这是一份nottingham诺丁汉大学MATH3026作业代写的成功案例

where $S$ is the infinite summation operator defined by
$$
\begin{aligned}
S x_{t}=\sum_{h=-\infty}^{t} x_{h} &=\left(1+B+B^{2}+\cdots\right) x_{t} \
&=(1-B)^{-1} x_{t}=\nabla^{-1} x_{t}
\end{aligned}
$$
Thus
$$
S=(1-B)^{-1}=\nabla^{-1}
$$
The operator $S^{2} x_{t}$ is similarly defined as
$$
\begin{aligned}
S^{2} x_{t} &=S x_{t}+S x_{t-1}+S x_{\mathrm{r}-2}+\cdots \
&=\sum_{i=-\infty}^{t} \sum_{k=-\infty}^{i} x_{h}
\end{aligned}
$$
Also
$$
S^{3} x_{t}=\sum_{j=-\infty}^{t} \sum_{i=-\infty}^{j} \sum_{h=-\infty}^{i} x_{h}
$$
and so on.

MATH3026 COURSE NOTES ：

(I) The $(0, l, l)$ process.
$$
\begin{aligned}
\nabla z_{t} &=a_{t}-\theta_{1} a_{t-1} \
&=\left(1-\theta_{1} B\right) a_{t}
\end{aligned}
$$
corresponding to $p=0, d=1, q=1, \phi(B)=1, \theta(B)=1-\theta_{1} B$.
(2) The $(0,2,2)$ process.
$$
\begin{aligned}
\nabla^{2} z_{1} &=a_{t}-\theta_{1} a_{t-1}-\theta_{2} a_{t-2} \
&=\left(1-\theta_{1} B-\theta_{2} B^{2}\right) a_{t}
\end{aligned}
$$
corresponding to $p=0, d=2, q=2, \phi(B)=1, \theta(B)=1-\theta_{1} B-\theta_{2} B^{2}$.
(3) The $(I, l, l)$ process.
$$
\nabla z_{t}-\phi_{1} \nabla z_{t-1}=a_{t}-\theta_{1} a_{t-1}
$$
or
$$
\left(1-\phi_{1} B\right) \nabla z_{t}=\left(1-\theta_{1} B\right) a_{\mathrm{r}}
$$
corresponding to $p=1, d=1, q=1, \phi(B)=1-\phi_{1} B, \theta(B)=1-\theta_{1} B$.