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.

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

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