now make a few comments on the white noise model considered in  (see also [@IH81] for further details and its link with the infinite Gaussian sequence model). First note that, for all $\varphi \in {\mathbb{L}^2}([0,1])$, almost surely, $$\int_{0}^1 \varphi(t) dX(t) = 
Y \int_{0}^1 f(t) \varphi(t) dt + (1-Y) \int_{0}^1 g(t) \varphi(t) dt + \int_{0}^1 \varphi(t) dW(t) \,.$$ Recall that, with a slight abuse of notation, we write $\langle \varphi,X \rangle := \int_{0}^1 \varphi(t) dX(t)$. The above almost sure equality implies that the conditional distribution of $\langle \varphi,X \rangle$ given $Y$ is Gaussian with expectation $\langle \varphi,f \rangle \mathds{1}_{Y=1} + \langle \varphi,g \rangle \mathds{1}_{Y=0}$ and variance $\|\varphi\|^2$. Therefore, the distribution of $\langle \varphi , X\rangle$ is a mixture of two Gaussian distributions: $$\frac{1}{2} \mathcal{N}\Bigl(\langle \varphi,f\rangle,\|\varphi\|^2\Bigr) + \frac{1}{2} \mathcal{N}\Bigl(\langle\varphi,g\rangle,\|\varphi\|^2\Bigr) \,.$$

 \
An important feature of the white noise model is that the coefficients $\bigl(\langle \varphi_j,W \rangle\bigr)_{j {\geqslant}1}$ associated with different frequencies of the standard Brownian motion are independent. This is because they are jointly Gaussian, with a diagonal infinite covariance matrix: $$\forall j \neq j',  \qquad \mathbb{E} \Big[\langle \varphi_j,W \rangle \langle \varphi_{j'},W \rangle\Big] = \int_{0}^1 \varphi_j(t) \varphi_{j'}(t) d t = 0 \,.$$ The above remarks imply together with  that $$\label{eq:Gaussiancoeffs}
\forall j 