of continuous functions on $[0,1]$. It is easy to check that the law of $X \vert Y$ is absolutely continuous with respect to $\mathbb{P}_0$ (see, [e.g.]{}, [@IW_1989]). Indeed, for any continuous trajectory $X$, the Girsanov formula implies that the density of ${\mathbb{P}}_f$ ([i.e.]{}, of $X|\{Y=1\}$) with respect to the reference measure $\mathbb{P}_0$ is given by $$q_f(X) := 
\frac{d \mathbb{P}_f}{d \mathbb{P}_0}(X) = \exp \left(\int_{0}^1 f(s) dX_s - \frac{1}{2}\|f\|^2 \right) \,.
\label{eq:qf}$$ Similarly, the density of ${\mathbb{P}}_g$ ([i.e.]{}, of $X|\{Y=0\}$) with respect to $\mathbb{P}_0$ is $$q_g(X) := \frac{d \mathbb{P}_g}{d \mathbb{P}_0}(X) = \exp \left(\int_{0}^1 g(s) dX_s - \frac{1}{2}\|g\|^2 \right) \,.
\label{eq:qg}$$ In the sequel we refer to $q_f$ and $q_g$ as the likelihood ratios of the models ${\mathbb{P}}_f$ and ${\mathbb{P}}_g$ versus ${\mathbb{P}}_0$. Now, using the Bayes formula, we can easily see that the regression function $\eta$ associated with (\[eq:model\]) is given by $$\begin{aligned}
\eta(X) &:=& \mathbb{E}[Y \vert X] = {\mathbb{P}}(Y=1 \vert X) \nonumber \\
&= &\frac{\frac{d {\mathbb{P}}_f}{d {\mathbb{P}}_0}(X)}{ \frac{d {\mathbb{P}}_f}{d {\mathbb{P}}_0}(X)+\frac{d {\mathbb{P}}_g}{d {\mathbb{P}}_0}(X)} \nonumber \\
& = & \frac{\exp\left(\int_{0}^1 (f(s)-g(s)) dX_s - \frac{1}{2}\|f\|^2 + \frac{1}{2}\|g\|^2\right)}{1+\exp\left(\int_{0}^1 (f(s)-g(s)) dX_s - \frac{1}{2}\|f\|^2 + \frac{1}{2}\|g\|^2\right)} \;.
\label{eq:eta}\end{aligned}$$ As an example, if we assume that we observe $dX(t)=f(t) dt$ with $X(0)=0$, then $\eta(X) = \frac{\exp \left( \frac{1}{2} \|f-g\|^2 