6
$\begingroup$

$\newcommand{\E}{\mathbb{E}}$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 $\operatorname{Var}(y_t)=\sigma^2\sum_{j=0}^\infty \varphi^2_j$

$\endgroup$
2
$\begingroup$

$$\text{Var}(y_t)=\text{Var}(\phi y_{t-1}) + \text{Var}(\varepsilon_{t}).$$ 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: $$\text{Var}(y)=\frac{\sigma^2}{1-\phi^2}.$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you! Could you please elaborate, why $var(y_t)=var(y_{t−1})$? $\endgroup$ – lovetimberland Nov 12 '16 at 18:37
  • 2
    $\begingroup$ and if I get the calculations right, it should be $var(y)=\frac{(σ^2)}{1-ϕ^2}$, no? $\endgroup$ – lovetimberland Nov 12 '16 at 19:35
  • $\begingroup$ you assume variance to be constant over time, so there is no heteroskedasticity, it is to say, different variances across time. This implies: $var(y_t)=var(y_{t-1})$ $\endgroup$ – adrian1121 Nov 13 '16 at 11:55
1
$\begingroup$

$$y_t = \varepsilon_{t} +\phi y_{t-1}= \varepsilon_{t} +\phi (\varepsilon_{t-1} +\phi y_{t-2}) = \sum_{j=0}^{\infty}\phi^j \varepsilon_{t-j},$$ 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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.