Brownian motion and posterior distribution I am a bit stuck on this question:
Suppose that $X_t = W_t + \alpha t$, where $W$ is a standard Brownian motion, and let $\mathcal{F}_t = \sigma ( X_u: 0 \leq u \leq t)$. The drift is constant in time but unknown: it takes one of the values $a$ or $-a$, where $a>0$. Each of the two possible values has equal (prior) probability.
Want to show that the dynamics of $X$ evolves as follows:
$$dX_t = d \hat{W}_t + a \tanh a X_t \, dt,$$
where $\hat{W}$ is a $\mathcal{F}_t$-Brownian motion.

Progress in my solution so far:
By simple calculations, it can be shown that the posterior for $\alpha$ is of the form $$ \mathbb{P} [\alpha =a | \mathcal{F}_t ] = \frac{ \exp\{a X_t\} }{\exp\{aX_t\} + \exp\{-a X_t\}}.$$
This shows that $\mathbb{E} (\alpha | \mathcal{F}_t) = a \tanh a X_t.$ I am stuck from here. I have tried applying Girsanov theorem to $\mathbb{E} ( \alpha | \mathcal{F}_t ) - \alpha$, but it doesn't really work. Are there any tricks you can think of?

 A: First of all, note that $\sigma(X_t)=\sigma({W}_t)$ since Borel sigma-algebras are translation invariant, so that $\mathcal{F}_t = \mathcal{F}^{W}_t$  for each $t\geqslant 0$. Now, for the filtered space $\left(\Omega,\mathcal{F}, (\mathcal{F}_{t})_{t\geqslant 0},\mathbb{P} \right)$ you have shown
$$
\mathbb{E}\left[\alpha \vert \mathcal{F}_t\right] = a \tanh aX_t\,.
$$
Consequently, consider a re-write of $\mathrm d X_t$ as
$$
\mathrm dX_t = \left(\mathrm dW_t + \left(\alpha - \mathbb{E}\left[\alpha \vert \mathcal{F}_t\right]\right)\mathrm dt\right) + \mathbb{E}\left[\alpha \vert \mathcal{F}_t\right]\mathrm dt,
$$ 
which suggests we define
$$
\mathrm d\hat{W}_t := \mathrm dW_t + \left(\alpha - \mathbb{E}\left[\alpha \vert \mathcal{F}_t\right]\right)\mathrm dt\,.\tag 1
$$
Note that, for the continuous process "$-\int\limits_{0}^{t}\left(\alpha - \mathbb{E}\left[\alpha \vert \mathcal{F}_s\right]\right) \mathrm ds$", we have,
$$
\int\limits_{0}^{t}\left(\alpha - \mathbb{E}\left[\alpha \vert \mathcal{F}_s\right]\right)^2\mathrm ds < \int\limits_{0}^{t}\left(\Big\vert\alpha\Big\vert + \Big\vert\mathbb{E}\left[\alpha \vert \mathcal{F}_s\right]\Big\vert\right)^2\mathrm ds < \int\limits_{0}^{t}\left(a + a\right)^2\mathrm ds = 4a^2t < \infty\,.
$$
This means Novikov's Condition is satisfied:
$$
\mathbb{E}\left[e^{\frac{1}{2}\int\limits_{0}^{t}\left(\alpha - \mathbb{E}\left[\alpha \vert \mathcal{F}_s\right]\right)^2\mathrm ds}\right] < e^{2a^2t} < \infty.
$$
Therefore, by a corollary of Girsanov's theorem using the preferred measure suggested by the theorem, $\hat{W}_t$ defined by $(1)$ is a $\mathcal{F}^{W}_t$-Brownian motion and, therefore, it is a $\mathcal{F}_t$-Brownian motion. 
