B_t+2\int_0^t\tanh(B_s)\,ds. $$
The coefficient \(2\) in the second integral of the exponent is \(\frac12\cdot 2^2\), as required by the stochastic exponential. The state dependence prevents replacing these path integrals by a deterministic expression, but no such replacement is needed: the proved energy bound ensures that \(Z\) is a true martingale with expectation one, and Girsanov’s theorem gives the stated Brownian motion under \(\mathbb{Q}\).
Using the New Brownian Motion in an SDE
The sign convention has a direct implication for drift coefficients. Suppose, under \(\mathbb{P}\), a process satisfies \(dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\). This notation separates the drift coefficient \(\mu\) from the diffusion coefficient \(\sigma\). Under the measure \(\mathbb{Q}\) above, \(dB_t=dB_t^{\mathbb{Q}}-\theta_t\,dt\), so substitution gives \(dX_t=(\mu(t,X_t)-\sigma(t,X_t)\theta_t)\,dt+\sigma(t,X_t)\,dB_t^{\mathbb{Q}}\). This is an algebraic change of driving Brownian motion; it does not by itself establish existence or uniqueness of an SDE solution.
For a one-dimensional Brownian motion with original drift \(a\) and unit diffusion, choosing the constant integrand \(\theta=a\) makes the new drift \(a-\theta=0\). On any finite horizon this choice satisfies the bounded-integrand theorem. In a state-dependent diffusion, cancelling a drift may require dividing by the diffusion coefficient; boundedness and integrability of the resulting integrand then need to be checked separately.
Common Sign Pitfall
A common error is to use \(+\int\theta\,dB\) in the density while keeping the plus sign in \(B^{\mathbb{Q}}=B+\int\theta\,dt\). For a concrete check, take \(\theta_t=1\) on \([0,1]\). The correct density is \(Z_1=e^{-B_1-1/2}\). Since \(B_1\sim N(0,1)\), its density-weighted mean is \(\mathbb{E}[Z_1B_1]=-1\); hence \(B_1+1\) has mean zero under \(\mathbb{Q}\), consistent with the Girsanov convention.
With the incorrect density \(e^{B_1-1/2}\), the same normal calculation gives \(\mathbb{E}[e^{B_1-1/2}B_1]=1\). Then \(B_1+1\) has mean \(2\), not zero, so it cannot be the claimed centered Brownian value. The positive-sign density corresponds instead to the opposite adjustment \(B_t-\int_0^t\theta_s\,ds\). Checking a one-time mean is a useful diagnostic, although the full Brownian conclusion comes from Girsanov’s theorem.
Check Your Understanding
Assume all integrands below are progressively measurable, and interpret the bounds up to equality \(dt\,d\mathbb{P}\)-almost everywhere.
- If \(|\theta_t|\le3\) on \([0,2]\), give an explicit upper bound for \(\mathbb{E}\exp(\frac12\int_0^2\theta_s^2\,ds)\).
- For the first worked example, compute \(\mathbb{E}[Z_2\mid\mathcal{F}_1]\) separately on \(\{B_1>0\}\) and \(\{B_1\le0\}\).
- If \(|\theta_t|\le k(t)\) and \(\int_0^T k(t)^2dt\) is finite, explain why both the \(\mathcal{L}^2\) condition and Novikov’s condition follow.
- For density \(\mathcal{E}(-\int\theta\,dB)\), write \(dB_t\) in terms of \(dB_t^{\mathbb{Q}}\) and \(dt\), then identify the changed drift in an SDE.
- In the sign-pitfall example, calculate the density-weighted mean of \(B_1\) under the positive-sign density and determine which adjustment centers it.