Girsanov for SDEs

Proof.

Step 1 (Pathwise rearrangement).
Wherever the hypothesis identity holds,

\[ \mathbf{Y}_t - \int_0^t b(\mathbf{Y}_s)\, ds = x + \int_0^t \boldsymbol{\gamma}\, ds + \mathbf{w}_t = x + \hat{\mathbf{w}}_t , \]

which is (1). No measure has been mentioned yet. This is algebra of paths.

Step 2 (The new noise and its filtration).
The Girsanov theorem applied to \(\boldsymbol{\gamma}\) makes \(\mathbb{Q}\) a probability measure with the same null sets as \(\mathbb{P}\), and makes \(\hat{\mathbf{w}}\) an \(n\)-dimensional Brownian motion relative to \(\{\mathcal{F}_t^{(n)}\}\) under \(\mathbb{Q}\), with each increment \(\hat{\mathbf{w}}_t - \hat{\mathbf{w}}_s\) independent of \(\mathcal{F}_s^{(n)}\). The filtration contains every \(\mathbb{P}\)-null set by construction, and every \(\mathbb{Q}\)-null set is \(\mathbb{P}\)-null by the equivalence, so it contains every \(\mathbb{Q}\)-null set. These are the two properties that the enlargement lemma supplies in the strong-solution setting, adaptedness over a filtration holding all null sets and independence of increments from the running past, now holding for \(\hat{\mathbf{w}}\) under \(\mathbb{Q}\).

Step 3 (The solution clauses under \(\mathbb{Q}\)).
We verify clauses (S1) through (S4) of the solution definition for \(\mathbf{Y}\), coordinate by coordinate in the systems convention, for the equation with drift \(b\) and unit noise coefficient, on the space \(\bigl( \Omega, \mathcal{F}_T^{(n)}, \mathbb{Q} \bigr)\), with initial value the constant \(x\), which is square integrable and independent of everything.

(S1) Adaptedness and path continuity are properties of the maps \(\omega \mapsto \mathbf{Y}_\cdot(\omega)\) alone and do not see the change of measure.

(S3) For every \(\omega\), the map \(s \mapsto b(\mathbf{Y}_s)\) is continuous, being a composition of continuous maps, hence integrable on the compact \([0, T]\). The constant integrand \(1\) is measurable and adapted with \(\mathbb{E}_{\mathbb{Q}}\bigl[ \int_0^T 1\, ds \bigr] = T \lt \infty\), so it lies in \(\mathcal{V}(0, T)\) read over the \(\mathbb{Q}\)-space.

(S4) The constant integrand is an elementary process, and on elementary processes the integral is defined as the telescoping sum of increments, so \(\int_0^t d\hat{w}^{(i)}_s = \hat{w}^{(i)}_t\) for each coordinate. Clause (S4) is then Step 1's identity, which holds almost surely simultaneously for every \(t\).

(S2) Lipschitz continuity gives linear growth, \(|b(\mathbf{y})| \leq |b(\mathbf{0})| + D\, |\mathbf{y}|\). Fix \(\omega\) on the full event of Step 1 and write \(\hat{W}^* = \sum_{i=1}^{n} \sup_{s \leq T} |\hat{w}^{(i)}_s|\), so that \(|\hat{\mathbf{w}}_t| \leq \hat{W}^*\) for all \(t\). From (1),

\[ |\mathbf{Y}_t| \leq \bigl( |x| + |b(\mathbf{0})|\, T + \hat{W}^* \bigr) + D \int_0^t |\mathbf{Y}_s|\, ds , \]

and \(t \mapsto |\mathbf{Y}_t(\omega)|\) is continuous, hence measurable with finite integral over \([0, T]\), so Gronwall's inequality, applied path by path with the additive constant in parentheses and the multiplier \(D\), gives \(|\mathbf{Y}_t| \leq \bigl( |x| + |b(\mathbf{0})|\, T + \hat{W}^* \bigr)\, e^{D T}\). Squaring, integrating in time, and using \((u + v + z)^2 \leq 3 (u^2 + v^2 + z^2)\),

\[ \mathbb{E}_{\mathbb{Q}}\Bigl[ \int_0^T |\mathbf{Y}_t|^2\, dt \Bigr] \leq 3\, T\, e^{2 D T} \Bigl( |x|^2 + |b(\mathbf{0})|^2 T^2 + \mathbb{E}_{\mathbb{Q}}\bigl[ (\hat{W}^*)^2 \bigr] \Bigr) , \]

and it remains to bound the last expectation. Each coordinate \(\hat{w}^{(i)}\) is a \(\mathbb{Q}\)-martingale with respect to \(\{\mathcal{F}_t^{(n)}\}\). It is adapted, each \(\hat{w}^{(i)}_t\) is integrable because its \(\mathbb{Q}\)-law is that of a Brownian coordinate, and for \(s \leq t\) the increment is independent of \(\mathcal{F}_s^{(n)}\) by Step 2, so the independence collapse gives \(\mathbb{E}_{\mathbb{Q}}\bigl[ \hat{w}^{(i)}_t - \hat{w}^{(i)}_s \mid \mathcal{F}_s^{(n)} \bigr] = \mathbb{E}_{\mathbb{Q}}\bigl[ \hat{w}^{(i)}_t - \hat{w}^{(i)}_s \bigr] = 0\), the mean vanishing because the law of the increment is centered Gaussian. Paths are continuous, so Doob's martingale inequality with \(p = 4\), read on the space \(\bigl( \Omega, \mathcal{F}_T^{(n)}, \mathbb{Q} \bigr)\) and with the dyadic supremum agreeing with the full supremum along continuous paths, gives

\[ \mathbb{Q}\Bigl[ \sup_{s \leq T} |\hat{w}^{(i)}_s| \geq \lambda \Bigr] \leq \frac{1}{\lambda^4}\, \mathbb{E}_{\mathbb{Q}}\bigl[ (\hat{w}^{(i)}_T)^4 \bigr] = \frac{3 T^2}{\lambda^4} , \]

the fourth moment being that of a centered Gaussian with variance \(T\) by the law identification. Writing \(\varphi_T\) for its density, the relation \(z\, \varphi_T(z) = -T\, \varphi_T'(z)\) and one integration by parts give \(\mathbb{E}_{\mathbb{Q}}[(\hat{w}^{(i)}_T)^4] = 3\, T\, \mathbb{E}_{\mathbb{Q}}[(\hat{w}^{(i)}_T)^2] = 3\, T^2\). The layer-cake identity \(S^2 = \int_0^\infty 2 \lambda\, \mathbf{1}_{\{\lambda \lt S\}}\, d\lambda\) and Tonelli's theorem convert the tail bound into a moment bound. Splitting the integral at \(\lambda = (3 T^2)^{1/4}\), where the two estimates \(\mathbb{Q}[\cdot] \leq 1\) and the display cross, we obtain

\[ \mathbb{E}_{\mathbb{Q}}\Bigl[ \sup_{s \leq T} |\hat{w}^{(i)}_s|^2 \Bigr] = \int_0^\infty 2 \lambda\, \mathbb{Q}\Bigl[ \sup_{s \leq T} |\hat{w}^{(i)}_s| \gt \lambda \Bigr]\, d\lambda \leq \sqrt{3}\, T + \sqrt{3}\, T = 2 \sqrt{3}\, T . \]

Then \((\hat{W}^*)^2 \leq n \sum_i \sup_{s \leq T} |\hat{w}^{(i)}_s|^2\) by the Cauchy-Schwarz inequality on the defining sum, so \(\mathbb{E}_{\mathbb{Q}}[(\hat{W}^*)^2] \leq 2 \sqrt{3}\, n^2 T \lt \infty\), and (S2) holds. Together with Step 2, all clauses of the weak solution definition are met, with initial law \(\delta_x\). This proves (2).

Step 4 (Uniqueness in law).
The strong solution \(X^x\) on the original space, driven by \(\mathbf{w}\) under \(\mathbb{P}\) over the enlarged filtration of a constant start, is a solution of the same equation with the same initial law \(\delta_x\). The coefficients satisfy (H1) and (H2). The constant noise coefficient trivially, and the drift because Lipschitz continuity supplies both the Lipschitz clause and, through \(|b(\mathbf{y})| \leq |b(\mathbf{0})| + D |\mathbf{y}|\), linear growth. The initial law \(\delta_x\) has finite second moment. The uniqueness in law lemma, read in the same systems convention, therefore applies to the pair of solutions \((X^x, \mathbf{w})\) under \(\mathbb{P}\) and \((\mathbf{Y}, \hat{\mathbf{w}})\) under \(\mathbb{Q}\). That lemma's proof is a sketch whose single unproved step is the induction carrying laws through the Picard iterates, and this page's reliance on it inherits exactly that one point of trust and no other. The conclusion is that the two solutions share their finite-dimensional distributions, which is (3).

Proof.

Step 1 (The tilt).
Set \(\mathbf{Y}_t = x + \mathbf{w}_t\) and \(\boldsymbol{\gamma}(s, \omega) = - a\bigl( \mathbf{Y}_s(\omega) \bigr)\). Each coordinate of \(\mathbf{Y}\) is adapted with continuous paths, hence progressively measurable by the continuity criterion, so the vector map \((s, \omega) \mapsto \mathbf{Y}_s(\omega)\) is jointly measurable on each strip \([0, t] \times \Omega\), and composing with the Borel map \(a\) preserves that measurability. Each coordinate of \(\boldsymbol{\gamma}\) is therefore progressively measurable, and \(|\boldsymbol{\gamma}| \leq C\). Let \(M\) be the exponential martingale of \(\boldsymbol{\gamma}\), let \(\mathbb{Q}\) be its measure from the Girsanov theorem, and set

\[ \hat{\mathbf{w}}_t = \int_0^t \boldsymbol{\gamma}(s, \omega)\, ds + \mathbf{w}_t = \mathbf{w}_t - \int_0^t a( x + \mathbf{w}_s )\, ds . \]

The theorem makes \(\mathbb{Q}\) a probability measure with the same null sets as \(\mathbb{P}\), and makes \(\hat{\mathbf{w}}\) an \(n\)-dimensional Brownian motion relative to \(\{\mathcal{F}_t^{(n)}\}\) under \(\mathbb{Q}\), its increments independent of the running past. As in Step 2 of the preceding proof, the filtration holds every \(\mathbb{Q}\)-null set, so the two properties of the enlargement lemma are in place.

Step 2 (The equation holds pathwise).
For every \(\omega\) and every \(t \in [0, T]\),

\[ x + \int_0^t a( \mathbf{Y}_s )\, ds + \hat{\mathbf{w}}_t = x + \int_0^t a( \mathbf{Y}_s )\, ds - \int_0^t a( \mathbf{Y}_s )\, ds + \mathbf{w}_t = \mathbf{Y}_t , \]

an identity of paths, holding surely.

Step 3 (The solution clauses under \(\mathbb{Q}\)).
The clauses of the solution definition are read coordinate by coordinate in the systems convention, as before. (S1) holds because adaptedness and continuity do not see the measure. For (S3), the path \(s \mapsto a(\mathbf{Y}_s(\omega))\) is Borel measurable for each \(\omega\), a Borel map composed with a continuous one, and is bounded by \(C\), so its time integral is finite surely, while the constant noise integrand lies in \(\mathcal{V}(0, T)\) over the \(\mathbb{Q}\)-space exactly as before. For (S4), the integral of the constant integrand against \(\hat{\mathbf{w}}\) telescopes to \(\hat{\mathbf{w}}_t\), and the clause is Step 2's identity. For (S2), Step 2 gives \(|\mathbf{Y}_t| \leq |x| + C T + \hat{W}^*\) with \(\hat{W}^* = \sum_i \sup_{s \leq T} |\hat{w}^{(i)}_s|\), and the bound \(\mathbb{E}_{\mathbb{Q}}\bigl[ (\hat{W}^*)^2 \bigr] \leq 2 \sqrt{3}\, n^2 T\) established in clause (S2) of the preceding proof applies verbatim, its only inputs being the Brownian clauses of Step 1. Hence

\[ \mathbb{E}_{\mathbb{Q}}\Bigl[ \int_0^T |\mathbf{Y}_t|^2\, dt \Bigr] \leq 3\, T \Bigl( |x|^2 + C^2 T^2 + 2 \sqrt{3}\, n^2 T \Bigr) \lt \infty . \]

The initial value is the constant \(x\), square integrable, independent of everything, with law \(\delta_x\). All clauses of the weak solution definition are met by the pair \((\mathbf{Y}, \hat{\mathbf{w}})\) on \(\bigl( \Omega, \mathcal{F}_T^{(n)}, \mathbb{Q} \bigr)\).