$\newcommand{\E}{\mathbb{E}}$How do I find the variance of an autoregressive AR(1) process $$y_t=\phi y_{t-1}+\varepsilon_{t}$$
where $\lvert {\phi}\rvert<1$ and knowing that
$$y_t=\sum_{j=0}^\infty \varphi_j\varepsilon_{t-j}$$
$\E(\varepsilon_t)=0,\ \E(\varepsilon^2_t) = \sigma^2,\ \E(\varepsilon_t\varepsilon_s)=0$ for $s\ne t,\ $ $\sum_{j=0}^\infty \varphi^2_j<\infty$ and $\operatorname{Var}(y_t)=\sigma^2\sum_{j=0}^\infty \varphi^2_j$
$$\text{Var}(y_t)=\text{Var}(\phi y_{t-1}) + \text{Var}(\varepsilon_{t}).$$ As we know, $E(\varepsilon_{t}^2)=\sigma^2$. Then we have: $$\text{Var}(y_t)=\text{Var}(\phi y_{t-1}) + \sigma^2.$$ Now using variance properties we take out $\phi$ from the variance: $$\text{Var}(y_t)=\phi^2\text{Var}(y_{t-1}) + \sigma^2.$$ Given that $\text{Var}(y_t)=\text{Var}(y_{t-1})$ we solve to get: $$\text{Var}(y)=\frac{\sigma^2}{1-\phi^2}.$$
$$y_t = \varepsilon_{t} +\phi y_{t-1}= \varepsilon_{t} +\phi (\varepsilon_{t-1} +\phi y_{t-2}) = \sum_{j=0}^{\infty}\phi^j \varepsilon_{t-j},$$ and If $|\phi|<1$, then $$\text{Var}(y_t) = \sum_{j=0}^{\infty}(\phi^j)^2 \text{Var}(\varepsilon_{t-j})=\sigma^2 (1+\phi^2+\phi^4+\dots) = \sigma^2\frac{1-\lim_{n \rightarrow \infty} \phi^{2n}}{1-\phi^2} = \frac{\sigma^2}{1-\phi^2}.$$
P.S. From here, we can conclude that $\text{Var}(y_t)$ isn't a function of time.
