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Convergence result for approximation error - stationary AR(1) and finite order MA
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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
$$phantom{qquadtext{with },,|phi|<1}x_t=sum_{s,=,0}^infty,phi^{,s}varepsilon_{t-s}qquadtext{with },,|phi|<1$$
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 $ntoinfty$ it follows that $n^{-1}Kto0$ while $|phi|<1$ and $varepsilon_t=O_p(,1,)$ ensures that as $mtoinfty$
$$left|,phi^{,m}sum_{s,=,m}^infty,phi^{,s-m}varepsilon_{t-s},right|to0$$
which is also true when summing $ntoinfty$ 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 $mtoinfty$ 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 $ntoinfty$ and $mtoinfty$. 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
probability limits convergence time-series
asked Jan 30 at 21:18
J.BeckJ.Beck
657
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add a comment |
$begingroup$
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
$$phantom{qquadtext{with },,|phi|<1}x_t=sum_{s,=,0}^infty,phi^{,s}varepsilon_{t-s}qquadtext{with },,|phi|<1$$
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 $ntoinfty$ it follows that $n^{-1}Kto0$ while $|phi|<1$ and $varepsilon_t=O_p(,1,)$ ensures that as $mtoinfty$
$$left|,phi^{,m}sum_{s,=,m}^infty,phi^{,s-m}varepsilon_{t-s},right|to0$$
which is also true when summing $ntoinfty$ 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 $mtoinfty$ 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 $ntoinfty$ and $mtoinfty$. 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
probability limits convergence time-series
asked Jan 30 at 21:18
J.BeckJ.Beck
657
$endgroup$
add a comment |
$begingroup$
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
$$phantom{qquadtext{with },,|phi|<1}x_t=sum_{s,=,0}^infty,phi^{,s}varepsilon_{t-s}qquadtext{with },,|phi|<1$$
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 $ntoinfty$ it follows that $n^{-1}Kto0$ while $|phi|<1$ and $varepsilon_t=O_p(,1,)$ ensures that as $mtoinfty$
$$left|,phi^{,m}sum_{s,=,m}^infty,phi^{,s-m}varepsilon_{t-s},right|to0$$
which is also true when summing $ntoinfty$ 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 $mtoinfty$ 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 $ntoinfty$ and $mtoinfty$. 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
probability limits convergence time-series
asked Jan 30 at 21:18
J.BeckJ.Beck
657
$endgroup$
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
$$phantom{qquadtext{with },,|phi|<1}x_t=sum_{s,=,0}^infty,phi^{,s}varepsilon_{t-s}qquadtext{with },,|phi|<1$$
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 $ntoinfty$ it follows that $n^{-1}Kto0$ while $|phi|<1$ and $varepsilon_t=O_p(,1,)$ ensures that as $mtoinfty$
$$left|,phi^{,m}sum_{s,=,m}^infty,phi^{,s-m}varepsilon_{t-s},right|to0$$
which is also true when summing $ntoinfty$ 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 $mtoinfty$ 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 $ntoinfty$ and $mtoinfty$. 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
probability limits convergence time-series
probability limits convergence time-series
asked Jan 30 at 21:18
J.BeckJ.Beck
657
asked Jan 30 at 21:18
J.BeckJ.Beck
657
asked Jan 30 at 21:18
J.BeckJ.Beck
657
asked Jan 30 at 21:18
J.BeckJ.Beck
657
asked Jan 30 at 21:18
J.BeckJ.Beck
657
657
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