If $X_t^2$ is stationary, is $X_t$ necessarily stationary? I came across a proof for one of the properties of the ARCH model which says that if $\mathbb{E}(X_t^2) < \infty$, then $\{X_t\}$ is stationary iff $\sum_{i=1}^pb_i < 1$ where the ARCH model is:
$X_t = \sigma_t\epsilon_t$ 
$\sigma_t^2 = b_0 + b_1X_{t-1}^2 + ... b_pX_{t-p}^2$
The main idea of the prove is to show that $X_t^2$ can be written as an AR(p) process and if $\sum_{i=1}^pb_i < 1$ is true, then all roots of the characteristic polynomial lie outside the unit circle and hence $\{X_t^2\}$ is stationary. It then says that hence $\{X_t\}$ is stationary. How does this follow?
 A: From the section given I understand how you might see that stationarity of $X^2_t$ implies stationarity of $X_t$ but actually it only implies a constant variance of $X_t$. 
The authors of that proof were using stationarity of $X^2_t$ to complete an argument they had started earlier by looking at unconditional moments of $X_t$
Recall the $2^{nd}$ order stationarity conditions:


*

*$E(X_t)<\infty$   $  \forall_{t\in Z}$

*$Var(X_t) = m$ $\forall_{t\in Z}$

*$Cov(X_t,X_{t+h})= \gamma_x(h)$    $\forall_{h\in Z}$


Condition 1 was proved by $E(X_t)=E(E(X_t|F_{t-1}))=0$
Condition 3 was proved by $E(X_tX_{t-1})=E(\sigma_t\epsilon_t\sigma_{t-1}\epsilon_{t-1})=E(E(\sigma_t\epsilon_t\sigma_{t-1}\epsilon_{t-1})|F_{t-1})=E(\sigma_{t}\sigma_{t-1}E(\epsilon_{t-1}\epsilon_{t})|F_{t-1}))=0$
But to prove the second condition they needed to prove a constant unconditional variance of $X_t$
$Var(X_{t})=Var(X_{t-1})=Var(X_{t-2})=...=m$
This is what leads to an assumption of stationarity of $X^2_{t}$ which you have mentioned uses its $AR(p)$ form. In brief:
\begin{align*}
 Var(X_{t})=&E(Var(X_{t})|F_{t-1}) + Var(E(X_t|F_{t-1}))\\
=&E(Var(u_t|F_{t-1}))\quad because\: the\: last\: term\: is\: 0\\ =&E(b_0 + b_1X_{t-1}^2 + ... b_pX_{t-p}^2)\\ =& b_0 + b_1E(X_{t-1}^2) + ... b_pE(X_{t-p}^2)\\ =&b_0 + b_1var(X_{t-1}) + ... b_pvar(X_{t-p})
\end{align*}
If X^2_t is stationary then the roots of the polynomial would lie out of the unit circle and $\Sigma b_i<1$ This makes it possible to write:
$$var(X_{t-1})=... = var(X_{t-p})= \frac{b_0}{1-b_1-...-b_p} \quad which\quad is \quad alas\quad constant!$$
