10
$\begingroup$

For the autoregressive AR(1) process $x_t = \delta + \phi x_{t-1} + \eta_t$, I am trying to prove that the variance is:

$\sigma_x^2 = \sigma_\eta^2/(1-\phi^2)$

And that the first-order covariance is:

$\gamma_{1,x} = \phi \sigma_x^2$.

I have tried many manipulations but I cannot succeed. I have the feeling that I didn't find the correct form yet in which I should write the process before I take expectations. Could anyone please help? Thanks in advance.

$\endgroup$
1
  • 3
    $\begingroup$ Technically, it doesn't have to actually be homework, just equivalent to homework. Please don't take offense, we get a lot of people posting their homework here hoping someone will do it for them. So it's a sensitive topic. $\endgroup$ Commented Jun 15, 2014 at 12:06

2 Answers 2

11
$\begingroup$

$Var(X_t)=var(\delta+\phi x_{t-1}+\eta_t)=0+var(\phi x_{t-1})+var(\eta_t)\\ =\phi^2var(x_{t-1})+\sigma^2_{\eta}=\frac{var(\eta_{t})}{1-\phi^2}=\frac{\sigma^2_{\eta}}{1-\phi^2}$

$\endgroup$
3
  • 1
    $\begingroup$ Please note that - as explained in the tag wiki info of the self-study tag, it's requested that for questions that are routine bookwork (the sort of question that might be set for an assignment or an exercise for example, whether it is assigned or not) that we don't provide full solutions, but lean more toward hints and guidance. $\endgroup$
    – Glen_b
    Commented Jun 15, 2014 at 14:11
  • $\begingroup$ (PS I am not suggesting that your answer should be changed or removed; it's a suggestion for similar questions in future) $\endgroup$
    – Glen_b
    Commented Jun 15, 2014 at 14:54
  • $\begingroup$ You are assuming that there is no any correlation between the three terms in the sum. This is not obvious sincd $x_{t-1}$ is a function of $\eta_t$. $\endgroup$ Commented Mar 6, 2023 at 14:44
10
$\begingroup$

\begin{align} \mathrm{Var}\left[X_t\right] &= \mathrm{E}\left[X_t\left(\delta+\phi X_{t-1}+\eta_t\right)\right]-\mathrm{E}^2\left[X_t\right]\\ &= \delta\mathrm{E}\left[X_t\right]+\phi\mathrm{E}\left[X_tX_{t-1}\right]+\mathrm{E}\left[X_t\eta_t\right]-\frac{\delta^2}{\left(1-\phi\right)^2}\\ &= \delta\mathrm{E}\left[X_t\right]+\phi\left(\mathrm{Cov}\left[X_t,X_{t-1}\right]+\mathrm{E}\left[X_t\right]\mathrm{E}\left[X_{t-1}\right]\right)+\mathrm{E}\left[X_t\eta_t\right]-\frac{\delta^2}{\left(1-\phi\right)^2}\\ &= -\frac{\phi\delta^2}{\left(1-\phi\right)^2}+\phi\left(\gamma_{1,x}+\frac{\delta^2}{\left(1-\phi\right)^2}\right)+\sigma_\eta^2\\ &= \phi^2\sigma_x^2+\sigma_\eta^2\\ &= \phi^2\mathrm{Var}\left[X_t\right]+\sigma_\eta^2\\ &\\ \mathrm{Var}\left[X_t\right] &= \frac{\sigma_\eta^2}{1-\phi^2}\\ \end{align}

$\endgroup$
2
  • 2
    $\begingroup$ Please note that - as explained in the tag wiki info of the self-study tag, it's requested that for questions that are routine bookwork (the sort of question that might be set for an assignment or an exercise for example, whether it is assigned or not) that we don't provide full solutions, but lean more toward hints and guidance. $\endgroup$
    – Glen_b
    Commented Jun 15, 2014 at 14:12
  • $\begingroup$ (PS I am not suggesting that your answer should be changed or removed; it's a suggestion for similar questions in future) $\endgroup$
    – Glen_b
    Commented Jun 15, 2014 at 14:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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