= Solution
The <autocovariance of an MA(1) process> comes directly from shared noise terms:
$$
\boxed{\gamma_0=\sigma^2(1+\theta^2),\qquad
\gamma_1=\gamma_{-1}=\sigma^2\theta,\qquad
\gamma_k=0\quad(|k|\ge2).}
$$
Work with the zero-mean <white noise> convention and $\sigma^2>0$. Put $\mathcal H_{t-1}=\operatorname{span}(X_1,\ldots,X_{t-1})$. Suppose the preceding <finite-past innovations> $U_1,\ldots,U_{t-1}$ are mutually orthogonal. Their triangular relation to the observations means that they span the same finite past. For $s\le t-2$, $U_s$ is a <linear combination> of $X_1,\ldots,X_s$, so $\operatorname{Cov}(X_t,U_s)=0$. Moreover,
$$
\operatorname{Cov}(X_t,U_{t-1})
=\operatorname{Cov}(X_t,X_{t-1}-\lambda_{t-1}U_{t-2})
=\gamma_1.
$$
Projection onto this <orthogonal basis> therefore has only its last coefficient nonzero:
$$
\boxed{\lambda_t=\frac{\gamma_1}{\varphi_{t-1}},\qquad
\widehat X_t=\lambda_tU_{t-1},\qquad
U_t=X_t-\widehat X_t.}
$$
The residual is orthogonal to the entire preceding span, proving the induction. Expanding its <variance> gives
$$
\boxed{\varphi_t=\gamma_0-2\lambda_t\gamma_1+\lambda_t^2\varphi_{t-1}
=\gamma_0-\gamma_1\lambda_t.}
$$
The initialization is $U_1=X_1$, $\varphi_1=\gamma_0$, $\lambda_1=0$ and $U_0=0$. Each finite <covariance matrix> is a <positive-definite matrix>: the last observation in any nonzero finite <linear combination> contains a noise term absent from earlier observations. Thus all the projection denominators are positive. These are the <finite-sample innovations of an MA(1) process>, requiring no Gaussian assumption.
Substituting the <covariances> into the recursion gives
$$
\lambda_t=\frac{\theta}{1+\theta^2-\theta\lambda_{t-1}}.
$$
The denominator has a positive limit when $\lambda_{t-1}\to\lambda\in[-1,1]$. Taking limits gives $(\lambda-\theta)(\theta\lambda-1)=0$. The permitted interval selects
$$
\boxed{\lambda=\begin{cases}
\theta,&|\theta|\le1,\\
1/\theta,&|\theta|>1.
\end{cases}}
$$
For $\theta=0$, every coefficient is zero; at $\theta=\pm1$, the roots coincide. This is the <limiting MA(1) innovations coefficient>. The limiting innovation <variance> is correspondingly $\sigma^2\max(1,\theta^2)$.
For the observed numerical case, $\gamma_0=2$ and $\gamma_1=-1$. The exact innovation calculations are
$$
\lambda_2=-\tfrac12,\quad
U_1=-\tfrac{13}{10},\quad
U_2=\tfrac45-\left(-\tfrac12\right)\left(-\tfrac{13}{10}\right)=\tfrac3{20},
\quad \varphi_2=\tfrac32,\quad \lambda_3=-\tfrac23.
$$
Therefore the predictor of $X_3$ based on the two observations and its error <variance> are
$$
\boxed{\widehat X_{3\mid2}=-\tfrac1{10},\qquad
\operatorname{Var}(X_3-\widehat X_{3\mid2})=\tfrac43.}
$$
For $X_4$, the available information is still only $\mathcal H_2$. Since its <covariances> with both $X_1$ and $X_2$ are zero, its <orthogonal projection> onto that span is zero:
$$
\boxed{\widehat X_{4\mid2}=0,\qquad
\operatorname{Var}(X_4-\widehat X_{4\mid2})=2.}
$$
This is <two-step prediction for an MA(1) process>. Applying the next one-step recursion would instead use an observed $X_3$, which is not available here.
Back to article page
