Forward/Backward Iteration and Stationary/Stability Suppose I have an AR(1) process of the form:
$$y_t = \phi y_{t-1} + \epsilon_t$$
where $\epsilon_t$ is a white noise process with mean zero and variance $\sigma^2$.
If $|\phi| < 1 $ , the model is called 'stable' and hence 'stationary' and I can solve it backward (i.e. backward iteration) to get : $y_t = \sum_{j=0}^{\infty} \phi^j \epsilon_{t-j}$.
If $|\phi |> 1 $ the model is not 'stable' and not 'stationary'. That is, the first two moments should not depend on time $t$. However, when I solve the above equation forward I get:
$y_t = - \sum_{j=0}^{\infty} \phi^{-(j+1)} \epsilon_{t+j+1}$
The sum here will converge and I can take the unconditional mean and variance $y_t$ to see that it does not depend on time, since $\epsilon$ is a white noise I can do this calculation quite easily.
So how come the condition that $|\phi |< 1 $ is given as the equivalent condition for weak stationarity. As I did here,  I can just solve the equation forward and get time independent mean and covariances - since the sum will converge and the errors are white noise.
I would appreciate any help.

A stochastic process is said to be weak (or covariance) stationary if its first and second moment are both constant and finite, and so time-independent, and if the autocovariance is a function of the lags only. This is exactly the case for your AR(1) process since $\phi$ is lower than one.
Indeed:
Condition 1. mean constant and finite, meaning that $E[y{_t}]= \mu_y$ $~~$ and $~~$ $\mu_y<\infty$
$E[y{_t}]= \mu_y =\sum_{j=0}^{\infty}\phi^jE[\epsilon_{t-j}] = 0    $    -> since $E[\epsilon_t]=0$, $\:$ $ \forall t $ $~$; Condition 1 satisfied.
Condition 2. variance constant and finite:
$E[y{_t^2}]$ $= \sum_{j=0}^{\infty}\phi^{2j}E[\epsilon^2_{t-j}]$= $\sigma^2_{\epsilon}$ $\sum_{j=0}^{\infty}\phi^{2j}$ = $\frac{\sigma_{\epsilon}^2}{(1-\phi^2)}$; $~~$ Condition 2 satisfied since $\ |\phi|<1$
As you can see the key point for having a finite variance (a necessary condition for weak stationarity) is that $\phi$ must be lower than 1 in module otherwise the infinite summation will not converge to a finite value
