Note. \(X_t = \alpha X_{t-1} + Z_t\) can be written as \((1 - \alpha B)X_t = Z_t\) using the backward shift operator. Therefore, \begin {align*} X_t &= \frac {Z_t}{1 - \alpha B}\\ &= (1 + \alpha B + \alpha ^2 B^2+ \alpha ^3 B^3 + ............ +...)Z_t\\ &= Z_t + \alpha Z_{t-1} + \alpha ^2 Z_{t-2} + .................. \\ \end {align*}

Thus \(E(X_t) = 0\) because \(X_t\) is a purely random process and

\begin {align*} var(X_t) &= (1 + \alpha ^2 + \alpha ^4 + .............)\sigma ^2_Z\\ &= \frac {\sigma ^2_Z}{1 - \alpha ^2} \end {align*}

provided \(|\alpha | < 1\)

The acv.f is given by \begin {align*} \varphi (k) &= E(X_t,X_{t+k})\\ &= E\Big [\Big (\sum \alpha ^i Z_{t-i}\Big )\Big (\sum \alpha ^i Z_{t+k-i}\Big )\Big ]\\ &= \sigma ^2_Z \sum ^{\infty }_{i=0} \alpha ^i \alpha ^{k+i}, \hspace {2cm} for \hspace {0.3cm} k\geq 0 \end {align*}

which converges for \(|\alpha | < 1\) to \begin {align*} \varphi (k) &= \frac {\alpha ^k \sigma ^2_Z}{1 - \alpha ^2}\\ &= \alpha ^k \sigma ^2_x\hspace {0.5cm}, \hspace {1.5cm} var(X_t) = \sigma ^2_x = \frac {\sigma ^2_Z}{1 - \alpha ^k} \end {align*}

for \(\hspace {0.2cm} k < 0\hspace {0.3cm}\), \(\hspace {0.3cm} \varphi (k) = \varphi (-k)\)
\(\varphi (k)\) does not depend on \(t\), an AR(1) is second order stationary provided \(|\alpha | < 1\). The ac.f is given by
\[\rho (k) = \alpha ^k, \hspace {1cm} k = 0, \pm 1, \pm 2, \cdots \cdots \]

For \(X_t = \alpha X_{t-1} + Z_t, \hspace {1cm}(AR(1)) \), we can multiply this by \(X_{t-k}\) to obtain
\[X_{t-k}X_t = \alpha X{t-k}\,\cdot \, X_{t-1} + X_{t-k}\,\cdot \, Z_t\] taking the expectation, we have \(\hspace {0.3cm}\boxed {\varphi (-k) = \alpha \varphi (-k+1)}\hspace {0.3cm}\) on the assumption that \(E(X_{t-k}\,\cdot \, Z_t) = 0\), for \(k>0\).

Since \(\varphi (-k) = \varphi (k)\) we have \(\varphi (k) = \alpha \varphi (k-1)\) for \(k > 0\).
Since \(\varphi (0) = \sigma ^2_X\), thus \(\varphi (k) = \alpha ^k \sigma ^2_X\) for \(k\geq 0\), thus \[\rho (k) = \frac {\varphi (k)}{\varphi (0)} = \frac {\alpha ^k \sigma ^2_X}{\sigma ^2_X} = \alpha ^k\hspace {0.2cm},\hspace {0.2cm} \text {for}\hspace {0.2cm} k\geq 0.\]

Since, \(|\rho (k)|<1\) we must have \(|\alpha |<1\).
If \(|\alpha |=1\), then \(|\rho (k)|=1\),for all k this is a degenerate case. Thus \(|\alpha |<1\) for a proper, stationary time series.