= Solution
For a stationary linear autoregression, <causal time series> means that the observation uses only current and past driving noise:
$$
Y_t=\sum_{j\geq0}\psi_j\varepsilon_{t-j},\qquad \sum_{j\geq0}|\psi_j|^2<\infty.
$$
This is a convergent in the sense of <mean-square convergence> <infinite moving-average representation>. The stronger usual stable-filter definition requires absolute summability; the argument below also handles the square-summable definition. Let $\Phi(z)=1-\sum_{k=1}^p\phi_kz^k$ and $\Psi(z)=\sum_{j\geq0}\psi_jz^j$. Substituting the filter into the recurrence and comparing coefficients of the orthogonal noise gives $\psi_0=1$ and $\psi_j=\sum_{k=1}^{\min(p,j)}\phi_k\psi_{j-k}$. Hence
$$
\Phi(z)\Psi(z)=1\qquad(|z|<1).
$$
The <Cauchy-Schwarz inequality> ensures that $\Psi$ is an <analytic function> in this disk. Thus $\Phi$ has no zero strictly inside it. A boundary zero is also impossible: a zero of multiplicity $m\geq1$ at $z_0=e^{i\omega_0}$ makes $|1/\Phi(re^{i\omega})|^2$ at least a constant times $[(1-r)^2+(\omega-\omega_0)^2]^{-m}$ near that point. Its integral over $\omega$ diverges as $r\uparrow1$. On the other hand, <orthogonality of complex exponentials> gives
$$
\frac1{2\pi}\int_{-\pi}^{\pi}|\Psi(re^{i\omega})|^2\,d\omega
=\sum_{j\geq0}|\psi_j|^2r^{2j}\leq\sum_{j\geq0}|\psi_j|^2<\infty,
$$
a contradiction. Therefore the <causality root criterion for an autoregressive model> is
$$
\boxed{\Phi(z)=0\ \Longrightarrow\ |z|>1.}
$$
Under absolute summability, the shorter boundary argument is continuity of $\Psi$ on the closed disk and the identity $\Phi\Psi=1$ there.
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