这是一份nottingham诺丁汉大学MATH4022作业代写的成功案例

For illustration, consider again the example
$$
(1-\phi B)(1-B) z_{l}=(1-\theta B) a_{l}
$$
The complementary function is the solution of the difference equation
$$
(1-\phi B)(1-B) C_{k}(t-k)=0
$$
that is
$$
C_{\mathrm{k}}(t-k)=b_{0}^{(k)}+b_{1}^{(k)} \phi^{\prime-k}
$$
where $b_{0}^{(k)}, b_{1}^{(k)}$ are coefficients which depend on the past history of the process and, it will be noted, change with the origin $k$.
Making use of the $\psi$ weights (4.2.7), a particular integral (4.2.13) is
$$
I_{h}(t-k)=\sum_{j=k+1}^{t}\left(A_{0}+A_{1} \phi^{t-j}\right) a_{j}
$$

MATH4022COURSE NOTES ：

For illustration, consider once more the model
$$
(1-\phi B)(1-B) z_{e}=(1-\theta B) a_{\mathrm{r}}
$$
We can write $z_{t}$ either as an infinite weighted sum of the $a_{j}$ ‘s
$$
z_{l}=\sum_{j=-\infty}^{t}\left(A_{0}+A_{1} \phi^{r-j}\right) a_{j}
$$
or in terms of the weighted finite sum as
$$
z_{t}=C_{k}(t-k)+\sum_{j=k \neq 1}^{t}\left(A_{0}+A_{1} \phi^{r-j}\right) a_{j}
$$
Furthermore, the complementary function is the truncated sum
$$
C_{k}(t-k)=\sum_{j=-\infty}^{k}\left(A_{0}+A_{t} \phi^{r-j}\right) a_{j}
$$
which can be written
$$
C_{k}(t-k)=b_{0}^{(k)}+b_{1}^{(k)} \phi^{r-k}
$$