# Approximating AR(1) by finite order MA process - convergence results

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