Absolutely summable linear filters preserve weak stationarity

Weak stationarity makes $\mathbb{E}[Z_t^2]=\gamma (0)$ for every $t$. Applying [F3] to $Z_t$ and the constant one gives $\mathbb{E}\mathrm{\mid }{Z}_t\mathrm{\mid }\le \sqrt{\gamma (0)}$, while applying it to the centered variables $Z_{t+h},Z_t$ gives $\mathrm{\mid }\gamma (h)\mathrm{\mid }\le \gamma (0)$. Consequently ${\sum }_{j\in \mathbb{Z}}\mathbb{E}\left[\mathrm{\mid }{c}_j{Z}_{t-j}\mathrm{\mid }\right]\le \sqrt{\gamma (0)}{\sum }_{j\in \mathbb{Z}}\mathrm{\mid }{c}_j\mathrm{\mid }<\mathrm{\infty }\mathrm{.}$ Tonelli [F5] on counting measure over $\mathbb{Z}$ times the probability measure shows that the random series of absolute values has finite expectation, hence is finite almost surely. Thus $Y_t$ is defined by an absolutely convergent sum.

For any finite $J\subset \mathbb{Z}$, the finite filter $Y_t^J={\sum }_{j\in J}{c}_j{Z}_{t-j}$ satisfies, by [F4], $\mathbb{E}[(Y_t^J{)}^2]={\sum }_{j,k\in J}{c}_j{c}_k\gamma (k-j)\le \gamma (0){\left({\sum }_{j\in J}\mathrm{\mid }{c}_j\mathrm{\mid }\right)}^2\mathrm{.}$ The equality is a finite covariance expansion; the inequality follows from $\mathrm{\mid }\gamma (k-j)\mathrm{\mid }\le \gamma (0)$ from step 1.1. Applying the same bound to a finite tail $J$ gives $\mathrm{\parallel }{\sum }_{j\in J}{c}_j{Z}_{t-j}{\mathrm{\parallel }}_2\le \sqrt{\gamma (0)}{\sum }_{j\in J}\mathrm{\mid }{c}_j\mathrm{\mid }$.

Enumerate $\mathbb{Z}$ as $0,1,-1,2,-2,\dots$ and let $Y_t^N$ be the partial sum through the first $N$ terms. The absolute almost-sure convergence from step 1.1 gives $Y_t-Y_t^N={\lim }_{M\to \mathrm{\infty }}(Y_t^M-Y_t^N)$ almost surely. Fatou [F7] and the finite-tail bound of step 2.1 therefore give $\mathbb{E}\mathrm{\mid }{Y}_t-Y_t^N{\mathrm{\mid }}^2\le \gamma (0){\left({\sum }_{j\text{ omitted after }N}\mathrm{\mid }{c}_j\mathrm{\mid }\right)}^2⟶0.$ Thus $Y_t^N\to Y_t$ in $L^2$ and $Y_t$ is square-integrable.

Each finite filter is centered. The $L^2$ convergence in step 3.1 implies $L^1$ convergence by [F3] applied to the difference and the constant one; hence $\mathbb{E}[Y_t]=0$. For finite $J$, covariance bilinearity [F4] gives $\mathrm{Cov}(Y_{t+h}^J,Y_t^J)={\sum }_{j,k\in J}{c}_j{c}_k\gamma (h-j+k)\mathrm{.}$ As $J$ increases through the enumeration, the left side converges to $\mathrm{Cov}(Y_{t+h},Y_t)$ by $L^2$ convergence and [F3]. The right side converges absolutely because ${\sum }_{j,k\in \mathbb{Z}}\mathrm{\mid }{c}_j{c}_k\gamma (h-j+k)\mathrm{\mid }\le \gamma (0){\left({\sum }_j\mathrm{\mid }{c}_j\mathrm{\mid }\right)}^2<\mathrm{\infty },$ with order of summation justified by [F6]. Its limit is independent of $t$, so $Y$ is weakly stationary with the stated autocovariance.

Let $c,d\in {\mathrm{\ell }}^1(\mathbb{Z})$. The triangle inequality and [F6] give ${\sum }_{m\in \mathbb{Z}}\mathrm{\mid }(c\ast d{)}_m\mathrm{\mid }\le {\sum }_{j,k\in \mathbb{Z}}\mathrm{\mid }{c}_j\mathrm{\mid }\mathrm{\mid }{d}_k\mathrm{\mid }=\left({\sum }_j\mathrm{\mid }{c}_j\mathrm{\mid }\right)\left({\sum }_k\mathrm{\mid }{d}_k\mathrm{\mid }\right)<\mathrm{\infty },$ so $c\ast d\in {\mathrm{\ell }}^1(\mathbb{Z})$. Tonelli [F5] and the uniform bound from step 1.1 give $\mathbb{E}\text{\hspace{0.17em}⁣}\left[{\sum }_{j,k\in \mathbb{Z}}\mathrm{\mid }{c}_j{d}_k{Z}_{t-j-k}\mathrm{\mid }\right]={\sum }_{j,k\in \mathbb{Z}}\mathrm{\mid }{c}_j{d}_k\mathrm{\mid }\mathbb{E}\mathrm{\mid }{Z}_{t-j-k}\mathrm{\mid }\le \sqrt{\gamma (0)}\left({\sum }_j\mathrm{\mid }{c}_j\mathrm{\mid }\right)\left({\sum }_k\mathrm{\mid }{d}_k\mathrm{\mid }\right)<\mathrm{\infty }\mathrm{.}$ Hence this double random series is absolutely convergent almost surely. Reindexing the absolutely convergent terms by $m=j+k$ gives ${\sum }_j{c}_j\left({\sum }_k{d}_k{Z}_{t-j-k}\right)={\sum }_m\left({\sum }_j{c}_j{d}_{m-j}\right)Z_{t-m}={\sum }_m(c\ast d{)}_m{Z}_{t-m}\mathrm{.}$ Both sides are the filter limits defined in the Statement, and the equality holds almost surely and in $L^2$ by the bounds of steps 2.1 and 3.1. This proves the composition formula.