Proving stationarity of AR(1) process
I would like to prove that the AR(1) process: $X_t=phi X_{t-1}+u_t$, where ${u_t}$ is white noise $(0,sigma^2)$ and $vertphivert<1$, is covariance stationary. One requirement is that $mathbb{E}(X_t)$ is a constant (in this case should be zero). I tried to prove it but had a hard time. Here are some of my thoughts: since we can write $X_t$ recursively as $X_t=sum_{k=0}^{infty}phi^{k}u_{t-k}$, its expectation should be
begin{align}
mathbb{E}(X_{t})&=int_{Omega}sum_{k=0}^{infty}phi^{k}u_{t-k}(omega)domega\&=int_{Omega}lim_{mrightarrowinfty}sum_{k=0}^{m}phi^{k}u_{t-k}(omega)domega\&triangleqint_{Omega}lim_{mrightarrowinfty}f_{m}(omega)domega
end{align}
Then we want to exchange $int_{Omega}$ with $sum_{k=0}^{infty}$, recalling the Dominated Convergence Theorem, we try to figure out whether $vert f_m(omega)vert$ can be bounded by some integrable functions. One possible choice is
begin{align}
vert f_{m}(omega)vert&=leftvert sum_{k=0}^{m}phi^{k}u_{t-k}(omega)rightvert \&leqsum_{k=0}^{m}vertphivert^{k}vert u_{t-k}(omega)vert\&triangleq g_{m}(omega)
end{align}
One sufficient condition is that $sup_tvert u_tvert<infty$, meaning that $u_t$ is a bounded random variable. But I don't think we need such strong condition for usual stationary AR(1) process.
Alternatively, note that $g_m(omega)$ is a sequence of non-negative, non-decreasing functions. If the point-wise limit of $g_{m}(omega)$ is $g(omega)$, then by Monotone Convergence Theorem we have that $g(omega)$ would be integrable.
begin{align}
int_{Omega}g(omega)domega&=lim_{mrightarrowinfty}int_{Omega}g_{m}(omega)domega\&=lim_{mrightarrowinfty}sum_{k=0}^{m}vertphivert^{k}mathbb{E}(vert u_{t-k}vert)\&leqlim_{mrightarrowinfty}sup_{t}mathbb{E}(vert u_{t}vert)sum_{k=0}^{m}vertphivert^{k}\&leqlim_{mrightarrowinfty}sigmasum_{k=0}^{m}vertphivert^{k}\&=frac{sigma}{1-vertphivert}<infty
end{align}
Then the previous Dominated Convergence Theorem applies and we are done.
My question is: do we know that the point-wise limit of $g_{m}(omega)$ exists? Or am I missing something, that the proof is totally different? Thanks!
stationary-processes
asked May 4 '15 at 16:30
SemiMetrics
263
add a comment |
I would like to prove that the AR(1) process: $X_t=phi X_{t-1}+u_t$, where ${u_t}$ is white noise $(0,sigma^2)$ and $vertphivert<1$, is covariance stationary. One requirement is that $mathbb{E}(X_t)$ is a constant (in this case should be zero). I tried to prove it but had a hard time. Here are some of my thoughts: since we can write $X_t$ recursively as $X_t=sum_{k=0}^{infty}phi^{k}u_{t-k}$, its expectation should be
begin{align}
mathbb{E}(X_{t})&=int_{Omega}sum_{k=0}^{infty}phi^{k}u_{t-k}(omega)domega\&=int_{Omega}lim_{mrightarrowinfty}sum_{k=0}^{m}phi^{k}u_{t-k}(omega)domega\&triangleqint_{Omega}lim_{mrightarrowinfty}f_{m}(omega)domega
end{align}
Then we want to exchange $int_{Omega}$ with $sum_{k=0}^{infty}$, recalling the Dominated Convergence Theorem, we try to figure out whether $vert f_m(omega)vert$ can be bounded by some integrable functions. One possible choice is
begin{align}
vert f_{m}(omega)vert&=leftvert sum_{k=0}^{m}phi^{k}u_{t-k}(omega)rightvert \&leqsum_{k=0}^{m}vertphivert^{k}vert u_{t-k}(omega)vert\&triangleq g_{m}(omega)
end{align}
One sufficient condition is that $sup_tvert u_tvert<infty$, meaning that $u_t$ is a bounded random variable. But I don't think we need such strong condition for usual stationary AR(1) process.
Alternatively, note that $g_m(omega)$ is a sequence of non-negative, non-decreasing functions. If the point-wise limit of $g_{m}(omega)$ is $g(omega)$, then by Monotone Convergence Theorem we have that $g(omega)$ would be integrable.
begin{align}
int_{Omega}g(omega)domega&=lim_{mrightarrowinfty}int_{Omega}g_{m}(omega)domega\&=lim_{mrightarrowinfty}sum_{k=0}^{m}vertphivert^{k}mathbb{E}(vert u_{t-k}vert)\&leqlim_{mrightarrowinfty}sup_{t}mathbb{E}(vert u_{t}vert)sum_{k=0}^{m}vertphivert^{k}\&leqlim_{mrightarrowinfty}sigmasum_{k=0}^{m}vertphivert^{k}\&=frac{sigma}{1-vertphivert}<infty
end{align}
Then the previous Dominated Convergence Theorem applies and we are done.
My question is: do we know that the point-wise limit of $g_{m}(omega)$ exists? Or am I missing something, that the proof is totally different? Thanks!
stationary-processes
asked May 4 '15 at 16:30
SemiMetrics
263
The almost sure convergence is frankly off-topic, simply note that as soon as $(u_t)$ is integrable with $(E|u_t|)$ bounded, the series $$sum_{k=0}^infty|phi|^kE|u_{t-k}|$$ converges.
– Did
Aug 20 '17 at 20:03
add a comment |
I would like to prove that the AR(1) process: $X_t=phi X_{t-1}+u_t$, where ${u_t}$ is white noise $(0,sigma^2)$ and $vertphivert<1$, is covariance stationary. One requirement is that $mathbb{E}(X_t)$ is a constant (in this case should be zero). I tried to prove it but had a hard time. Here are some of my thoughts: since we can write $X_t$ recursively as $X_t=sum_{k=0}^{infty}phi^{k}u_{t-k}$, its expectation should be
begin{align}
mathbb{E}(X_{t})&=int_{Omega}sum_{k=0}^{infty}phi^{k}u_{t-k}(omega)domega\&=int_{Omega}lim_{mrightarrowinfty}sum_{k=0}^{m}phi^{k}u_{t-k}(omega)domega\&triangleqint_{Omega}lim_{mrightarrowinfty}f_{m}(omega)domega
end{align}
Then we want to exchange $int_{Omega}$ with $sum_{k=0}^{infty}$, recalling the Dominated Convergence Theorem, we try to figure out whether $vert f_m(omega)vert$ can be bounded by some integrable functions. One possible choice is
begin{align}
vert f_{m}(omega)vert&=leftvert sum_{k=0}^{m}phi^{k}u_{t-k}(omega)rightvert \&leqsum_{k=0}^{m}vertphivert^{k}vert u_{t-k}(omega)vert\&triangleq g_{m}(omega)
end{align}
One sufficient condition is that $sup_tvert u_tvert<infty$, meaning that $u_t$ is a bounded random variable. But I don't think we need such strong condition for usual stationary AR(1) process.
Alternatively, note that $g_m(omega)$ is a sequence of non-negative, non-decreasing functions. If the point-wise limit of $g_{m}(omega)$ is $g(omega)$, then by Monotone Convergence Theorem we have that $g(omega)$ would be integrable.
begin{align}
int_{Omega}g(omega)domega&=lim_{mrightarrowinfty}int_{Omega}g_{m}(omega)domega\&=lim_{mrightarrowinfty}sum_{k=0}^{m}vertphivert^{k}mathbb{E}(vert u_{t-k}vert)\&leqlim_{mrightarrowinfty}sup_{t}mathbb{E}(vert u_{t}vert)sum_{k=0}^{m}vertphivert^{k}\&leqlim_{mrightarrowinfty}sigmasum_{k=0}^{m}vertphivert^{k}\&=frac{sigma}{1-vertphivert}<infty
end{align}
Then the previous Dominated Convergence Theorem applies and we are done.
My question is: do we know that the point-wise limit of $g_{m}(omega)$ exists? Or am I missing something, that the proof is totally different? Thanks!
stationary-processes
asked May 4 '15 at 16:30
SemiMetrics
263
I would like to prove that the AR(1) process: $X_t=phi X_{t-1}+u_t$, where ${u_t}$ is white noise $(0,sigma^2)$ and $vertphivert<1$, is covariance stationary. One requirement is that $mathbb{E}(X_t)$ is a constant (in this case should be zero). I tried to prove it but had a hard time. Here are some of my thoughts: since we can write $X_t$ recursively as $X_t=sum_{k=0}^{infty}phi^{k}u_{t-k}$, its expectation should be
begin{align}
mathbb{E}(X_{t})&=int_{Omega}sum_{k=0}^{infty}phi^{k}u_{t-k}(omega)domega\&=int_{Omega}lim_{mrightarrowinfty}sum_{k=0}^{m}phi^{k}u_{t-k}(omega)domega\&triangleqint_{Omega}lim_{mrightarrowinfty}f_{m}(omega)domega
end{align}
Then we want to exchange $int_{Omega}$ with $sum_{k=0}^{infty}$, recalling the Dominated Convergence Theorem, we try to figure out whether $vert f_m(omega)vert$ can be bounded by some integrable functions. One possible choice is
begin{align}
vert f_{m}(omega)vert&=leftvert sum_{k=0}^{m}phi^{k}u_{t-k}(omega)rightvert \&leqsum_{k=0}^{m}vertphivert^{k}vert u_{t-k}(omega)vert\&triangleq g_{m}(omega)
end{align}
One sufficient condition is that $sup_tvert u_tvert<infty$, meaning that $u_t$ is a bounded random variable. But I don't think we need such strong condition for usual stationary AR(1) process.
Alternatively, note that $g_m(omega)$ is a sequence of non-negative, non-decreasing functions. If the point-wise limit of $g_{m}(omega)$ is $g(omega)$, then by Monotone Convergence Theorem we have that $g(omega)$ would be integrable.
begin{align}
int_{Omega}g(omega)domega&=lim_{mrightarrowinfty}int_{Omega}g_{m}(omega)domega\&=lim_{mrightarrowinfty}sum_{k=0}^{m}vertphivert^{k}mathbb{E}(vert u_{t-k}vert)\&leqlim_{mrightarrowinfty}sup_{t}mathbb{E}(vert u_{t}vert)sum_{k=0}^{m}vertphivert^{k}\&leqlim_{mrightarrowinfty}sigmasum_{k=0}^{m}vertphivert^{k}\&=frac{sigma}{1-vertphivert}<infty
end{align}
Then the previous Dominated Convergence Theorem applies and we are done.
My question is: do we know that the point-wise limit of $g_{m}(omega)$ exists? Or am I missing something, that the proof is totally different? Thanks!
stationary-processes
stationary-processes
asked May 4 '15 at 16:30
SemiMetrics
263
asked May 4 '15 at 16:30
SemiMetrics
263
asked May 4 '15 at 16:30
SemiMetrics
263
asked May 4 '15 at 16:30
SemiMetrics
263
asked May 4 '15 at 16:30
SemiMetrics
263
263
The almost sure convergence is frankly off-topic, simply note that as soon as $(u_t)$ is integrable with $(E|u_t|)$ bounded, the series $$sum_{k=0}^infty|phi|^kE|u_{t-k}|$$ converges.
– Did
Aug 20 '17 at 20:03
add a comment |
The almost sure convergence is frankly off-topic, simply note that as soon as $(u_t)$ is integrable with $(E|u_t|)$ bounded, the series $$sum_{k=0}^infty|phi|^kE|u_{t-k}|$$ converges.
– Did
Aug 20 '17 at 20:03
The almost sure convergence is frankly off-topic, simply note that as soon as $(u_t)$ is integrable with $(E|u_t|)$ bounded, the series $$sum_{k=0}^infty|phi|^kE|u_{t-k}|$$ converges.
– Did
Aug 20 '17 at 20:03
The almost sure convergence is frankly off-topic, simply note that as soon as $(u_t)$ is integrable with $(E|u_t|)$ bounded, the series $$sum_{k=0}^infty|phi|^kE|u_{t-k}|$$ converges.
– Did
Aug 20 '17 at 20:03
add a comment |
1 Answer
1
active
oldest
votes
Actually, the pointwise limit of $(g_n)$ does exists almost surely. Indeed, pick $ rin (|phi|,1)$, and notice that
$$mathbb P{omegamid vertphivert^{k}vert u_{t-k}(omega)vertgt r^k }leqslant frac{|phi|^k}{r^k}mathbb E[|u_{t-k} |]leqslant frac{|phi|^k}{r^k}sigma . $$
Then conclude by the Borel-Cantelli lemma.
answered May 7 '15 at 22:07
Davide Giraudo
125k16150259
1
Thank you very much, Davide Giraudo! Can I proof it like this: by Markov's inequality, we have $Pr.(vertphivert^kvert u_{t-k}vert>A)leqfrac{vertphivert^{2k}mathbb{E}(vert u_{t-k}vert^2)}{A^2}$, where $A>0$ is a constant. These probabilities converge since $mathbb{E}(u_t)=sigma^2$ for all $t$. Then by Borel-Cantelli lemma, we got the almost sure convergence of $g_m(omega)$ series.
– SemiMetrics
May 14 '15 at 2:51
Yes (and this is actually simpler).
– Davide Giraudo
May 14 '15 at 8:08
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1266671%2fproving-stationarity-of-ar1-process%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Actually, the pointwise limit of $(g_n)$ does exists almost surely. Indeed, pick $ rin (|phi|,1)$, and notice that
$$mathbb P{omegamid vertphivert^{k}vert u_{t-k}(omega)vertgt r^k }leqslant frac{|phi|^k}{r^k}mathbb E[|u_{t-k} |]leqslant frac{|phi|^k}{r^k}sigma . $$
Then conclude by the Borel-Cantelli lemma.
answered May 7 '15 at 22:07
Davide Giraudo
125k16150259
1
Thank you very much, Davide Giraudo! Can I proof it like this: by Markov's inequality, we have $Pr.(vertphivert^kvert u_{t-k}vert>A)leqfrac{vertphivert^{2k}mathbb{E}(vert u_{t-k}vert^2)}{A^2}$, where $A>0$ is a constant. These probabilities converge since $mathbb{E}(u_t)=sigma^2$ for all $t$. Then by Borel-Cantelli lemma, we got the almost sure convergence of $g_m(omega)$ series.
– SemiMetrics
May 14 '15 at 2:51
Yes (and this is actually simpler).
– Davide Giraudo
May 14 '15 at 8:08
add a comment |
Actually, the pointwise limit of $(g_n)$ does exists almost surely. Indeed, pick $ rin (|phi|,1)$, and notice that
$$mathbb P{omegamid vertphivert^{k}vert u_{t-k}(omega)vertgt r^k }leqslant frac{|phi|^k}{r^k}mathbb E[|u_{t-k} |]leqslant frac{|phi|^k}{r^k}sigma . $$
Then conclude by the Borel-Cantelli lemma.
answered May 7 '15 at 22:07
Davide Giraudo
125k16150259
1
Thank you very much, Davide Giraudo! Can I proof it like this: by Markov's inequality, we have $Pr.(vertphivert^kvert u_{t-k}vert>A)leqfrac{vertphivert^{2k}mathbb{E}(vert u_{t-k}vert^2)}{A^2}$, where $A>0$ is a constant. These probabilities converge since $mathbb{E}(u_t)=sigma^2$ for all $t$. Then by Borel-Cantelli lemma, we got the almost sure convergence of $g_m(omega)$ series.
– SemiMetrics
May 14 '15 at 2:51
Yes (and this is actually simpler).
– Davide Giraudo
May 14 '15 at 8:08
add a comment |
Actually, the pointwise limit of $(g_n)$ does exists almost surely. Indeed, pick $ rin (|phi|,1)$, and notice that
$$mathbb P{omegamid vertphivert^{k}vert u_{t-k}(omega)vertgt r^k }leqslant frac{|phi|^k}{r^k}mathbb E[|u_{t-k} |]leqslant frac{|phi|^k}{r^k}sigma . $$
Then conclude by the Borel-Cantelli lemma.
answered May 7 '15 at 22:07
Davide Giraudo
125k16150259
Actually, the pointwise limit of $(g_n)$ does exists almost surely. Indeed, pick $ rin (|phi|,1)$, and notice that
$$mathbb P{omegamid vertphivert^{k}vert u_{t-k}(omega)vertgt r^k }leqslant frac{|phi|^k}{r^k}mathbb E[|u_{t-k} |]leqslant frac{|phi|^k}{r^k}sigma . $$
Then conclude by the Borel-Cantelli lemma.
answered May 7 '15 at 22:07
Davide Giraudo
125k16150259
edited May 14 '15 at 8:24
answered May 7 '15 at 22:07
Davide Giraudo
125k16150259
answered May 7 '15 at 22:07
Davide Giraudo
125k16150259
answered May 7 '15 at 22:07
Davide Giraudo
125k16150259
125k16150259
1
Thank you very much, Davide Giraudo! Can I proof it like this: by Markov's inequality, we have $Pr.(vertphivert^kvert u_{t-k}vert>A)leqfrac{vertphivert^{2k}mathbb{E}(vert u_{t-k}vert^2)}{A^2}$, where $A>0$ is a constant. These probabilities converge since $mathbb{E}(u_t)=sigma^2$ for all $t$. Then by Borel-Cantelli lemma, we got the almost sure convergence of $g_m(omega)$ series.
– SemiMetrics
May 14 '15 at 2:51
Yes (and this is actually simpler).
– Davide Giraudo
May 14 '15 at 8:08
add a comment |
1
Thank you very much, Davide Giraudo! Can I proof it like this: by Markov's inequality, we have $Pr.(vertphivert^kvert u_{t-k}vert>A)leqfrac{vertphivert^{2k}mathbb{E}(vert u_{t-k}vert^2)}{A^2}$, where $A>0$ is a constant. These probabilities converge since $mathbb{E}(u_t)=sigma^2$ for all $t$. Then by Borel-Cantelli lemma, we got the almost sure convergence of $g_m(omega)$ series.
– SemiMetrics
May 14 '15 at 2:51
Yes (and this is actually simpler).
– Davide Giraudo
May 14 '15 at 8:08
1
1
Thank you very much, Davide Giraudo! Can I proof it like this: by Markov's inequality, we have $Pr.(vertphivert^kvert u_{t-k}vert>A)leqfrac{vertphivert^{2k}mathbb{E}(vert u_{t-k}vert^2)}{A^2}$, where $A>0$ is a constant. These probabilities converge since $mathbb{E}(u_t)=sigma^2$ for all $t$. Then by Borel-Cantelli lemma, we got the almost sure convergence of $g_m(omega)$ series.
– SemiMetrics
May 14 '15 at 2:51
Thank you very much, Davide Giraudo! Can I proof it like this: by Markov's inequality, we have $Pr.(vertphivert^kvert u_{t-k}vert>A)leqfrac{vertphivert^{2k}mathbb{E}(vert u_{t-k}vert^2)}{A^2}$, where $A>0$ is a constant. These probabilities converge since $mathbb{E}(u_t)=sigma^2$ for all $t$. Then by Borel-Cantelli lemma, we got the almost sure convergence of $g_m(omega)$ series.
– SemiMetrics
May 14 '15 at 2:51
Yes (and this is actually simpler).
– Davide Giraudo
May 14 '15 at 8:08
Yes (and this is actually simpler).
– Davide Giraudo
May 14 '15 at 8:08
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1266671%2fproving-stationarity-of-ar1-process%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
The almost sure convergence is frankly off-topic, simply note that as soon as $(u_t)$ is integrable with $(E|u_t|)$ bounded, the series $$sum_{k=0}^infty|phi|^kE|u_{t-k}|$$ converges.
– Did
Aug 20 '17 at 20:03