When ARMA is well defined?

Given the steady state output of the recursive equation defined by the transfer function $W(z)$ fed by the stochastic process $v(t)$: $$y(t)=W(z)v(t)$$ Then $y(t)$ is well defined iff:

1. $v(t)$ is stationary
2. $W(z)$ is assintotically stable

So ARMA process $y(t)=W(z)e(t)$ with $e(t) \sim WN(0,\lambda^2)$ is well defined if $W(z)$ is assintotically stable (because the white noise is stationary by definition).

Idea of proof

1. Suppose $A(z)=1$: $$y(t)=C(z)v(t)=c_0v(t)+c_1v(t-1)+...+c_nv(t-n)$$ the result is true by direct verification.
2. Suppose: $$y(t)=\frac{1}{1-az^{-1}}v(t)=\frac{z}{z-a}v(t)$$ then there is a single pole $z=a$ and a direct computation of the steady state solution reveals that if $|a|<1$ then $y(t)$ is well defined and stationary.
3. Consider now the general case: $$y(t)=\frac{C(z)}{A(z)}v(t)=\frac{C(z)}{(z-p_1)(z-p_2)...(z-p_m)}v(t)=C(z)\frac{1}{(z-p_1)}\frac{1}{(z-p_2)}...\frac{1}{(z-p_m)}v(t)$$ it is a series of simple operators which correspond to case 1 or 2.