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I really need help with this problem and would appreciate any input:

When $t$ denotes the time-period, terms $\alpha$, $\phi_1$, and $\theta_1$ are constants, $a_t$ represents error-terms that are NID(0, $\sigma^2$) if a variable r is modeled as ARMA(1,1) process,

$$r_t=\alpha + \phi_1 r_{t-1} + \theta_1 a_{t-1} + a_{t}$$

What is variance of the $r_t$?

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It is better to write your model like this: $$r_t=\alpha + \phi_1 r_{t-1} + \theta_1 a_{t-1} + a_{t}$$ where, {$a_t$} is white noise series.

First, multiply the model by $a_t$ and take expectation: \begin{align} E(r_t a_t)&=\alpha E(a_t)+ \phi_1 E(r_{t-1}a_t)+\theta_1 E(a_{t-1}a_t)+ E(a_t^2)\\ &=E(a_t^2)=\sigma^2 \end{align}

Here, we make use of the fact that $E(a_t r_{t-1})=0$, $E(a_t)=0$ and $E(a_{t-1}a_t)=0$.

Taking the variance of original equation, we have \begin{align} Var(r_t)&=\phi_1^2 Var(r_{t-1})+\theta_1^2\sigma^2+\sigma^2+2 \phi_1 \theta_1 E(r_{t-1} a_{t-1}) \\ &=\phi_1^2 Var(r_{t-1})+\theta_1^2\sigma^2+\sigma^2+2 \phi_1 \theta_1 \sigma^2 \end{align} If $r_t$ is weakly stationary, then $Var(r_t)=Var(r_{t-1})$, so $$Var(r_t)=\frac{(1+2\phi_1 \theta_1 + \theta_1^2)\sigma^2}{1-\phi_1^2}$$

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    $\begingroup$ In your second equation, you are a bit quick. You skip two terms that are equal to zero; out of three terms that are equal to zero you comment on only one in the subsequent line. Further derivation is also quite brief. Otherwise it's a nice answer. $\endgroup$ – Richard Hardy Feb 17 '16 at 10:13
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    $\begingroup$ Now the second line in the derivation of $\mathbb{E}$ is redundant. $\endgroup$ – Richard Hardy Feb 17 '16 at 10:32
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    $\begingroup$ Hi, thanks for your answer, it's very helpful. I'm not quite sure how to type equations in the way you have done (it's my first time posting here), however, I wanted to ask whether there is a more systematic way of solving for the autocovariance of lag 0,1,2,...,h of such a process? Basically, I know how to calculate the function for an ARMA(1,1) without a constant but I can't find any textbook with a derivation with constant... $\endgroup$ – Vladmir Putin Feb 17 '16 at 10:33
  • $\begingroup$ @RichardHardy Thanks for your valuable inputs. I think it is OK now. $\endgroup$ – Neeraj Feb 17 '16 at 10:44
  • $\begingroup$ Hi, this answer is actually incorrect. I read the book you suggested and from where I believe you provided the derivation however the whole point was to keep the constant term rather than setting it equal to zero. As I said in the original question the answer is 1.161 times sigma-squared which is not what I am getting when the constant is set to zero.. $\endgroup$ – Vladmir Putin Feb 18 '16 at 2:03

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