\bQ $, whereas both $ \F^{\eta^1}_{-\infty } $ and $ \F^{\eta^2}_{-\infty } $ are trivial.*]{}

Consider the STYE on the direct product group $ G \times G $ $$(\eta^1_k,\eta^2_k) = (\eta^1_{k-1},\eta^2_{k-1}) \cdot (\xi^1_k,\xi^2_k) , 
\quad k \in -\N$$ with the noise law $ {\widetilde}{\mu}=({\widetilde}{\mu}_k : k \in -\N) $ given by $$\int_{G \times G} \phi(g,h) {\widetilde}{\mu}_k(dg \times dh) = \int_G \phi(g,g) \mu_k(dg)$$ for any bounded measurable function $ \phi $ on $ G \times G $ and for $ k \in -\N $. Denote the set of solutions by $ {\widetilde}{\sP}_{{\widetilde}{\mu}} $. We denote the marginal laws of $ (\eta^1,\eta^2) $ under the measure $ \bQ $ by $ {\widetilde}{\bP} $. Then $ {\widetilde}{\bP} $ and the regular conditional probabilities $ {\widetilde}{\bP}(\cdot \mid \eps=g) $ belong to $ {\widetilde}{\sP}_{{\widetilde}{\mu}} $. Note that implies that $${\widetilde}{\bP}(\cdot) = \int_G {\widetilde}{\bP}(\cdot \mid \eps=g) \bQ( \eps \in dg ) .$$ In this integral expression of the solution $ {\widetilde}{\bP} $, the integrand $ {\widetilde}{\bP}(\cdot \mid \eps=g) $ belongs to $ {\rm ex}({\widetilde}{\sP}_{{\widetilde}{\mu}}) $ for $ \bQ( \eps \in dg ) $-almost every $ g \in G $. In fact, by and , we see that $ \F^{\eta^1,\eta^2}_{-\infty } $ is $ {\widetilde}{\bP}(\cdot \mid \eps=g) 