An MA($q$) process is invertible when its innovations admit a causal absolutely summable linear representation in present and past observations. With the backshift operator $B$, MA(1) satisfies
$$
X_t=(1+\theta B)\varepsilon_t.
$$
If $|\theta|<1$, the geometric series converges absolutely:
$$
\boxed{\varepsilon_t=(1+\theta B)^{-1}X_t
=\sum_{j=0}^\infty(-\theta)^jX_{t-j}.}
$$
Thus the process is invertible.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 218 / 6 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-218/6/b/solution