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I am currently struggling with a result pertaining to the finite order MA approximation of a simple AR$\,(\,1\,)$ process defined on a double sided time-index set $\,T=\mathbb{Z}$. I would be very grateful if someone could help me understand the requirements on sample size and approximation order to establish the convergence result stated in $(\,*\,)$ below.

In particular, let a stationary AR$\,(\,1\,)$ process be written as $$\phantom{\qquad\text{with }\,\,|\phi|<1}x_t=\sum_{s\,=\,0}^\infty\,\phi^{\,s}\varepsilon_{t-s}\qquad\text{with }\,\,|\phi|<1$$ and approximated by an $m$-th order MA $$x_t^m=\sum_{s\,=\,0}^{m-1}\,\phi^{\,s}\varepsilon_{t-s}$$ for some given $m\,\in\,\mathbb{N}$. Moreover, I consider a Lipschitz function $g$ such that $$|\,g(\,x\,)-g(\,y\,)\,|\,\leq\,K\,|\,x-y\,|$$ for some Lipschitz constant $K.$

Ultimately, I would like to show (and understand) that for a sample $\big(\,x_t\,:\,t=1,\,...,\,n\,\big)$ and a corresponding sample $\big(\,x_t^m\,:\,t=1,\,...,\,n\,\big)$ \begin{equation}\left|\,\dfrac{1}{n}\,\sum_{t=1}^n\,g(\,x_t\,)\,-\,\dfrac{1}{n}\,\sum_{t=1}^n\,g(\,x_t^m\,)\,\right|\,=\,o_p(\,1\,)\tag{$\,*\,$}\end{equation}

I suppose that, by Lipschitz continuity, I can write $$\left|\,\dfrac{1}{n}\,\sum_{t=1}^n\,g(\,x_t\,)\,-\,\dfrac{1}{n}\,\sum_{t=1}^n\,g(\,x_t^m\,)\,\right|\,\leq\,\dfrac{1}{n}\,\sum_{t=1}^nK\,\big|\,x_t\,-\,x_t^m\,\big|.$$ Then, seeing as $$x_t-x_t^m\,=\,\sum_{s\,=\,0}^\infty\,\phi^{\,s}\varepsilon_{t-s}-\sum_{s\,=\,0}^{m-1}\,\phi^{\,s}\varepsilon_{t-s}\,=\,\sum_{s\,=\,m}^\infty\,\phi^{\,s}\varepsilon_{t-s}=\phi^{\,m}\sum_{s\,=\,m}^\infty\,\phi^{\,s-m}\varepsilon_{t-s}$$it would seem to follow that $$\left|\,\dfrac{1}{n}\,\sum_{t=1}^n\,g(\,x_t\,)\,-\,\dfrac{1}{n}\,\sum_{t=1}^n\,g(\,x_t^m\,)\,\right|\,\leq\,\dfrac{K}{n}\,\sum_{t=1}^n\,\left|\,\phi^{\,m}\sum_{s\,=\,m}^\infty\,\phi^{\,s-m}\varepsilon_{t-s}\,\right|.$$

If I were to argue in a recklessly cavalier manner, I'd say that it's sort of obvious that as $n\to\infty$ it follows that $n^{-1}K\to0$ while $|\phi|<1$ and $\varepsilon_t=O_p(\,1\,)$ ensures that as $m\to\infty$ $$\left|\,\phi^{\,m}\sum_{s\,=\,m}^\infty\,\phi^{\,s-m}\varepsilon_{t-s}\,\right|\to0$$ which is also true when summing $n\to\infty$ such terms.
While I am well aware that I made a couple of huge mistakes with the above line of reasoning, however, I do have a hard time pinpointing those mistakes exactly $\,$-$\,$ for example, I see that with $s$ running up to $\infty$ and letting $m\to\infty$ I'd finally end up with a term that has $\phi^{\infty-\infty}$ which is not defined. My supposition at this point would be that $m$ need to run to $\infty$ at a slower rate than $n.$

I'd very much appreciate, if someone could walk through the reasoning required to establish that the absolute difference of sample averages in $(\,*\,)$ is $o_p(\,1\,)$ $\,\,$-$\,\,$ especially when it comes to the right way (in terms of order and rate) to let $n\to\infty$ and $m\to\infty$. I'd suspect that there is a case to be made for $m$ to be a function of $n$, thus ensuring that $m$ tends to $\infty$ sufficiently slowly.

Thank you so very much.

Best wishes,
Jon

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