Is $X_t=X_{t-1}^{\alpha} + \varepsilon_t$ stationary for $\alpha<1$?

Question

Let {$\varepsilon_t$} be iid. Then, we have time series defined by $$X_t=cX_{t-1}^{\alpha} + \varepsilon_t,$$ with $0<\alpha<1$ and $c\in\mathbb{R}$ and let $\varepsilon_t$ be non-negative. Is it strictly stationary?

If we have $\alpha=1$ we obtain classic AR(1) process, where we need $c<1$ for stationarity. For lower $\alpha$ it seems that $X_t$ is "smaller" and should be also stationary, but I have a hard time proving that. Also, do we need then some restriction for $c$ in such case?

Answer 1

If someone is interested, this problem is actually a special case of nonlinear autoregressive process.

They are defined as $X_t=f(X_{t-1}) + \varepsilon_t$, where some condition for stationarity (even ergodicity...) is: $f$ is some measurable function satisfying for some $c\in[0,1)$ and $K>0$ the following: $\|f(x)\|\leq c\|x\|+K$ for all $x\in\mathbb{R}$ and $\mathbb{E}|\varepsilon_t|<\infty$. Reference is here, and some generalization for heavy tailed noise (if we don't want the existence of moment assumption) is here.

Answer 2

One can see that it is not stationary in the special case that variance of the shock is zero. You have

$$X_1=cX_0^{\alpha}$$

$$X_2=cX_1^{\alpha}=c^{1+\alpha}X_0^{2\alpha}$$

and generally

$$X_t=c^{1+\alpha(t-1)}X_0^{t\alpha}$$

This can converge to different values depending on $X_0$. When $X_0=0$ it stays at zero. However, for example when $\alpha=0.01$, $c=0.99$ and $X_0=0.01$ it converges to $\approx0.99$. Hence it does not have a well-defined unconditional mean. I believe this argument could be generalized to the case with a strictly positive variance.

Answer 3

The answer will depend on your parameters and your choice of noise. I'll give partial answer, using the following reference: Peigné, Woess, Stochastic dyamical systems with weak contractivity properties. I. Strong and local contractivity, which contains many helpful other references (Goldie...).

First, I will exclude the value $0$ of the domain, so that $(X_n)$ takes its values in $\mathbb{R}_+^*$. I denote by $x := X_0$, and set:

$$Y_n^{\ln (x)} := \ln (X_n^x),$$.

$$\Psi_\epsilon (y) := \ln(ce^{\alpha y} + \epsilon).$$

Note that $0 \leq \Psi'_\varepsilon \leq \alpha < 1$ for all $\varepsilon$.

Then $Y_{n+1}^y = \Psi_{\epsilon_n} (Y_n^y)$, and $(\Psi_{\epsilon_n})_{n \geq 0}$ is an i.i.d. family of contraction mappings. The sequence $(Y_n)$ is a Stochastic Dynamical System (SDS). In addition, since all the mappings are $\alpha$-contracting, we get $\lim_{n \to + \infty} |Y_n^x-Y_n^y| = 0$ almost surely for all $x$, $y$. Using the terminology of the article, this SDS is strongly contractive.

Under another hypothesis, that the SDS be recurrent, Theorem 2.13 tells you that, up to multiplication by a constant, there is a unique invariant Radon measure $\nu$, for which the Markov chain $(Y_n)$ is ergodic, and supported exactly on the smallest nonempty closed subset $L$ such that $\mathbb{P} (\Psi_\epsilon (L) \subset L) = 1$. There are further conditions to guarantee that $\nu(L) < +\infty$ (positive recurrence). Which, I think, is more than what you ask for.

Now, let me go back to this recurrence criterion. It asserts that

$$\mathbb{P} (\liminf_{n \to + \infty} |Y_n^y| < +\infty) = 1$$

for one (or, equivalently here, every) $y$. Since the $\Psi_\varepsilon$ are bounded from below by $\ln (\varepsilon)$, only the upper bound is non-trivial. And here, it will depend on the distribution of the noise $\varepsilon$: if it is very (very, very, very) heavy-tailed, I can imagine that the Markov chain $(Y_n)$ is not recurrent. Anyway, let us get a upper bound: for positive $y$,

$$\Psi_\varepsilon(y) \leq \ln (c+\varepsilon) + \alpha y,$$

whence, if $(Y_m^y, \ldots, Y_{n+m}^y)$ are positive,

$$Y_{n+m}^y \leq \alpha^n Y_m^y + \sum_{k = 0}^{n-1} \alpha^k \ln(c+\varepsilon_{n+m-k}).$$

In particular, if $\lim_{n \to + \infty} Y_n^y = +\infty$, then

$$\lim_{n \to + \infty} \sum_{k = 1}^n \alpha^k \ln(c+\varepsilon_{n-k}) = +\infty.$$

But $\sum_{k = 1}^n \alpha^k \ln(c+\varepsilon_{n-k})$ has the same distribution as $\sum_{k = 0}^{n-1} \alpha^k \ln(c+\varepsilon_k)$. In particular, if $\mathbb{E} (\ln(c+\varepsilon)) < +\infty$, then $\sum_{k = 0}^{n-1} \alpha^k \ln(c+\varepsilon_k)$ converges in $\mathbb{L}^1$, so almost surely

$$\liminf_{n \to + \infty} \sum_{k = 1}^n \alpha^k \ln(c+\varepsilon_{n-k}) < +\infty,$$

and the Markov chain is recurrent. Hence, all you need for the existence and uniqueness of an invariant measure is that $\varepsilon >0$ almost surely and $\mathbb{E} (\ln(1+\varepsilon)) < +\infty$.