$$Y_t=\beta_1Y_{t-1}+\epsilon_t$$ $$\text{with}$$ $$\epsilon_t=\beta_2\epsilon_{t-1}+u_t, ~~~~~~~~~~\epsilon_t \sim i.i.d.(0, \sigma^2)$$ $$\text{where}$$ $$\beta_1 \ne 0, ~~~~~~~~~ |\beta_2|<1$$

So far I know that both $Y_t=\beta_1Y_{t-1}+\epsilon_t$ and $\epsilon_t=\beta_2\epsilon_{t-1}+u_t$ are AR(1) processes but the sum of two AR(1) gives an ARMA(2,1) model, this isn't our case.

From this point, which steps do I need to follow in order to prove that $Y_t$ is an AR(2) model?


1 Answer 1


Using lag operators the first equation can be rewritten as

$$ (1-\beta_1 L) Y_t = \epsilon_t $$


$$ (1- \beta_2 L) \epsilon_t = u_t$$

Hence since the latter is causal (because $|\beta_2| <1$) $$ (1-\beta_1 L) Y_t = \frac{u_t}{1- \beta_2 L} $$


$$ (1-\beta_1 L)(1- \beta_2 L) Y_t = (1-(\beta_1+\beta_2)L+\beta_1 \beta_2 L^2)Y_t= u_t$$

Hence $Y_t$ is an AR(2) process with coefficients $a_1=\beta_1+\beta_2, a_2= -\beta_1\beta_2$


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