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For the Gaussian Brownian drift filter $Y_t=\alpha t+W_t$, with independent prior $\alpha \sim N(m_0,{\tau }_0^{-1})$, the observation law has densityrelative to driftless Wiener measure. The law having fixed observation drift $k$ has density $e^{k{Y}_t-k^{2}t/2}$ relative to the same reference. Their ratio is therefore a positive mean-one density on every finite horizon. In the observation filtration it satisfies $d{D}_t=-D_t(m_t-k)d{\hat{W}}_t$, where $m_t$ is the posterior mean and $\hat{W}$ the innovation process. This density-ratio argument establishes a true martingale without an unsupported global Novikov condition for the unbounded posterior mean.

Observe $Y_t=\lambda t+W_t$, with an independent Gaussian random vector $\lambda$ of mean $m_0$ and covariance matrix $V_0$. Its posterior is Gaussian with mean $m_t=(I+t{V}_0{)}^{-1}(m_0+V_0{Y}_t)$ and covariance matrix $V_t=V_0(I+t{V}_0{)}^{-1}$. These formulas also hold for a singular positive semidefinite matrix $V_0$. When $V_0$ is invertible, completing the square in the Bayes' theorem gives $V_t^{-1}=V_0^{-1}+tI$ and $m_t=V_t(V_0^{-1}{m}_0+Y_t)$. The observed innovation process ${\hat{W}}_t=Y_t-{\int }_0^t{m}_{s}ds$ is a Brownian motion in the observation filtration, and $d{m}_t=V_{t}d{\hat{W}}_t$.

In a linear Gaussian state-space model, let the predicted state mean and covariance be $m_t^{-},P_t^{-}$. The observation innovation process has conditional mean zero and covariance ${\Sigma }_t=A{P}_t^{-}{A}^{\top }+R$, with innovation ${ϵ}_t=y_t-A{m}_t^{-}$. The conditional multivariate normal distribution gives the predictive observation density. Multiplying these conditional densities by the chain rule gives the displayed likelihood function; equivalently the innovations are independent Gaussian vectors. The Kalman filter supplies $m_t^{-},P_t^{-}$ recursively, so one evaluates the exact observed-data likelihood function without integrating all latent states jointly.

A causal square-integrable linear time series can be written $X_t=\mu +{\sum }_{j\ge 0}{\psi }_j{\epsilon }_{t-j}$. Square summability of the coefficients ensures mean-square convergence for white noise. An innovation representation additionally identifies its driving noise as the innovation process.

Assume, as in the Bayesian model, that the unknown drift $\alpha$ is independent of the driving Brownian motion, with known positive prior precision ${\tau }_0$. Recover the observed signal from the positive stock price:For a fixed time $t$, the centered observed path $Y_s-(s/t)Y_t$ is a Brownian bridge independent of $(\alpha ,Y_t)$: it is independent of the prior and has zero covariance with $W_t$ in their joint Gaussian law. Thus the whole observed history has the same posterior information about $\alpha$ as its endpoint. This proves Brownian endpoint sufficiency for a constant drift rather than assuming it.

The likelihood of that endpoint, as a function of $\alpha$, is proportional to $\exp (\alpha Y_t-{\alpha }^{2}t/2)$. Multiplying by the normal distribution prior and completing the square gives the Gaussian Brownian drift filterDefine the observed innovation process ${\hat{W}}_t=Y_t-{\int }_0^t{\hat{\alpha }}_{s}ds$. It is a continuous observed-filtration martingale: for $s<u$, independence of future noise and the tower property of conditional expectation give $\mathbb{E}[\alpha -{\hat{\alpha }}_u\mid {\mathcal{F}}_s^S]=0$, so the conditional mean of its increment vanishes. Its quadratic variation is $t$, since subtracting the finite-variation drift from $Y$ does not change quadratic variation. The Lévy characterization of Brownian motion makes $\hat{W}$ an observed-filtration Brownian motion. Differentiating the posterior mean givesConversely, the last equation determines $\hat{\alpha }$ from $\hat{W}$ with deterministic coefficients and known initial value, and $Y=\hat{W}+\int \hat{\alpha }dt$. Thus the observed and innovation filtrations coincide, supplying the natural Brownian information needed for replication. Consequently the observed stock dynamics are $d{S}_t/S_t=\sigma d{\hat{W}}_t+\sigma {\hat{\alpha }}_{t}dt$. Its observed market price of risk is ${\kappa }_t={\hat{\alpha }}_t-r/\sigma$, and its normalized state-price density solvesThe posterior mean is unbounded, so bounded-coefficient Novikov condition reasoning from Question 2 is not available automatically. A direct finite-horizon argument supplies true pricing. Relative to driftless Wiener measure for $Y$, the Bayesian mixture has observation densityA law with observation drift $k=r/\sigma$ instead has density $e^{k{Y}_t-k^{2}t/2}$. ThereforeThis is the finite-horizon pricing density for Gaussian drift learning. The ratio changes the physical observed law into a genuine probability law, hence has mean one. Indeed, differentiation of the explicit likelihood, using $d{Y}_t=d{\hat{W}}_t+{\hat{\alpha }}_{t}dt$, givesThe Itô formula therefore gives $d{D}_t=-D_t{\kappa }_{t}d{\hat{W}}_t$, agreeing with the stochastic exponential above. Under this pricing law the stock drift is $\sigma k=r$.

For logarithmic utility, terminal marginal utility is $1/w_T$. The state-price budget constraint and the marginal condition imply $w_T^{\ast }=1/(y{\zeta }_T)$ and $1/y=w_0$. Since ${\zeta }_T{w}_T^{\ast }=w_0$ is constant, its conditional-expectation pricing process is constant as well. Thus the entire optimal wealth process isIndeed, for any other nonnegative admissible terminal wealth $X$, the elementary inequality $\log (X/w_T^{\ast })\le X/w_T^{\ast }-1$ givesThis directly verifies optimality. Expected candidate log wealth is finite because $\mathbb{E}{\hat{\alpha }}_t^2={\hat{\alpha }}_0^2+{\tau }_0^{-1}-({\tau }_0+t{)}^{-1}$ and its time integral is finite on every fixed horizon. Applying the Itô formula to $1/\zeta$ givesThe observed self-financing portfolio with risky dollar fraction ${\pi }_t$ has diffusion coefficient ${\pi }_t\sigma$. Matching the coefficients proves the log-optimal investment with Gaussian drift learning ruleThe remaining dollars are held in the bank account; the fraction may be negative or exceed one because the question imposes neither short-sale nor borrowing restrictions.

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.

Treat the two-point prior for $\alpha$ as independent of the driving Brownian motion. On a partition $0=t_0<\cdots <t_n=t$, conditionally on $\alpha =\pm a$, the increments of $X$ are independent normal distributions with means $\pm a\Delta t_j$ and variances $\Delta t_j$. Their likelihood ratio isThus every partition containing $t$ gives the same Bayesian posterior, depending only on the endpoint. Refine partitions through a dense set of observation times. Continuity makes their increasing observation fields generate the full path field ${\mathcal{X}}_t$, and convergence of bounded conditional expectations preserves that Bayesian posterior. Equal prior probabilities and the Bayes' theorem consequently giveIts conditional mean drift isThis is the binary Brownian drift filter.

Define the observed innovation processIt is continuous and adapted to ${\mathcal{X}}_t$. For $s<t$, independence of future Brownian motion increments from the past and the latent drift givesFor every $u\ge s$, the tower property of conditional expectation givesSince $|h|\le a$, conditional integration is justified. Subtracting the integrated identities shows $E[{\hat{W}}_t-{\hat{W}}_s\mid {\mathcal{X}}_s]=0$. Thus $\hat{W}$ is a square-integrable martingale. Subtracting a continuous finite variation process does not change quadratic variation, so $[\hat{W}{]}_t=[X{]}_t=[W{]}_t=t$. The Lévy characterization of Brownian motion now proves that $\hat{W}$ is a Brownian motion in the observation filtration. Therefore

Use the standard linear Gaussian state-space model convention: the initial state is independent of both mutually independent iid noise sequences. This independence from the initial state is needed for the usual forecast recursion. Let ${\mathcal{Y}}_t=\sigma (y_1,\dots ,y_t)$ and write $m_{t-1}=x_{t-1}^{t-1}$ and $C_{t-1}=P_{t-1}^{t-1}$. Taking conditional expectations in the state equation givesTo justify the covariance, the forecast error is $\Phi (x_{t-1}-m_{t-1})+w_t$. Its two terms have zero cross-covariance because $w_t$ is independent of the previous state and observations. Their covariances therefore add. In this Gaussian model the conditional error covariance depends on the parameters and time but not on the realized data, so it also equals the unconditional forecast-error covariance used in the notation of the question.

For completeness, derive the filtering step rather than leave the algorithm unspecified. Set $m_t^{-}=x_t^{t-1}$ and $C_t^{-}=P_t^{t-1}$. Conditionally on the previous data, the observation has mean $A{m}_t^{-}$ and covarianceDefine the innovation process and gain byThe state forecast error and innovation are jointly normal, with cross-covariance $C_t^{-}{A}^{\top }$. The residual $x_t-m_t^{-}-K_t{ϵ}_t$ is uncorrelated with ${ϵ}_t$ by the definition of $K_t$, hence independent of it by joint normality. Its mean is zero and its covariance is $C_t^{-}-C_t^{-}{A}^{\top }{\Sigma }_t^{-1}A{C}_t^{-}$. This proves the conditional multivariate normal distribution updateInitialize $x_0^0={\mu }_0,P_0^0={\Sigma }_0$. For $t=1,\dots ,n$, predict using the first boxed pair, store that forecast, form ${ϵ}_t,{\Sigma }_t,K_t$, and update using the second boxed pair. The updated state then starts the next iteration. This is the Kalman filter, and supplies the entire forecast sequence in one forward pass.

The requested unconditional innovation moments areIn fact ${ϵ}_t\mid {\mathcal{Y}}_{t-1}\sim N_q(0,{\Sigma }_t)$, with ${\Sigma }_t$ deterministic for fixed parameters. Thus the innovation is independent of the past observations, and hence the innovations are mutually independent. Equivalently, orthogonality to the past together with joint normality gives this independence.

The chain rule for densities now yields the Gaussian innovation likelihood, treating the specified initial mean and covariance as known:Both ${ϵ}_t$ and ${\Sigma }_t$ are functions of the trial parameters, so the filtering recursion must be run again for each candidate $\Theta$. The ordinary density formula assumes ${\Sigma }_t$ positive definite; for example positive definite $R$ ensures this. Singular observation laws instead require densities on their Gaussian support.

Evaluate the log-likelihood numerically by the forward recursion, and maximize it over admissible $\Phi ,A,Q,R$, enforcing the covariance constraints. A Cholesky decomposition ${\Sigma }_t=L_t{L}_t^{\top }$ gives $\log |{\Sigma }_t|=2{\sum }_j\log (L_t{)}_{jj}$ and the quadratic term $\Vert L_t^{-1}{ϵ}_t{\Vert }^2$, avoiding an explicit matrix inverse. Covariance parameterizations or constrained optimization keep $Q,R$ positive semidefinite and the innovation covariance nonsingular. An attained maximizer is a maximum-likelihood estimator; in general this is a numerical optimization problem and neither uniqueness nor global convergence of a local optimizer is automatic.

Here the innovation process consists of linear innovations, the linear one-step prediction errors: ${\epsilon }_t=X_t-{\mathrm{proj}}_{{\mathcal{H}}_{t-1}}{X}_t$, where ${\mathcal{H}}_{t-1}$ is the closed linear span of the past $X$'s in $L^2$. For Gaussian processes this is also the conditional expectation prediction error. For general non-Gaussian strong white noise these two notions can differ.

The given ${\eta }_t$ is not the linear innovation process. The moving-average factor has its zero at $1/3$, inside the unit disk, and is noninvertible as a causal moving-average filter. The identityis the moving-average root reflection that places this zero outside the unit disk. Thus the causal invertible representation has innovation variance $9$, rather than the given variance $1$.

For an explicit verification, defineThe filter has constant squared modulus $9$, so $\epsilon$ is weak white noise with variance $9$. The new moving-average factor has root $3$ and is invertible, while its autoregressive factor is causal. Hence the past spans of $X$ and $\epsilon$ agree, and $X_t-{\epsilon }_t$ belongs to that past span. Orthogonality of ${\epsilon }_t$ to past $\epsilon$ therefore identifies it as the linear innovation process. The variance difference proves that it cannot be ${\eta }_t$. If the original noise is Gaussian, the new linear innovations are independent Gaussian variables; without Gaussianity they need only be uncorrelated random variables.

The root-reflected, invertible autoregressive moving-average model isThis follows by multiplying the filter identity for $\epsilon$ by $1-B/3$. Its moving-average root $3$ and autoregressive root $4$ are both outside the unit disk. Thus it is the representation in terms of the innovation process, rather than merely another noise representation with the same spectrum.

Expanding the stable autoregressive inverse gives the infinite moving-average representation in linear innovations:ThereforeFor comparison, the causal representation in the originally supplied noise isBoth converge in $L^2$, since their coefficients are square summable. The second is causal, but its driving noise is not the linear innovation process.