I have an extremely basic question of time series on covariance stationarity. Consider $$ Y_t=\gamma Y_{t-2}+\epsilon_t $$ where $\epsilon_t$ is such that $E(\epsilon_t)=0$ $\forall t$, $Var(\epsilon_t)=E(\epsilon_t^2)=\sigma^2>0$ $\forall t$, $cov(\epsilon_t, \epsilon_s)=E(\epsilon_t \epsilon_s)=0$ $\forall s\neq t$ (i.e. $\epsilon_t$ is a white noise).
Suppose that $\gamma=1$ and $(Y_0, Y_1)=(0,0)$. I want to show that $\{Y_t\}_t$ is not covariance stationary.
I know that $\{Y_t\}_t$ is covariance stationary if $E(Y_t)$ does not depend on $t$ and $cov (Y_t, Y_{t-k})$ is independent of $t$ for $k=0,1,...$, $\forall t$.
Let's consider $Var(Y_t)$ for $t=3,4$ $$ Var(Y_3)= Var(\gamma \underbrace{Y_1}_0+\epsilon_3)=Var(\epsilon_3)=\sigma^2 $$ and $$ Var(Y_4)= Var(\gamma Y_2+\epsilon_4)=Var(\gamma(\gamma\underbrace{Y_0}_0+\epsilon_2) +\epsilon_4)=Var(\gamma \epsilon_2+\epsilon_4)= \sigma^2(\gamma^2+1)\underbrace{=}_{\gamma=1} 2\sigma^2\neq Var(Y_3) $$ Hence $\{Y_t\}_t$ is not covariance stationary.
Question: a sufficient condition for covariance stationarity is to have the roots of the polynomial equation $$ 1-\gamma L^2=0 $$ greater than $1$ in absolute value, i.e. $$ \begin{cases} |\frac{1}{\sqrt{\gamma}}|>1\\ |- \frac{1}{\sqrt{\gamma}}|>1\\ \end{cases} $$ $\leftrightarrow$ $|\gamma|<1$
Now, if I consider $\gamma=0.5$, we can see that the sufficient condition of covariance stationarity is satisfied but still $Var(Y_4)\neq Var(Y_3)$. What am I doing wrong?
The problem is that you initiate your process at time zero, while an $\operatorname{AR}$ process does not have an initialization time. Instead, what we would do is call the process with a start an approximation. Results exist about choosing optimal choice of the initial time point value (which in your case indeed is 0). The process is simply not covariance stationary close to an initialization point. This is treated in many texts about the subject under the headline stability.
Instead use the results about the polynomial as you do in your second method. From there it follows that for your case of $\gamma = 1$, the process is indeed not stationary (it is a random walk) and your second example with $\gamma = 0.5$, it is stationary.