Variance of ARMA (1,1) Model

Question

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:

$$(1-a_1L)y_t=a_0+(1+b_1L)\epsilon_t$$

Re-arranging:

$$y_t=\frac{1}{1-a_1L}(a_0+(1+b_1L)\epsilon_t)$$

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

$$=\sum^\infty_{j=0}a_1^ja_0+\sum^\infty_{j=0}a_1^jL^j(\epsilon_t+b_1\epsilon_{t-1})$$

$$=\frac{a_0}{1-a_1}+\sum^\infty_{j=0}a_1^j(\epsilon_{t-j}+b_1\epsilon_{t-1-j})$$

Taking Variance,

$$Var(y_t)=Var(\frac{a_0}{1-a_1}+\sum^\infty_{j=0}a_1^j(\epsilon_{t-j}+b_1\epsilon_{t-1-j}))$$

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

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

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

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

@Neeraaj

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

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

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

2. 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.