# Do measurable maps preserve stationary ergodicity?

In a recent effort to establish stationary ergodicity for a certain stochastic process, I just happened to come across a statement, which I find to be little bit confounding.

Given two measurable spaces $$(\,E,\mathcal{E}\,)$$ and $$(\,\tilde{E},\tilde{\mathcal{E}}\,)$$, let $$(v_t)_{\,t\in\mathbb{Z}}$$ be a stationary ergodic sequence of $$E$$-valued random elements and define a measurable function $$f:{E}^\mathbb{N}\to \tilde{E}$$. Then the sequence $$(\tilde{v}_t)_{\,t\in\mathbb{Z}}$$ defined by $$\tilde{v}_t=f(v_t,v_{t-1},...)\qquad \text{for all }\quad t\in\mathbb{Z}$$ is stationary ergodic. (Straumann and Mikosch, 2006, p. 2455 - https://projecteuclid.org/euclid.aos/1169571804)

I do have to admit that I am somewhat irritated by this statement. In particular, let $$f$$ be prescribed by $$(v_t,v_{t-1},...)\,\mapsto\,\sum_{j=0}^\infty\rho^jv_{t-j}.$$ Then, I'd argue that $$(\tilde{v}_t)_{\,t\in\mathbb{Z}}$$ is not necessarily stationary (as would follow from the statement above). In fact, stationarity of $$\tilde{v}_t$$ would depend on whether $$|\rho|<1$$ or not.

I'd very much appreciate, if someone could tell me where I am going wrong, what I am missing, or whether the statement is not entirely accurate.

For reference, the statement is supposed to build on proposition 4.3 (p. 26) in Ulrich Krengel's (1985) monograph Ergodic Theorems.

• $f$ must be a measurable map. If $\rho > 1$, it's not even clear you have a map, $\omega$ by $\omega$. – Michael May 20 '19 at 21:09