# Finding variance of AR process

$\newcommand{\E}{\mathbb{E}}$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: $$\text{Var}(y)=\frac{\sigma^2}{1-\phi^2}.$$
• Thank you! Could you please elaborate, why $var(y_t)=var(y_{t−1})$? Commented Nov 12, 2016 at 18:37
• and if I get the calculations right, it should be $var(y)=\frac{(σ^2)}{1-ϕ^2}$, no? Commented Nov 12, 2016 at 19:35
• you assume variance to be constant over time, so there is no heteroskedasticity, it is to say, different variances across time. This implies: $var(y_t)=var(y_{t-1})$ Commented Nov 13, 2016 at 11:55
$$y_t = \varepsilon_{t} +\phi y_{t-1}= \varepsilon_{t} +\phi (\varepsilon_{t-1} +\phi y_{t-2}) = \sum_{j=0}^{\infty}\phi^j \varepsilon_{t-j},$$ 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.