“…In Section 6.1 we will explain how this Poisson problem emerges in the context of averaging. For the time being it suffices to say that the operator L x is the generator of the process Y x,y t := Y 1,x,y t which solves the SDE dY x,y t = g(x, Y x,y t )dt + √ 2a(x, Y x,y t ) dB t , (9) and is obtained from (3) by setting = 1; clearly, for each x fixed, the asymptotic behaviour as t → ∞ of ( 9) is the same as the asymptotic behaviour as → 0 of (3) and the invariant measure µ x (dy) (of either processes) is precisely the only probability measure such that R d (L x f )(x, y)µ x (dy) = 0 , (10) for every f in the domain of L x . Under our assumptions (see Section 2) such an invariant measure exists and, since we will take L x to be elliptic, it is unique and it has a smooth density (for every x ∈ R n ); with abuse of notation, we still denote the density by µ x (y).…”