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Fix $t>0$. For $0\le s\le t$, the residual observed path isThis Brownian bridge is independent of $(\lambda ,Y_t)$. Indeed it is independent of $\lambda$, while its covariance with $W_t$ is $sI-(s/t)tI=0$; zero covariance implies independence in the joint Gaussian law. The same argument for every finite collection of bridge times, followed by continuity and generation from rational times, gives independence of the entire bridge. Since the full observed path is determined by its endpoint and this independent bridge, Brownian endpoint sufficiency for a constant drift gives

First suppose the prior covariance matrix $V_0$ is positive definite. The Bayes' theorem multiplies the prior Gaussian density by the conditional observation density, whose dependence on $\lambda$ is $\exp ({\lambda }^T{Y}_t-t|\lambda {|}^2/2)$. Thus the posterior exponent, up to terms not depending on $\lambda$, isCompleting the square gives the Gaussian Brownian drift filter:For invertible $V_0$ these are equivalently $V_t=(V_0^{-1}+tI{)}^{-1}$ and ${\lambda }_t=V_t(V_0^{-1}{\lambda }_0+Y_t)$. At $t=0$ they reduce to the prior parameters.

The boxed formulas remain valid for a singular positive semidefinite matrix $V_0$. To see this directly without inverse prior matrices, the joint Gaussian random vector $(\lambda ,Y_t)$ hasThe latter matrix is invertible for $t>0$. The conditional multivariate normal distribution therefore has meanand covariance matrixAlternatively, subtract this mean from $\lambda$; the resulting Gaussian random vector has zero covariance with $Y_t$ and is therefore independent of it, which proves these conditional parameters.

Write the proposed innovation process in the observable formIt is adapted to the observation filtration and continuous. To prove the martingale property, take $0\le u<v$. Future increments of $W$ are independent of the joint information generated by $\lambda$ and the past of $W$, soAlso the tower property of conditional expectation gives $\mathbb{E}[{\lambda }_s\mid {\mathcal{G}}_u]={\lambda }_u$ for $s\ge u$. Conditional integration is legitimate because $\mathbb{E}|{\lambda }_s{|}^2\le \mathbb{E}|\lambda {|}^2<\infty$. Subtracting the integral proves $\mathbb{E}[{\hat{W}}_v-{\hat{W}}_u\mid {\mathcal{G}}_u]=0$. Removing a continuous finite variation process does not change quadratic covariation, henceThe Lévy characterization of multidimensional Brownian motion now applies: a continuous zero-start martingale with this bracket matrix is a standard $d$-dimensional Brownian motion relative to its given filtration. The observable innovation is a Wiener process in the observation filtration.

Differentiating the posterior formulas yields $V_t^{\prime }=-V_t^2$ andFor the final expression, the printed inverse matrices require the extra assumption that $V_0$ is positive definite. Under that assumption put $A_t=V_t^{-1}=V_0^{-1}+tI$. The Itô formula givessince $A_t{V}_t=I$ and $\mathrm{tr}(A_t{V}_t^2)=\mathrm{tr}{V}_t$. On the other hand,The trace terms cancel in the innovation exponential for Gaussian drift filtering, leavingThus $Z_0=1$ and $Z$ is a positive stochastic exponential, hence a nonnegative local martingale. To obtain the requested supermartingale conclusion explicitly, choose localizing stopping times ${\tau }_n\uparrow \infty$ for which $Z^{{\tau }_n}$ is a true martingale. The Conditional Fatou lemma givesThe same argument at $u=0$ gives $\mathbb{E}{Z}_v\le 1$, supplying integrability. The process is a supermartingale in the observation filtration. No unproved exponential-integrability condition on the random drift is required.

For a singular prior the literal inverse-matrix expression in the PDF is undefined, but there is a natural extension with the same conclusion. Define, for fixed $y$,For positive definite $V_0$, a Gaussian integral shows this reciprocal is exactly the printed expression. For any positive semidefinite $V_0$, the integral is positive and finite, $L_t=-{\Delta }_{y}L/2$, and ${\nabla }_y\log L(t,Y_t)={\lambda }_t$, by the same posterior weighting. The Itô formula consequently gives $d\log L(t,Y_t)={\lambda }_t^{T}d{Y}_t-|{\lambda }_t{|}^{2}dt/2={\lambda }_t^{T}d{\hat{W}}_t+|{\lambda }_t{|}^{2}dt/2$. Its reciprocal satisfies the boxed stochastic differential equation and the same supermartingale proof. This completes the singular-covariance extension while explicitly separating it from the undefined printed formula.