The Girsanov Theorem

Proof.

Step 0 (Integrals against \(\nu\) and against \(f\, d\mu\) agree).
For a non-negative measurable \(g\), the identity \(\int g\, d\nu = \int g f\, d\mu\) holds: it is the hypothesis when \(g\) is an indicator, extends to simple functions by linearity, and passes to arbitrary non-negative \(g\) by taking monotone limits of simple functions on both sides. Splitting an integrable \(g\) into positive and negative parts extends it to all of \(L^1(\nu)\). In particular \(\mathbb{E}_\mu[f |X|] = \mathbb{E}_\nu[|X|] \lt \infty\), so \(\mathbb{E}_\mu[f X \mid \mathcal{H}]\) is defined.

Step 1 (The left side satisfies the defining property of the right side).
Write \(W = \mathbb{E}_\nu[X \mid \mathcal{H}] \cdot \mathbb{E}_\mu[f \mid \mathcal{H}]\), a product of \(\mathcal{H}\)-measurable random variables and hence \(\mathcal{H}\)-measurable. Fix \(H \in \mathcal{H}\). The random variable \(Z = \mathbf{1}_H\, \mathbb{E}_\nu[X \mid \mathcal{H}]\) is \(\mathcal{H}\)-measurable, and \(Z f \in L^1(\mu)\), because conditional Jensen applied under \(\nu\) with the convex function \(|\cdot|\), followed by the tower property under \(\nu\), gives

\[ \mathbb{E}_\mu\bigl[ |Z| f \bigr] \leq \mathbb{E}_\nu\bigl[ \bigl| \mathbb{E}_\nu[X \mid \mathcal{H}] \bigr| \bigr] \leq \mathbb{E}_\nu\bigl[ \mathbb{E}_\nu[\, |X| \mid \mathcal{H}] \bigr] = \mathbb{E}_\nu[|X|] \lt \infty , \]

where the first inequality is Step 0 applied to \(|Z| = |Z| \cdot 1\). By take-out under \(\mu\), \(\mathbb{E}_\mu[Z f \mid \mathcal{H}] = Z\, \mathbb{E}_\mu[f \mid \mathcal{H}] = \mathbf{1}_H W\), and taking total expectations,

\[ \begin{align*} \int_H W\, d\mu &= \mathbb{E}_\mu\bigl[ \mathbb{E}_\mu[ Z f \mid \mathcal{H}] \bigr] = \mathbb{E}_\mu[ Z f ] = \int_H \mathbb{E}_\nu[X \mid \mathcal{H}]\, f\, d\mu \\\\ &= \int_H \mathbb{E}_\nu[X \mid \mathcal{H}]\, d\nu = \int_H X\, d\nu = \int_H X f\, d\mu = \int_H f X\, d\mu , \end{align*} \]

where the second line uses Step 0 twice and, in its middle equality, the defining property of \(\mathbb{E}_\nu[X \mid \mathcal{H}]\) with the set \(H \in \mathcal{H}\). Running the same chain with \(|X|\) in place of \(X\) shows that the \(\mathcal{H}\)-measurable random variable \(\mathbb{E}_\nu[\,|X| \mid \mathcal{H}] \cdot \mathbb{E}_\mu[f \mid \mathcal{H}]\) has \(\mu\)-integral \(\mathbb{E}_\nu[|X|] \lt \infty\), and it dominates \(|W|\) \(\mu\)-almost everywhere. Conditional Jensen under \(\nu\) gives the domination off a \(\nu\)-null exceptional set, which may be taken in \(\mathcal{H}\) since both sides are \(\mathcal{H}\)-measurable, and on that set the factor \(\mathbb{E}_\mu[f \mid \mathcal{H}]\), non-negative by monotonicity, integrates to the \(\nu\)-measure zero of the set and so vanishes \(\mu\)-a.e. there, making both sides zero. So \(W\) is \(\mu\)-integrable, \(\mathcal{H}\)-measurable, and integrates to \(\int_H f X\, d\mu\) over every \(H \in \mathcal{H}\). These are exactly the properties that characterize \(\mathbb{E}_\mu[f X \mid \mathcal{H}]\) up to \(\mu\)-null modification, and the identity follows.

Step 2 (The version does not matter).
Two versions of \(\mathbb{E}_\nu[X \mid \mathcal{H}]\) differ only on a \(\nu\)-null set \(H_0 \in \mathcal{H}\). On such a set, \(\int_{H_0} \mathbb{E}_\mu[f \mid \mathcal{H}]\, d\mu = \nu(H_0) = 0\) by the defining property, and since the integrand is non-negative it vanishes \(\mu\)-a.e. on \(H_0\). The product \(W\) is therefore unchanged \(\mu\)-a.s. when the version changes, and the same argument applies to versions of the two remaining conditional expectations, whose exceptional sets are already \(\mu\)-null.

Proof.

Step 1 (Restriction preserves absolute continuity).
Let \(H \in \mathcal{M}_t\) with \(\mathbb{P}(H) = 0\). Since \(\mathcal{M}_t \subseteq \mathcal{M}_T\), the set \(H\) lies in \(\mathcal{M}_T\), where absolute continuity holds, so \(\mathbb{Q}(H) = 0\). Probability measures are finite, hence \(\sigma\)-finite, so the Radon-Nikodym theorem applies on \((\Omega, \mathcal{M}_t)\) and delivers a non-negative \(\mathcal{M}_t\)-measurable density \(Z_t\), unique up to \(\mathbb{P}\)-null modification.

Step 2 (The martingale identity).
Fix \(0 \leq s \leq t \leq T\) and \(H \in \mathcal{M}_s\). Then \(H\) belongs to both \(\mathcal{M}_s\) and \(\mathcal{M}_t\), and each density computes the same number:

\[ \int_H Z_t\, d\mathbb{P} = \mathbb{Q}(H) = \int_H Z_s\, d\mathbb{P} . \]

The random variable \(Z_s\) is \(\mathcal{M}_s\)-measurable, integrable with \(\mathbb{E}_{\mathbb{P}}[Z_s] = \mathbb{Q}(\Omega) = 1\), and by the display it integrates like \(Z_t\) over every set in \(\mathcal{M}_s\). By the defining property of conditional expectation, \(Z_s = \mathbb{E}_{\mathbb{P}}[ Z_t \mid \mathcal{M}_s ]\) \(\mathbb{P}\)-a.s. This is condition (M3) of the martingale definition, adaptedness (M1) holds because each \(Z_t\) is \(\mathcal{M}_t\)-measurable by construction, and integrability (M2) was just computed. Taking \(t = T\) in the identity gives \(Z_s = \mathbb{E}_{\mathbb{P}}[ Z_T \mid \mathcal{M}_s ]\).

Proof.

Step 1 (Localization).
Boundedness and progressive measurability place each \(a_j\) and \(\tfrac{1}{2}|\mathbf{a}|^2\) in the coefficient classes of the vector Itô process definition with \(m = 1\), so \(V\) is a one-dimensional Itô process with everywhere-continuous paths. For \(R \gt 0\) let \(\tau_R\) be the first exit time of \(V\) from the open interval \((-R, R)\), a stopping time by the first exit theorem. On the compact interval \([0, T]\) every continuous path of \(V\) is bounded, so for each \(\omega\) there is a threshold beyond which \(\tau_R \geq T\). The stopped variable \(M_{t \wedge \tau_R}\) is eventually equal to \(M_t\) as \(R \to \infty\), for every \(\omega\) and every \(t \in [0, T]\). Note also that \(V_{s \wedge \tau_R} \in [-R, R]\) for all \(s\): before the exit this holds by definition, and at the exit it holds because continuity forces the exit position onto the boundary of the interval.

Step 2 (The stopped process has truncated coefficients).
By the stopped Itô integral theorem, applied on each horizon \(t \leq T\) and read component by component against each driving coordinate \(w^{(j)}\) with the filtration \(\{\mathcal{F}_s^{(n)}\}\), as the multi-dimensional integral prescribes, each \(\mathbf{1}_{[0, \tau_R)}\, a_j\) lies in \(\mathcal{V}(0, T)\) and

\[ V_{t \wedge \tau_R} = - \sum_{j=1}^{n} \int_0^t \mathbf{1}_{[0, \tau_R)}(s)\, a_j\, dw_s^{(j)} - \int_0^t \mathbf{1}_{[0, \tau_R)}(s)\, \tfrac{1}{2} |\mathbf{a}|^2\, ds , \]

the \(ds\)-term because stopping an ordinary time integral at \(t \wedge \tau_R\) is the same as integrating the truncated integrand, an identity verified path by path. So \(V^R := V_{\cdot \wedge \tau_R}\) is again an Itô process, with coefficients cut off at \(\tau_R\).

Step 3 (Itô's formula below the stop).
Apply the general Itô formula to \(V^R\) with \(g(t, x) = e^x\), which is smooth on \([0, \infty) \times \mathbb{R}\). The membership condition asks that \(e^{V^R_s} \bigl( -a_j\, \mathbf{1}_{[0, \tau_R)}(s) \bigr)\) lie in \(\mathcal{V}(0, T)\) for each \(j\), and it does. The process is progressively measurable, and Step 1 bounds it by \(C e^{R}\). The \(ds\)-integrand produced by the formula vanishes identically,

\[ e^{V^R_s} \Bigl( - \tfrac{1}{2} |\mathbf{a}|^2\, \mathbf{1}_{[0, \tau_R)} + \tfrac{1}{2} |\mathbf{a}|^2\, \mathbf{1}_{[0, \tau_R)} \Bigr) = 0 , \]

the first term from the drift of \(V^R\) and the second from half the squared row of its \(d\mathbf{w}\)-coefficients. Since \(e^{V^R_s}\, \mathbf{1}_{[0, \tau_R)}(s) = M_s\, \mathbf{1}_{[0, \tau_R)}(s)\), the formula collapses to

\[ M_{t \wedge \tau_R} = 1 - \sum_{j=1}^{n} \int_0^t \mathbf{1}_{[0, \tau_R)}(s)\, M_s\, a_j\, dw_s^{(j)} . \]

Each integrand is bounded, hence in \(\mathcal{V}(0, T)\), so each integral is a martingale by the martingale property of Itô integrals, and a finite sum of martingales is a martingale. Therefore \(M_{\cdot \wedge \tau_R}\) is a martingale with \(\mathbb{E}[M_{t \wedge \tau_R}] = 1\).

Step 4 (A second-moment bound, uniform in \(R\)).
Let \(N\) denote the exponential process built from the integrand \(2\mathbf{a}\) in place of \(\mathbf{a}\). Expanding the exponents,

\[ M_t^2 = e^{2 V_t} = N_t \cdot \exp\Bigl( \int_0^t |\mathbf{a}|^2\, ds \Bigr) \leq e^{C^2 T}\, N_t , \]

and, for \(s \leq \tau_R\), the same expansion gives \(N_s = \exp\bigl( 2 V_s - \int_0^s |\mathbf{a}|^2\, ds \bigr) \leq e^{2R}\). Steps 2 and 3 therefore apply verbatim to \(N\) with the same stopping time \(\tau_R\). The only property of \(\tau_R\) those steps used is that the resulting integrands are bounded on \([0, \tau_R)\), which the display just secured. Hence \(\mathbb{E}[N_{t \wedge \tau_R}] = 1\), and

\[ \mathbb{E}\bigl[ M_{t \wedge \tau_R}^2 \bigr] \leq e^{C^2 T}\, \mathbb{E}\bigl[ N_{t \wedge \tau_R} \bigr] = e^{C^2 T} \quad \text{for every } R . \]

Letting \(R \to \infty\) along the pointwise convergence of Step 1, Fatou's lemma yields \(\mathbb{E}[M_t^2] \leq e^{C^2 T}\) for every \(t \in [0, T]\).

Step 5 (From stopped to global martingale).
Write \(\Delta_R = M_{t \wedge \tau_R} - M_t\), so \(\Delta_R \to 0\) pointwise. For every threshold \(K \gt 0\), the pointwise inequality \(|\Delta_R| \leq \min(|\Delta_R|, K) + \Delta_R^2 / K\) holds, because on \(\{|\Delta_R| \gt K\}\) the second summand alone dominates. Taking expectations, \(\mathbb{E}[\min(|\Delta_R|, K)] \to 0\) by the dominated convergence theorem with the constant dominator \(K\), while \(\mathbb{E}[\Delta_R^2] \leq 2\, \mathbb{E}[M_{t \wedge \tau_R}^2] + 2\, \mathbb{E}[M_t^2] \leq 4 e^{C^2 T}\) by Step 4. Hence \(\limsup_R \mathbb{E}|\Delta_R| \leq 4 e^{C^2 T} / K\) for every \(K\), and \(\mathbb{E}|\Delta_R| \to 0\).

Now fix \(0 \leq s \leq t \leq T\) and \(A \in \mathcal{F}_s^{(n)}\). Integrating the stopped martingale identity of Step 3 over \(A\) gives \(\mathbb{E}[\mathbf{1}_A\, M_{t \wedge \tau_R}] = \mathbb{E}[\mathbf{1}_A\, M_{s \wedge \tau_R}]\), and since \(|\mathbb{E}[\mathbf{1}_A\, M_{t \wedge \tau_R}] - \mathbb{E}[\mathbf{1}_A\, M_t]| \leq \mathbb{E}|\Delta_R| \to 0\), with the same estimate at time \(s\), the limits agree: \(\mathbb{E}[\mathbf{1}_A\, M_t] = \mathbb{E}[\mathbf{1}_A\, M_s]\) for every \(A \in \mathcal{F}_s^{(n)}\). The variable \(M_s\) is \(\mathcal{F}_s^{(n)}\)-measurable, being a continuous function of adapted integrals, and integrable, so the defining property identifies \(M_s = \mathbb{E}[M_t \mid \mathcal{F}_s^{(n)}]\) a.s. The three conditions of the martingale definition hold on \([0, T]\), and taking \(s = 0\) gives \(\mathbb{E}[M_t] = M_0 = 1\). This proves (1).

Step 6 (Moments).
For real \(p \geq 1\), the process \(p\, \mathbf{a}\) is progressively measurable and bounded by \(p\, C\), so part (1) applies to its exponential process \(N^{(p)}\), giving \(\mathbb{E}[N^{(p)}_t] = 1\). The exponents rearrange as

\[ M_t^p = \exp\Bigl( - \int_0^t p\, \mathbf{a} \cdot d\mathbf{w} - \frac{p}{2} \int_0^t |\mathbf{a}|^2\, ds \Bigr) = N^{(p)}_t \cdot \exp\Bigl( \frac{p^2 - p}{2} \int_0^t |\mathbf{a}|^2\, ds \Bigr) \leq e^{\frac{1}{2} p (p-1) C^2 T}\, N^{(p)}_t , \]

and taking expectations proves (2).

Step 7 (The unstopped identity).
First the membership claim. Each \(M a_j\) is a product of progressively measurable processes, and by Tonelli's theorem together with the bound of Step 4,

\[ \mathbb{E}\Bigl[ \int_0^T M_s^2\, a_j^2\, ds \Bigr] \leq C^2 \int_0^T \mathbb{E}\bigl[ M_s^2 \bigr]\, ds \leq C^2\, T\, e^{C^2 T} \lt \infty , \]

so \(M a_j \in \mathcal{V}(0, T)\). Next, the truncated integrands of Step 3 converge to \(M a_j\) in the \(\mathcal{V}\)-norm: the random variable \(\int_0^t \mathbf{1}_{[\tau_R, \infty)}(s)\, M_s^2\, a_j^2\, ds\) vanishes for \(R\) large by Step 1, is dominated by the integrable variable \(\int_0^T M_s^2\, a_j^2\, ds\), and so has expectation tending to zero by dominated convergence. By the Itô isometry, applied per component on the horizon \(t\), each truncated integral converges in \(L^2(\mathbb{P})\) to \(\int_0^t M_s\, a_j\, dw_s^{(j)}\), and the finite sum over \(j\) converges as well. The left side of Step 3's identity converges to \(M_t\) pointwise. An \(L^2\)-convergent sequence has an almost surely convergent subsequence, and along it both limits must agree, so for each \(t\), almost surely,

\[ M_t = 1 - \sum_{j=1}^{n} \int_0^t M_s\, a_j\, dw_s^{(j)} . \]

Both sides are continuous in \(t\) in their chosen realizations, so the identity extends from a countable dense set of times to all of \([0, T]\) simultaneously, off a single null set. This proves (3).

Proof.

The product is progressively measurable, hence measurable and adapted. For the integrability condition, the pointwise inequality \(M_s^2 \Phi_s^2 \leq \tfrac{1}{2} M_s^4 + \tfrac{1}{2} \Phi_s^4\) together with Tonelli's theorem and the moment bound of the exponential martingale lemma with \(p = 4\) gives

\[ \mathbb{E}\Bigl[ \int_0^T h^2 M_s^2 \Phi_s^2\, ds \Bigr] \leq \frac{(C')^2\, T}{2} \Bigl( e^{6 C^2 T} + \sup_{s \in [0, T]} \mathbb{E}\bigl[ \Phi_s^4 \bigr] \Bigr) \lt \infty . \]

Proof.

Step 1 (A probability measure equivalent to \(\mathbb{P}\)).
The Itô integrals defining \(V_T\) are finite real numbers for every \(\omega\) in their continuous realizations, so \(M_T(\omega) = e^{V_T(\omega)} \gt 0\) everywhere. Non-negativity gives \(\mathbb{Q} \geq 0\), the exponential martingale lemma gives \(\mathbb{Q}(\Omega) = \mathbb{E}_{\mathbb{P}}[M_T] = 1\), and for disjoint sets \(A_1, A_2, \ldots\) the partial sums of \(\sum_k \mathbf{1}_{A_k} M_T\) increase to \(\mathbf{1}_{\cup_k A_k} M_T\), so countable additivity follows from the monotone convergence theorem. Absolute continuity \(\mathbb{Q} \ll \mathbb{P}\) is built into the definition. Conversely, if \(\mathbb{Q}(A) = 0\) then the non-negative variable \(\mathbf{1}_A M_T\) has zero expectation, hence vanishes \(\mathbb{P}\)-a.s., and since \(M_T \gt 0\) everywhere this forces \(\mathbb{P}(A) = 0\). This proves (1).

Step 2 (The density process is \(M\) itself).
Fix \(t \in [0, T]\) and \(A \in \mathcal{F}_t^{(n)}\). By the defining property of conditional expectation and the martingale property of \(M\),

\[ \mathbb{Q}(A) = \mathbb{E}_{\mathbb{P}}\bigl[ \mathbf{1}_A\, M_T \bigr] = \mathbb{E}_{\mathbb{P}}\bigl[ \mathbf{1}_A\, \mathbb{E}_{\mathbb{P}}[ M_T \mid \mathcal{F}_t^{(n)} ] \bigr] = \mathbb{E}_{\mathbb{P}}\bigl[ \mathbf{1}_A\, M_t \bigr] . \]

The variable \(M_t\) is non-negative and \(\mathcal{F}_t^{(n)}\)-measurable, so it is a density of the restricted \(\mathbb{Q}\) with respect to the restricted \(\mathbb{P}\), and by the uniqueness half of the Radon-Nikodym theorem it is the density, up to \(\mathbb{P}\)-null modification. This proves (2), and it identifies the abstract density process of the pair \((\mathbb{P}, \mathbb{Q})\), whose martingale property the density process lemma establishes in general, with the concrete exponential \(M\).

Step 3 (Moments of the shifted motion).
Each coordinate \(w^{(i)}_s\) is a centered Gaussian variable with variance \(s\), by property (W1) of the Brownian motion definition. Writing \(\varphi_s\) for its density, the relation \(z\, \varphi_s(z) = -s\, \varphi_s'(z)\) and one integration by parts give the recursion

\[ \mathbb{E}\bigl[ (w^{(i)}_s)^{2k} \bigr] = \int_{\mathbb{R}} z^{2k}\, \varphi_s(z)\, dz = (2k - 1)\, s \int_{\mathbb{R}} z^{2k - 2}\, \varphi_s(z)\, dz = (2k - 1)\, s\, \mathbb{E}\bigl[ (w^{(i)}_s)^{2k - 2} \bigr] , \]

so by induction every even moment is finite and bounded uniformly for \(s \in [0, T]\). Since \(|Y^{(i)}_s| \leq C T + |w^{(i)}_s|\) and \((x + y)^{2k} \leq 2^{2k - 1} ( x^{2k} + y^{2k} )\) by convexity, every even moment of \(Y^{(i)}_s\) up to order eight is bounded uniformly on \([0, T]\). Two consequences will be needed below. First, for each \(i, j\),

\[ \begin{align*} \sup_{s \in [0, T]} \mathbb{E}\Bigl[ \bigl( Y^{(i)}_s \bigr)^4 \Bigr] &\lt \infty , \\\\ \sup_{s \in [0, T]} \mathbb{E}\Bigl[ \bigl( Y^{(i)}_s Y^{(j)}_s - \delta_{ij}\, s \bigr)^4 \Bigr] &\lt \infty , \end{align*} \]

the second because \((x y - c)^4 \leq 8\, ( x^4 y^4 + c^4 ) \leq 8 \bigl( \tfrac{1}{2} x^8 + \tfrac{1}{2} y^8 + c^4 \bigr)\) with \(|c| \leq T\). Second, for \(X\) standing for either \(Y^{(i)}_t\) or \(Y^{(i)}_t Y^{(j)}_t - \delta_{ij}\, t\), the second moment \(\mathbb{E}[X^2]\) is finite by the even-moment bounds just recorded, so \(\mathbb{E}_{\mathbb{Q}}[|X|] = \mathbb{E}_{\mathbb{P}}[M_T |X|] \leq \tfrac{1}{2} \mathbb{E}[M_T^2] + \tfrac{1}{2} \mathbb{E}[X^2] \lt \infty\). Both variables are \(\mathbb{Q}\)-integrable.

Step 4 (The compensated products are \(\mathbb{P}\)-martingales).
The process \(\mathbf{Y}\) is an \(n\)-dimensional Itô process with drift \(\mathbf{a}\), coefficient matrix the identity, and \(\mathbf{Y}_0 = \mathbf{0}\), while the exponential martingale lemma exhibits \(M\) as an Itô process with zero drift and \(d\mathbf{w}\)-row \(-M \mathbf{a}^\top\). Fix \(i\) and set \(K_t = M_t\, Y^{(i)}_t\). The stochastic product rule asks that, for each \(j\), the process \(Y^{(i)} \cdot ( -M a_j ) + M\, \delta_{ij}\) lie in \(\mathcal{V}(0, T)\), which the membership lemma grants: take \(h = -a_j\), \(\Phi = Y^{(i)}\) for the first summand and \(h = 1\), \(\Phi = \delta_{ij}\) for the second, and note that \(\mathcal{V}(0, T)\) is closed under sums since \((f + g)^2 \leq 2 f^2 + 2 g^2\). The product rule then yields

\[ dK_t = M_t \bigl( a_i\, dt + dw^{(i)}_t \bigr) + Y^{(i)}_t \bigl( - M_t\, \mathbf{a} \cdot d\mathbf{w}_t \bigr) + \sum_{j=1}^{n} ( - M_t\, a_j )\, \delta_{ij}\, dt = M_t\, dw^{(i)}_t - Y^{(i)}_t M_t\, \mathbf{a} \cdot d\mathbf{w}_t , \]

the two \(dt\)-terms cancelling exactly. Since \(K_0 = 0\), the process \(K\) is a sum of Itô integrals of \(\mathcal{V}(0, T)\)-integrands and hence a martingale under \(\mathbb{P}\), by the martingale property of Itô integrals.

Now fix \(i, j\) and treat the product \(Y^{(i)} Y^{(j)}\). The product rule applies with membership condition \(Y^{(i)} \delta_{jk} + Y^{(j)} \delta_{ik} \in \mathcal{V}(0, T)\) for each \(k\), granted directly by \(\mathbb{E}\bigl[ (Y^{(i)}_s)^2 \bigr] \leq 2 C^2 T^2 + 2 s\) and Tonelli, and produces the bracket \(\sum_k \delta_{ik} \delta_{jk}\, dt = \delta_{ij}\, dt\):

\[ d\bigl( Y^{(i)} Y^{(j)} \bigr) = Y^{(i)}\, dY^{(j)} + Y^{(j)}\, dY^{(i)} + \delta_{ij}\, dt . \]

Absorbing the constant \(-\delta_{ij}\) into the drift coefficient, the process \(G_t = Y^{(i)}_t Y^{(j)}_t - \delta_{ij}\, t\) is an Itô process with drift \(Y^{(i)} a_j + Y^{(j)} a_i\) and \(d\mathbf{w}\)-row \(Y^{(i)} \mathbf{e}_j^\top + Y^{(j)} \mathbf{e}_i^\top\), where \(\mathbf{e}_i\) denotes the \(i\)-th standard basis vector. Set \(\widetilde{K}_t = M_t\, G_t\). The product rule's membership condition reads, for each \(k\), \(G \cdot ( - M a_k ) + M \bigl( Y^{(i)} \delta_{jk} + Y^{(j)} \delta_{ik} \bigr) \in \mathcal{V}(0, T)\), and the membership lemma grants every summand through Step 3's fourth-moment bounds, with \(\Phi = G\) and \(\Phi = Y^{(i)}, Y^{(j)}\) respectively. The bracket is \(\sum_k ( - M a_k ) \bigl( Y^{(i)} \delta_{jk} + Y^{(j)} \delta_{ik} \bigr)\, dt = - M \bigl( Y^{(i)} a_j + Y^{(j)} a_i \bigr)\, dt\), which cancels \(M\) times the drift of \(G\), so

\[ d\widetilde{K}_t = M_t \Bigl( Y^{(i)}_t\, dw^{(j)}_t + Y^{(j)}_t\, dw^{(i)}_t \Bigr) - G_t\, M_t\, \mathbf{a} \cdot d\mathbf{w}_t . \]

Again \(\widetilde{K}_0 = 0\), the integrands lie in \(\mathcal{V}(0, T)\), and \(\widetilde{K}\) is a martingale under \(\mathbb{P}\).

Step 5 (Transfer to \(\mathbb{Q}\) by Bayes' rule).
Fix \(0 \leq s \leq t \leq T\) and let \(X\) stand for either \(Y^{(i)}_t\) or \(Y^{(i)}_t Y^{(j)}_t - \delta_{ij}\, t\). Apply Bayes' rule on \(\bigl( \Omega, \mathcal{F}_T^{(n)} \bigr)\) with \(\mu\) the restriction of \(\mathbb{P}\), \(\nu = \mathbb{Q}\), density \(f = M_T\), and \(\mathcal{H} = \mathcal{F}_s^{(n)}\). The \(\mathbb{Q}\)-integrability of \(X\) is Step 3, so

\[ \mathbb{E}_{\mathbb{Q}}\bigl[ X \mid \mathcal{F}_s^{(n)} \bigr] \cdot \mathbb{E}_{\mathbb{P}}\bigl[ M_T \mid \mathcal{F}_s^{(n)} \bigr] = \mathbb{E}_{\mathbb{P}}\bigl[ M_T\, X \mid \mathcal{F}_s^{(n)} \bigr] . \]

The second factor on the left is \(M_s\), by the martingale property of \(M\). For the right side, the tower property through \(\mathcal{F}_t^{(n)}\) and take-out of the \(\mathcal{F}_t^{(n)}\)-measurable factor \(X\), whose product with \(M_T\) is integrable by \(|M_T X| \leq \tfrac{1}{2} M_T^2 + \tfrac{1}{2} X^2\) and Step 3, give

\[ \mathbb{E}_{\mathbb{P}}\bigl[ M_T X \mid \mathcal{F}_s^{(n)} \bigr] = \mathbb{E}_{\mathbb{P}}\Bigl[ X\, \mathbb{E}_{\mathbb{P}}\bigl[ M_T \mid \mathcal{F}_t^{(n)} \bigr] \Bigm| \mathcal{F}_s^{(n)} \Bigr] = \mathbb{E}_{\mathbb{P}}\bigl[ X M_t \mid \mathcal{F}_s^{(n)} \bigr] , \]

and \(X M_t\) is exactly \(K_t\), respectively \(\widetilde{K}_t\), so by Step 4 its conditional expectation at time \(s\) is \(K_s\), respectively \(\widetilde{K}_s\), that is, \(M_s\, Y^{(i)}_s\), respectively \(M_s \bigl( Y^{(i)}_s Y^{(j)}_s - \delta_{ij}\, s \bigr)\). Combining the displays and dividing by \(M_s\), which is strictly positive at every \(\omega\),

\[ \begin{align*} \mathbb{E}_{\mathbb{Q}}\bigl[ Y^{(i)}_t \mid \mathcal{F}_s^{(n)} \bigr] &= Y^{(i)}_s , \\\\ \mathbb{E}_{\mathbb{Q}}\bigl[ Y^{(i)}_t Y^{(j)}_t - \delta_{ij}\, t \Bigm| \mathcal{F}_s^{(n)} \bigr] &= Y^{(i)}_s Y^{(j)}_s - \delta_{ij}\, s , \end{align*} \]

first \(\mathbb{P}\)-a.s. and then \(\mathbb{Q}\)-a.s., since every \(\mathbb{P}\)-null set is \(\mathbb{Q}\)-null by (1). Each process is adapted and \(\mathbb{Q}\)-integrable, so under \(\mathbb{Q}\) the coordinates \(Y^{(i)}\) and the compensated products \(Y^{(i)} Y^{(j)} - \delta_{ij}\, t\) are martingales with respect to \(\{\mathcal{F}_t^{(n)}\}\) on \([0, T]\).

Step 6 (Conclusion by the Lévy characterization).
The process \(\mathbf{Y}\) is continuous, starts at the origin, and is adapted to \(\{\mathcal{F}_t^{(n)}\}\). Step 5 verifies conditions (i) and (ii) of the Lévy characterization with respect to \(\{\mathcal{F}_t^{(n)}\}\) and \(\mathbb{Q}\), and the characterization delivers exactly the conclusion of (3): the law of \(\mathbf{Y}\) under \(\mathbb{Q}\) is that of an \(n\)-dimensional standard Brownian motion on \([0, T]\), with each increment independent of \(\mathcal{F}_s^{(n)}\).