Proving stationarity of AR(1) process












5












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Did
    Aug 20 '17 at 20:03


















5












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Did
    Aug 20 '17 at 20:03
















5












5








5


4



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked May 4 '15 at 16:30









SemiMetricsSemiMetrics

263




263












  • $begingroup$
    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.
    $endgroup$
    – Did
    Aug 20 '17 at 20:03




















  • $begingroup$
    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.
    $endgroup$
    – Did
    Aug 20 '17 at 20:03


















$begingroup$
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.
$endgroup$
– Did
Aug 20 '17 at 20:03






$begingroup$
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.
$endgroup$
– Did
Aug 20 '17 at 20:03












1 Answer
1






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    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.
    $endgroup$
    – SemiMetrics
    May 14 '15 at 2:51










  • $begingroup$
    Yes (and this is actually simpler).
    $endgroup$
    – Davide Giraudo
    May 14 '15 at 8:08











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$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    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.
    $endgroup$
    – SemiMetrics
    May 14 '15 at 2:51










  • $begingroup$
    Yes (and this is actually simpler).
    $endgroup$
    – Davide Giraudo
    May 14 '15 at 8:08
















0












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    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.
    $endgroup$
    – SemiMetrics
    May 14 '15 at 2:51










  • $begingroup$
    Yes (and this is actually simpler).
    $endgroup$
    – Davide Giraudo
    May 14 '15 at 8:08














0












0








0





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited May 14 '15 at 8:24

























answered May 7 '15 at 22:07









Davide GiraudoDavide Giraudo

126k16150261




126k16150261








  • 1




    $begingroup$
    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.
    $endgroup$
    – SemiMetrics
    May 14 '15 at 2:51










  • $begingroup$
    Yes (and this is actually simpler).
    $endgroup$
    – Davide Giraudo
    May 14 '15 at 8:08














  • 1




    $begingroup$
    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.
    $endgroup$
    – SemiMetrics
    May 14 '15 at 2:51










  • $begingroup$
    Yes (and this is actually simpler).
    $endgroup$
    – Davide Giraudo
    May 14 '15 at 8:08








1




1




$begingroup$
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.
$endgroup$
– SemiMetrics
May 14 '15 at 2:51




$begingroup$
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.
$endgroup$
– SemiMetrics
May 14 '15 at 2:51












$begingroup$
Yes (and this is actually simpler).
$endgroup$
– Davide Giraudo
May 14 '15 at 8:08




$begingroup$
Yes (and this is actually simpler).
$endgroup$
– Davide Giraudo
May 14 '15 at 8:08


















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