= Solution
Fix $t>0$. For $0\leq s\leq t$, the residual observed path is
$$
Y_s-\frac{s}{t}Y_t=W_s-\frac{s}{t}W_t.
$$
This <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
$$
\mathcal L(\lambda\mid\mathcal G_t)=\mathcal L(\lambda\mid Y_t).
$$
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 $\exp(\lambda^TY_t-t|\lambda|^2/2)$. Thus the posterior exponent, up to terms not depending on $\lambda$, is
$$
-\frac12\lambda^T(V_0^{-1}+tI)\lambda+\lambda^T(V_0^{-1}\lambda_0+Y_t).
$$
Completing the square gives the <Gaussian Brownian drift filter>:
$$
\boxed{\lambda_t=(I+tV_0)^{-1}(\lambda_0+V_0Y_t),\qquad V_t=V_0(I+tV_0)^{-1},\qquad\mathcal L(\lambda\mid\mathcal G_t)=N(\lambda_t,V_t).}
$$
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> $(\lambda,Y_t)$ has
$$
\operatorname{Cov}(\lambda,Y_t)=tV_0,\qquad\operatorname{Cov}(Y_t)=t(I+tV_0).
$$
The latter matrix is invertible for $t>0$. The <conditional multivariate normal distribution> therefore has mean
$$
\lambda_0+V_0(I+tV_0)^{-1}(Y_t-t\lambda_0)=\lambda_t
$$
and <covariance matrix>
$$
V_0-tV_0(I+tV_0)^{-1}V_0=V_0(I+tV_0)^{-1}=V_t.
$$
Alternatively, 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 form
$$
\widehat W_t=Y_t-\int_0^t\lambda_s\,ds.
$$
It is adapted to the observation <filtration> and continuous. To prove the <martingale> property, take $0\leq u<v$. Future increments of $W$ are independent of the joint information generated by $\lambda$ and the past of $W$, so
$$
\mathbb E[Y_v-Y_u\mid\mathcal G_u]=(v-u)\mathbb E[\lambda\mid\mathcal G_u]=(v-u)\lambda_u.
$$
Also the <tower property of conditional expectation> gives $\mathbb E[\lambda_s\mid\mathcal G_u]=\lambda_u$ for $s\geq u$. Conditional integration is legitimate because $\mathbb E|\lambda_s|^2\leq\mathbb E|\lambda|^2<\infty$. Subtracting the integral proves $\mathbb E[\widehat W_v-\widehat W_u\mid\mathcal G_u]=0$. Removing a continuous <finite variation> process does not change <quadratic covariation>, hence
$$
[\widehat W^i,\widehat W^j]_t=[W^i,W^j]_t=\delta_{ij}t.
$$
The <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>. \b[The observable innovation is a Wiener process in the observation filtration.]
Differentiating the posterior formulas yields $V_t'=-V_t^2$ and
$$
d\lambda_t=V_t(dY_t-\lambda_tdt)=V_t\,d\widehat W_t.
$$
For 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> gives
$$
d(\lambda_t^TA_t\lambda_t)=2\lambda_t^T\,d\widehat W_t+\left(|\lambda_t|^2+\operatorname{tr}V_t\right)dt,
$$
since $A_tV_t=I$ and $\operatorname{tr}(A_tV_t^2)=\operatorname{tr}V_t$. On the other hand,
$$
d\log\det(I+tV_0)=\operatorname{tr}V_t\,dt.
$$
The trace terms cancel in the <innovation exponential for Gaussian drift filtering>, leaving
$$
d\log Z_t=-\lambda_t^T\,d\widehat W_t-\frac12|\lambda_t|^2dt,\qquad\boxed{dZ_t=-Z_t\lambda_t^T\,d\widehat W_t.}
$$
Thus $Z_0=1$ 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> gives
$$
\mathbb E[Z_v\mid\mathcal G_u]\leq\liminf_n\mathbb E[Z_{v\wedge\tau_n}\mid\mathcal G_u]=\liminf_nZ_{u\wedge\tau_n}=Z_u.
$$
The same argument at $u=0$ gives $\mathbb EZ_v\leq1$, supplying integrability. \b[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 $y$,
$$
L(t,y)=\mathbb E_{a\sim N(\lambda_0,V_0)}\exp\left(a^Ty-\frac t2|a|^2\right),\qquad Z_t=L(t,Y_t)^{-1}.
$$
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, $L_t=-\Delta_yL/2$, and $\nabla_y\log L(t,Y_t)=\lambda_t$, by the same posterior weighting. The <Itô formula> consequently gives $d\log L(t,Y_t)=\lambda_t^T\,dY_t-|\lambda_t|^2dt/2=\lambda_t^T\,d\widehat W_t+|\lambda_t|^2dt/2$. 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.
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