How to determine covariant stationary values?

Hi: Here's an attempt at a heuristic solution.

Dividing both sides by $(1 - rho L)z_{t}$ ( using $rho$ instead of B and $epsilon_t$ instead of $u_{t}$ because that notation is easier for me ), gives

$z_{t} = frac{a}{1-rho L} + frac{epsilon_t}{1 - rho L} + frac{epsilon_{t-1}}{ 1 - rho L} $

= $frac{a}{1-rho L} + sum_{i=0}^{infty} rho^{i} epsilon_{t-i} + sum_{i=0}^{infty} rho^{i} epsilon_{t-i-1}$

This means that we have a similiar expression for $z_{t+1}$.

$ z_{t+1} = frac{a}{1-rho L} + sum_{i=0}^{infty} rho^{i} epsilon_{t-i+1} + sum_{i=0}^{infty} rho^{i} epsilon_{t-i}$

To calculate $cov(z_{t}, z_{t+1})$, the first term on the RHS doesn't play a role since it's not random. Now, since the $epsilon_{t}$ are independent, one only needs to find the terms in the two expressions where the same $epsilon_{t}$ occurs because that will result in a non-zero covariance term. The second infinite sum in $z_{t+1}$ coincides exactly with the first infinite sum in $z_{t}$ so all of those terms coincide without any shifting. Next, the terms in the second infinite sum of $z_{t}$, shifted over two time periods, coincide exactly with the terms in the first infinite sum of $z_{t+1}$ so all of those terms coincide.

So, now it should be clear that $cov(y_{t}, y_{t+1})$ will be equal to $cov(y_{t+k}, y_{t+k+1})$ because, with the terms in each expression be shifted over by k time units, the term terms that coincide won't change. Of course, this is a heuristic argument but hopefully you can work out the details.