Given that $e_t$ is a white noise process and $Y_t = e_t + 0.5(-1)^te_{t-1}$, is $\{Y_t\}$ a stationary process?
I can show that \begin{align*} \mu_Y(t) &= E[Y_e]\\ &=E[e_t + 0.5(-1)^te_{t-1}]\\ &=E[e_t] + 0.5E[(-1)^te_{t-1}]\\ &= 0 \end{align*} and \begin{align*} \gamma_{Y,t,s} &= Cov(Y_t,Y_s)\\ &=E[Y_tY_S] - E[Y_t]E[Y_s]\\ &=E[(e_t + 0.5(-1)^te_{t-1}) (e_s + 0.5(-1)^se_{s-1})]\\ &=E[e_t]E[e_s]+ 0.5(-1)^sE[e_t]E[e_{s-1}] + 0.5(-1)^tE[e_{t-1}]E[e_s] + 0.25(-1)^{t+s}E[e_{t-1}]E[e_{s-1}]\\ &=0 \end{align*} I am not sure if this step is correct: $$0.5E[(-1)^te_{t-1}] = 0.5(-1)^tE[e_{t-1}]$$ I think it is correct because the randomness comes from $e_{t-1}$.
But if it indeed is, then $\{Y_t\}$ is weakly stationary. But how can I determine whether $\{Y_t\}$ is stationary, i.e the joint distribution of $Y_{t_1},\ldots,Y_{t_n}$ is the same of the joint distribution of $Y_{t_1-k},\ldots,Y_{t_n-k}$?
Thanks all in advance.