= Solution
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> is
$$
\frac{L_+}{L_-}
=\prod_{j=1}^n\exp\left[\frac{(\Delta X_j+a\Delta t_j)^2-(\Delta X_j-a\Delta t_j)^2}{2\Delta t_j}\right]
=\exp\left(2a\sum_{j=1}^n\Delta X_j\right)
=e^{2aX_t}.
$$
Thus 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 give
$$
\boxed{q_t=P(\alpha=a\mid\mathcal X_t)=\frac{e^{aX_t}}{e^{aX_t}+e^{-aX_t}}.}
$$
Its conditional mean <drift> is
$$
E[\alpha\mid\mathcal X_t]=a(2q_t-1)=a\tanh(aX_t)=h(X_t).
$$
This is the <binary Brownian drift filter>.
Define the observed <innovation process>
$$
\widehat W_t=X_t-\int_0^t h(X_u)du.
$$
It is continuous and adapted to $\mathcal X_t$. For $s<t$, independence of future <Brownian motion> increments from the past and the latent <drift> gives
$$
E[X_t-X_s\mid\mathcal X_s]=(t-s)E[\alpha\mid\mathcal X_s].
$$
For every $u\geq s$, the tower property of <conditional expectation> gives
$$
E[h(X_u)\mid\mathcal X_s]
=E\bigl[E[\alpha\mid\mathcal X_u]\mid\mathcal X_s\bigr]
=E[\alpha\mid\mathcal X_s].
$$
Since $|h|\leq a$, conditional integration is justified. Subtracting the integrated identities shows $E[\widehat W_t-\widehat W_s\mid\mathcal X_s]=0$. Thus $\widehat W$ is a square-integrable <martingale>. Subtracting a continuous <finite variation> process does not change <quadratic variation>, so $[\widehat W]_t=[X]_t=[W]_t=t$. The <Lévy characterization of Brownian motion> now proves that $\widehat W$ is a <Brownian motion> in the observation <filtration>. Therefore
$$
\boxed{dX_t=d\widehat W_t+a\tanh(aX_t)dt.}
$$
Back to article page
