= Solution
Write $X_t=\sum_{j\geq0}\psi_jZ_{t-j}$, with innovation <variance> four. The generating function of this <linear process> is
$$
\Psi(z)=\frac{1+0.5z}{1+0.30z-0.28z^2}.
$$
Equating coefficients gives $\psi_0=1$, $\psi_1=0.2$ and $\psi_j=-0.3\psi_{j-1}+0.28\psi_{j-2}$ for $j\geq2$. Thus the first five coefficients, including lag zero, are
$$
\boxed{(\psi_0,\psi_1,\psi_2,\psi_3,\psi_4)=(1,\ 0.2,\ 0.22,\ -0.01,\ 0.0646).}
$$
For all remaining lags, partial fractions give
$$
\Psi(z)=\frac{9/11}{1-0.4z}+\frac{2/11}{1+0.7z},\qquad
\boxed{\psi_j=\frac9{11}(0.4)^j+\frac2{11}(-0.7)^j\quad(j\geq0).}
$$
The coefficients are absolutely summable, so the series converges in mean square and defines the causal moving-average expansion. In the unit-<variance> convention of part (c), $X_t=\sum_{j\geq0}c_jW_{t-j}$ with $c_j=2\psi_j$; its first five coefficients are $(2,0.4,0.44,-0.02,0.1292)$. This is the <two-geometric-coefficient expansion of a causal ARMA(2,1) process>.
<White noise> orthogonality gives, for any integer $h$,
$$
\boxed{\gamma(h)=4\sum_{j=0}^\infty\psi_j\psi_{j+|h|}=\sum_{j=0}^\infty c_jc_{j+|h|},\qquad
\rho(h)=\frac{\sum_{j\geq0}\psi_j\psi_{j+|h|}}{\sum_{j\geq0}\psi_j^2}.}
$$
To see this directly, expand the <covariance> of the two convergent series. Only matching noise indices contribute. <Cauchy-Schwarz inequality> makes the coefficient-product sum finite, justifying the <covariance> limit.
There is also a closed expression. Set $a=9/11$, $d=2/11$, $r=0.4$, $s=-0.7$. For $h\geq0$,
$$
\gamma(h)=4\left[\frac{a^2r^h}{1-r^2}+\frac{d^2s^h}{1-s^2}+\frac{ad(r^h+s^h)}{1-rs}\right],
$$
and negative lags follow by symmetry. Dividing by the expression at zero gives the same <autocorrelation function>.
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