\right)}{1+\exp \left( \frac{1}{2} \|f-g\|^2 \right)}$, which is larger than or equal to $1/2$ and gets closer to $1$ when $\|f-g\|$ increases. Roughly speaking, this means in that example that the distribution $\mathbb{P}_f$ is more likely than the distribution $\mathbb{P}_g$, which is consistent with the definition of the model given by .

 \
The Bayes classifier $\Phi^\star$ of the classification problem is then given by $$\label{eq:bayes_rule}
\Phi^{\star}(X) := 
\mathds{1}_{\left\lbrace \eta(X){\geqslant}\frac12 \right\rbrace} =
\mathds{1}_{\left\lbrace \int_{0}^1 (f(s)-g(s)) dX_s {\geqslant}\frac{1}{2}\|f\|^2 - \frac{1}{2}\|g\|^2  \right\rbrace} \;.$$ It is well known that the Bayes classifier $\Phi^{\star}$ corresponds to the optimal classifier of the considered binary classification problem (see, [e.g.]{}, [@DGL96]) in the sense that it satisfies . In particular, for any other classifier $\Phi$, the excess risk of classification is given by $$\label{eq:excess_risk}
\mathcal{R}_{f,g}(\Phi)-\mathcal{R}_{f,g}(\Phi^{\star}) = \mathbb{E} \left[ \left|2 \eta(X)-1\right| \mathds{1}_{\Phi(X) \neq \Phi^{\star}(X)}\right] \,.$$

In our statistical setting, the functions $f$ and $g$ are unknown so that it is impossible to compute the oracle Bayes classifier . However, we can construct an approximation of it using the sample $(X_i,Y_i)_{1 {\leqslant}i {\leqslant}n}$. In Section \[s:upper\] we design a plug-in estimator combined with a projection step, and analyze its excess risk under a smoothness assumption on $f$ and $g$. The 