Manipulation of Time Series AR(2) process

Question

Consider the following Time Series: $$X(t+1)−X(t) = β(X(t)−X(t−1))−α(X(t)−μt)+e(t),$$ $t\in\mathbb Z$ where $\alpha$, $\beta$ are positive constants and $e$ is a white noise.

Consider the process: $$Z(t) = X(t)− μt − \frac{(β−1)μ}{(1+α)} $$

Show that $Z$ is the AR(2) process given by $$Z(t)=(β−α)Z(t−1)−βZ(t−2)+e(t), t∈\mathbb Z,$$

I have tried calculating $X(t-1)$ and $X(t-2)$ from the second equation and substituting into the first but it gets quite messy. I am having trouble eliminating the $X(t+1)$ term.

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Answer 1

Rewrite the equation for $X$ as $$\tag{*} X(t)-X(t-1)=\beta\left(X(t-1)-X(t-2)\right)-\alpha\left(X(t-1)-\mu(t-1)\right)+e(t-1). $$ Then we replace $X(t)$ by $$ Z(t)+\mu t+C $$ and similarly for $X(t-1)$ and $X(t-2)$. We have $$ X(t)-X(t-1)=Z(t)-Z(t-1)+\mu; $$ $$ X(t-1)-X(t-2)=Z(t-1)-Z(t-2)+\mu $$ $$ X(t-1)-\mu(t-1)=Z(t-1)+C $$ hence plugging these terms in (*) gives $$ Z(t)-Z(t-1)+\mu= \beta(Z(t-1)-Z(t-2)+\mu)-\alpha(Z(t-1)+C)+e(t-1) $$ and rearranging $$ Z(t)=(\beta-\alpha+1)Z(t-1)- \beta Z(t-2)+e(t-1)-\mu-\alpha C+\mu \beta $$ hence with $C$ such that $\mu+\alpha C-\mu\beta=0$, that is, $C=\mu(\beta-1)/\alpha$, we get the wanted formula, up to the differences:

• the noise term is $e(t-1)$ instead of $e(t)$;
• the coefficient in front of $Z(t-1)$ is $\beta-\alpha+1$ instead of $\beta-\alpha$.