Does autocorrelation imply stationarity? Let 
$$
\begin{aligned}
y_t &= a + bx_t + u_t, \\ 
u_t &= \phi u_{t-1} + e_t
\end{aligned}
$$
where $ e_t$ follows a White Noise process.
Let Breusch-Godfrey LM test statistic be strictly greater than the relevant $\chi^2$ critical value.
I had a similar question in my problem set where it asks me to determine the stationarity of $y_t$. In the solution it concluded that since there exist autocorrelation in the given model, it implies that $\phi < 1$.
I am suspicious of this argument and want to know if there is a hidden assumption that was made before. Any idea will be appreciated.
 A: Under weak stationarity, it must be that the covariance function $C(Y_t,Y_{t+\tau})$ depends only on $\tau$, the mean function $E(Y_t)$ is constant for all $t$, and these moments exist (are finite!).
Assume exogeneity of $U_t$ with respect to $X_\tau$ for all $\tau$. Then you can write
$C(Y_t,Y_{t+\tau}) = b^2C(X_t,X_{t+\tau}) + C(U_t,U_{t+\tau})$. Assume $X_t$ is a stationary process, you get
$C(Y_t,Y_{t+\tau}) = k_X(\tau) + C(U_t,U_{t+\tau})$. Substitution gives
$C(Y_t,Y_{t+\tau}) = k_X(\tau) + C(U_t,\phi^{\tau}U_t + \sum^{\tau-1}_{l=0}\phi^{l-1}e_{t+\tau-l}) = k_X(\tau) + C(U_t,\phi^{\tau}U_t)$,
since $e_t$ is white noise. The last term is crucial for the weak stationarity of $Y_t$. You have
$C(U_t,\phi^{\tau}U_t) = \phi^{\tau}C(U_t,U_t) = \phi^{\tau}V(U_t)$.
How can you obtain $V(U_t)$? Transform $U_t = \phi L U_{t} +e_t$ into $U_t(1-\phi L) =  e_t$. Hence,
$V(U_t) = V(\frac{e_t}{1-\phi L}) = V(\sum_{l=0}^{\infty}\phi^{l}e_{t-l})$, where the second equality holds only if $|\phi|\leq 1$.
Now, from the white noise assumption, you can simplify this last expression into
$\sum_{l=0}^{\infty} \phi^{2l} V(e_{t}) = \frac{1}{1-\phi^2}\sigma_e$. For $\phi=1$, you obtain that $V(U_t)$ is not finite, hence $C(Y_t,Y_{t+\tau})$ is not finite, and the process $Y_t$ is not stationary.
This shows that autocorrelated process may not be stationary.
However, it is also true that uncorrelated processes may not be stationary.
Consider the simple model: $Y_t = t + e_t$.
$E(Y_t) = t$, hence depends on $t$ which violates the stationarity assumption.
