Assume that $ G $ is not trivial. Then, for any noise law $ \mu $, the case [[$ ({\bf C0}) $]{}]{} in the table never occurs.

In the sequel, we always assume that $ G $ is not trivial. Now the problem is how to characterize the trichotomy [[$ ({\bf C1}) $]{}]{}-[[$ ({\bf C3}) $]{}]{}.

The key fact to the proof of Theorem \[YorTh\] is the one-to-one and onto correspondence of each solution to what we call an entrance law.

Let $ \mu = (\mu_k , k \in -\N) $ be given.

A family $ \lambda = (\lambda_k , k \in -\N) $ of probability laws $ \lambda_k $ on $ G $ is called an [*entrance law*]{} for the noise law $ \mu $ if the following recurrence relation holds: $$\lambda_k = \lambda_{k-1} * \mu_k , 
\quad k \in -\N. 
\label{consistency}$$ Here $ \mu * \lambda $ for two measures $ \mu $ and $ \lambda $ on $ G $ stands for the convolution of $ \mu $ and $ \lambda $: $$\int_G \phi(g) \mu * \lambda (dg) = \int_G \int_G \phi(gh) \mu(dg) \lambda(dh)$$ for any bounded measurable function $ \phi $ on $ G $.

For 