Construction of a (jointly?) stationary and ergodic vector sequence.
$begingroup$
In a recent effort to understand stationary ergodic processes, I stumbled upon a paper that leaves me somewhat puzzled. I would be very grateful for any pointers.
The line of reasoning is as follows:
Given a set of assumptions, it is deduced that a random sequence ${f_t:tinmathbb{Z}}$ is stationary ergodic.
For any measurable map $h$, stationary ergodicity of ${f_t:tinmathbb{Z}}$ implies that ${h(f_t):tinmathbb{Z}}$ is stationary ergodic.
Moreover, there is a sequence of innovations ${u_t:tinmathbb{Z}}$ - also presumed stationary ergodic.
Given these facts, the paper goes on to assert (verbatim)
Together with ${u_t}$ being SE (Assumption 3), it follows that
${(u_t,h(f_t))}$ is a stationary and ergodic vector sequence.
The proof concludes with the insight that: Continuity, and thus by extension, measurability of another function $g,:,mathbb{R}^2tomathbb{R}$ in turn implies that ${g(u_t,h(f_t))}$ is stationary ergodic.
At this point my struggle is with the statement that “${(u_t,h(f_t))}$ is a stationary and ergodic vector sequence”.
While it was shown that ${u_t}$ and ${f_t}$ are stationary ergodic, I do not see how that necessarily implies that
${(u_t,h(f_t))}$ is jointly stationary ergodic. However, I am under the impression that this is a requirement for ${g(u_t,h(f_t))}$ to be stationary ergodic.
I would very much appreciate if anyone could tell me where I am going wrong, or wether I am missing any crucial piece of information - such as ${u_t}$ is assumed to be iid (?).
Thank you so very much.
Best,
Jon
real-analysis probability-theory measure-theory stochastic-processes ergodic-theory
asked Jan 3 at 16:53
J.BeckJ.Beck
636
$endgroup$
add a comment |
$begingroup$
In a recent effort to understand stationary ergodic processes, I stumbled upon a paper that leaves me somewhat puzzled. I would be very grateful for any pointers.
The line of reasoning is as follows:
Given a set of assumptions, it is deduced that a random sequence ${f_t:tinmathbb{Z}}$ is stationary ergodic.
For any measurable map $h$, stationary ergodicity of ${f_t:tinmathbb{Z}}$ implies that ${h(f_t):tinmathbb{Z}}$ is stationary ergodic.
Moreover, there is a sequence of innovations ${u_t:tinmathbb{Z}}$ - also presumed stationary ergodic.
Given these facts, the paper goes on to assert (verbatim)
Together with ${u_t}$ being SE (Assumption 3), it follows that
${(u_t,h(f_t))}$ is a stationary and ergodic vector sequence.
The proof concludes with the insight that: Continuity, and thus by extension, measurability of another function $g,:,mathbb{R}^2tomathbb{R}$ in turn implies that ${g(u_t,h(f_t))}$ is stationary ergodic.
At this point my struggle is with the statement that “${(u_t,h(f_t))}$ is a stationary and ergodic vector sequence”.
While it was shown that ${u_t}$ and ${f_t}$ are stationary ergodic, I do not see how that necessarily implies that
${(u_t,h(f_t))}$ is jointly stationary ergodic. However, I am under the impression that this is a requirement for ${g(u_t,h(f_t))}$ to be stationary ergodic.
I would very much appreciate if anyone could tell me where I am going wrong, or wether I am missing any crucial piece of information - such as ${u_t}$ is assumed to be iid (?).
Thank you so very much.
Best,
Jon
real-analysis probability-theory measure-theory stochastic-processes ergodic-theory
asked Jan 3 at 16:53
J.BeckJ.Beck
636
$endgroup$
add a comment |
$begingroup$
In a recent effort to understand stationary ergodic processes, I stumbled upon a paper that leaves me somewhat puzzled. I would be very grateful for any pointers.
The line of reasoning is as follows:
Given a set of assumptions, it is deduced that a random sequence ${f_t:tinmathbb{Z}}$ is stationary ergodic.
For any measurable map $h$, stationary ergodicity of ${f_t:tinmathbb{Z}}$ implies that ${h(f_t):tinmathbb{Z}}$ is stationary ergodic.
Moreover, there is a sequence of innovations ${u_t:tinmathbb{Z}}$ - also presumed stationary ergodic.
Given these facts, the paper goes on to assert (verbatim)
Together with ${u_t}$ being SE (Assumption 3), it follows that
${(u_t,h(f_t))}$ is a stationary and ergodic vector sequence.
The proof concludes with the insight that: Continuity, and thus by extension, measurability of another function $g,:,mathbb{R}^2tomathbb{R}$ in turn implies that ${g(u_t,h(f_t))}$ is stationary ergodic.
At this point my struggle is with the statement that “${(u_t,h(f_t))}$ is a stationary and ergodic vector sequence”.
While it was shown that ${u_t}$ and ${f_t}$ are stationary ergodic, I do not see how that necessarily implies that
${(u_t,h(f_t))}$ is jointly stationary ergodic. However, I am under the impression that this is a requirement for ${g(u_t,h(f_t))}$ to be stationary ergodic.
I would very much appreciate if anyone could tell me where I am going wrong, or wether I am missing any crucial piece of information - such as ${u_t}$ is assumed to be iid (?).
Thank you so very much.
Best,
Jon
real-analysis probability-theory measure-theory stochastic-processes ergodic-theory
asked Jan 3 at 16:53
J.BeckJ.Beck
636
$endgroup$
In a recent effort to understand stationary ergodic processes, I stumbled upon a paper that leaves me somewhat puzzled. I would be very grateful for any pointers.
The line of reasoning is as follows:
Given a set of assumptions, it is deduced that a random sequence ${f_t:tinmathbb{Z}}$ is stationary ergodic.
For any measurable map $h$, stationary ergodicity of ${f_t:tinmathbb{Z}}$ implies that ${h(f_t):tinmathbb{Z}}$ is stationary ergodic.
Moreover, there is a sequence of innovations ${u_t:tinmathbb{Z}}$ - also presumed stationary ergodic.
Given these facts, the paper goes on to assert (verbatim)
Together with ${u_t}$ being SE (Assumption 3), it follows that
${(u_t,h(f_t))}$ is a stationary and ergodic vector sequence.
The proof concludes with the insight that: Continuity, and thus by extension, measurability of another function $g,:,mathbb{R}^2tomathbb{R}$ in turn implies that ${g(u_t,h(f_t))}$ is stationary ergodic.
At this point my struggle is with the statement that “${(u_t,h(f_t))}$ is a stationary and ergodic vector sequence”.
While it was shown that ${u_t}$ and ${f_t}$ are stationary ergodic, I do not see how that necessarily implies that
${(u_t,h(f_t))}$ is jointly stationary ergodic. However, I am under the impression that this is a requirement for ${g(u_t,h(f_t))}$ to be stationary ergodic.
I would very much appreciate if anyone could tell me where I am going wrong, or wether I am missing any crucial piece of information - such as ${u_t}$ is assumed to be iid (?).
Thank you so very much.
Best,
Jon
real-analysis probability-theory measure-theory stochastic-processes ergodic-theory
real-analysis probability-theory measure-theory stochastic-processes ergodic-theory
asked Jan 3 at 16:53
J.BeckJ.Beck
636
asked Jan 3 at 16:53
J.BeckJ.Beck
636
asked Jan 3 at 16:53
J.BeckJ.Beck
636
asked Jan 3 at 16:53
J.BeckJ.Beck
636
asked Jan 3 at 16:53
J.BeckJ.Beck
636
636
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