Let us have the following AutoRegression process, expressed in term of lag operators:
$$ \begin{pmatrix} 1-0.4L & 0.1L \\ 0.1L & 1-0.4L \end{pmatrix} \begin{pmatrix} x_t \\ y_t \end{pmatrix} = \begin{pmatrix} 10 + e_{1t} \\ 1 + e_{2t} \end{pmatrix}$$
I have checked that $1-0.8L+0.15L^2$ has roots $\frac{10}{3}$ and $2$ so the process is stationary. I know how to find mean for a one-dimensional process, which is just applying "inverse operator" to the constant term. You just omit $L$s
I tried to do the same $$ \frac{1}{0.35} \begin{pmatrix} 0.6 & -0.1 \\ -0.1 & 0.6 \end{pmatrix} \begin{pmatrix} 10 \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{10}{7} \\ \frac{40}{35} \end{pmatrix}$$
Is this method still legit?
I think you are considering a vector autoregressive (VAR) process. VAR($p$) is written as:
$$\mathbf{\pi}(L)\mathbf{Y}_t = \mathbf{c} +\epsilon_t$$
where $\mathbf{\pi}(L)=\mathbf{I}_n-\pi_1L-...-\pi_pL^p$, and VAR($p$) is said to be stable if $\det\left(\mathbf{\pi}(L)=\mathbf{I}_n-\pi_1z-...-\pi_pz^p\right)$ have modulus greater than one. If it is stationary then: $$\mathbf{\mu} = (\mathbf{I}_n-\pi_1-...-\pi_p)^{-1}\mathbf{c}$$
So, for bivariate VAR($1$) model: $\mathbf{Y}_t=\pi\mathbf{Y}_{t-1}+\epsilon_t$ $$\begin{matrix} y_{1t} \\ y_{2t} \end{matrix} = \begin{matrix} \pi_{11} & \pi_{12} \\ \pi_{21} & \pi_{22} \end{matrix}\times\begin{matrix} y_{1t-1} \\ y_{2t-1} \end{matrix}+\begin{matrix} \epsilon_{1t} \\ \epsilon_{2t} \end{matrix}$$ Then $\det(\mathbf{I}_2-\pi z)=0$ becomes: $$(1-\pi_{11})(1-\pi_22 z)-\pi_{12}\pi_{21}z^2=0$$ Notice if $\pi_{12}=\pi_{21}=0$ then it reduces to univariate stability condition. I will leave it here, as extensive material is available online on this matter.
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