I'm trying to find the variance of an ARMA(1,1) model of the following form: $$y_t=a_0+a_1y_{t-1}+\epsilon_t+b_1\epsilon_{t-1}$$

where $\epsilon_t$ is a white noise process. I have found it more convenient to write this model in terms of $\epsilon_t$'s:

Writing using lag operators:




$$=\sum^\infty_{j=0}a_1^jL^j(a_0+(1+b_1L)\epsilon_t) $$



Taking Variance,


$$\sum^\infty_{j=0}a_1^jVar(\epsilon_{t-j})+b_1\sum^\infty_{j=0}a_1^jVar(\epsilon_{t-j-1})+2Cov(\sum^\infty_{j=0}a_1^j\epsilon_{t-j}, b_1\sum^\infty_{j=0}a_1^j\epsilon_{t-1-j})$$

$$=\sum^j_{j=0}a_1^j\sigma^2+b_1\sum^\infty_{j=0}a_1^j\sigma^2+\sum^\infty_{j=0}a_1^j.2Cov(\epsilon_{t-j}, b_1\epsilon_{t-1-j})$$


This answer seems intuitive however it differs from ARMA (1,1) Variance Calculation


$$Var(y_t)=\frac{(1+2a_1 b_1 + b_1^2)\sigma^2}{1-a_1^2}$$

  • $\begingroup$ (A) $y_{t-1}$ is not independent of $\epsilon_{t-1}$ and I suspect that leads to the $2a_1 b_1$ term $\endgroup$
    – Henry
    Commented Apr 12, 2023 at 14:42
  • $\begingroup$ (B) $Var(kX) = k^2 Var(X)$ and I suspect that leads to the $b_1^2$ and $a_1^2$ terms being squares $\endgroup$
    – Henry
    Commented Apr 12, 2023 at 14:43

1 Answer 1


Two errors in your solution:

  1. As @Henry pointed out in the comment, you forgot to square the coefficients when you took them out of the variance. Your second line after "Taking Variance" should be

$$ \sum^\infty_{j=0}a_1^{2j}Var(\epsilon_{t-j})+b_1^2\sum^\infty_{j=0}a_1^{2j}Var(\epsilon_{t-j-1}) + \ldots $$

  1. In the same expression, the covariance contains some non-zero terms. We have: $$ \mathrm{Cov}\left(\sum^\infty_{j=0}a_1^j\epsilon_{t-j}, b_1\sum^\infty_{j=0}a_1^j\epsilon_{t-1-j}\right) = \sum^\infty_{j=1} \mathrm{Cov}\left(a_1^j\epsilon_{t-j}, b_1a_1^{j-1}\epsilon_{t-j}\right)= \sum^\infty_{j=1} b_1 a_1^{2j-1}\sigma^2 $$ This gives you the missing $a_1b_1$ term.
  • $\begingroup$ Thank you for your answer, mathemtically this makes sense to me, only thing is, how come the covariance between $\epsilon_{t-j}$ and $\epsilon_{t-j-1}$ isn't zero? Since $\epsilon_t$ is white noise, then would you not expect there to be no covariance? Thank you $\endgroup$ Commented Apr 12, 2023 at 18:03
  • $\begingroup$ The second term in the left sum ($a_1\epsilon_{t-1}$) and the first term in the right sum ($b_1\epsilon_{t-1}$) are correlated. Similarly for the subsequent terms. $\endgroup$ Commented Apr 12, 2023 at 20:46

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