I am trying to understand the following recursion,
$$Y_{t}=(1.2)Y_{t-1}-(0.4)Y_{t-2}+Z_{t} $$ $t \in \mathbb{Z}$
Where ${Z_{t}}$ are iid $N(0,\sigma^{2}$))
My trouble is that the $1.2$ factor seems to make it blow up as we iterate the recursion, thus giving me difficulty in getting a form for $Y_{t}.$
I do see that we will have a term,
$$\sum_{g=0}^{\infty} (1.1)^{g} Z_{t-g}$$
but I don't know how to account for the rest of the recursion that goes on forever since t includes negative integers as well. Along with accounting for the (1.2) factor which will go to infinity.
Also something that seems interesting is that if we try to calculate the mean of both sides it seems to imply that $\mu=0$
This is not any different from the deterministic case. You have
$$Y_t-1.2 Y_{t-1}+0.4Y_{t-2}=Z_t$$
where $Z_t$ doesn't depend explicitly on $Y_t$. The general homogeneous solution is
$$H_t=c_1 r_1^t + c_2 r_2^t$$
where $r_1,r_2$ are solutions to $\lambda^2-1.2\lambda+0.4=0$, which are easily found. The particular solution is the convolution of $Z_t$ with a particular homogeneous solution $P_t$. I think it will be the one with $P_1=1,P_0=0$, whose coefficients are the solution to
$$\begin{bmatrix} r_1 & r_2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$
which is the first column of the inverse of the coefficient matrix, i.e. $\begin{bmatrix} \frac{1}{r_1-r_2} \\ \frac{-1}{r_1-r_2} \end{bmatrix}$. ($r_1$ and $r_2$ are indeed different in this case).
The general solution to the inhomogeneous problem is the sum of a particular solution and the general homogeneous solution. Note that this is generally not bounded in either direction in time, since $0<r_1<1<r_2$.