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 $\vert\phi\vert<1$, 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)d\omega\\&=\int_{\Omega}\lim_{m\rightarrow\infty}\sum_{k=0}^{m}\phi^{k}u_{t-k}(\omega)d\omega\\&\triangleq\int_{\Omega}\lim_{m\rightarrow\infty}f_{m}(\omega)d\omega \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&=\left\vert \sum_{k=0}^{m}\phi^{k}u_{t-k}(\omega)\right\vert \\&\leq\sum_{k=0}^{m}\vert\phi\vert^{k}\vert u_{t-k}(\omega)\vert\\&\triangleq g_{m}(\omega) \end{align} One sufficient condition is that $\sup_t\vert u_t\vert<\infty$, 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)d\omega&=\lim_{m\rightarrow\infty}\int_{\Omega}g_{m}(\omega)d\omega\\&=\lim_{m\rightarrow\infty}\sum_{k=0}^{m}\vert\phi\vert^{k}\mathbb{E}(\vert u_{t-k}\vert)\\&\leq\lim_{m\rightarrow\infty}\sup_{t}\mathbb{E}(\vert u_{t}\vert)\sum_{k=0}^{m}\vert\phi\vert^{k}\\&\leq\lim_{m\rightarrow\infty}\sigma\sum_{k=0}^{m}\vert\phi\vert^{k}\\&=\frac{\sigma}{1-\vert\phi\vert}<\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!
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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
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Actually, the pointwise limit of $(g_n)$ does exists almost surely. Indeed, pick $ r\in (|\phi|,1)$, and notice that $$\mathbb P\{\omega\mid \vert\phi\vert^{k}\vert u_{t-k}(\omega)\vert\gt r^k \}\leqslant \frac{|\phi|^k}{r^k}\mathbb E[|u_{t-k} |]\leqslant \frac{|\phi|^k}{r^k}\sigma . $$ Then conclude by the Borel-Cantelli lemma.
Davide Giraudo
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1Thank you very much, Davide Giraudo! Can I proof it like this: by Markov's inequality, we have $Pr.(\vert\phi\vert^k\vert u_{t-k}\vert>A)\leq\frac{\vert\phi\vert^{2k}\mathbb{E}(\vert u_{t-k}\vert^2)}{A^2}$, 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 02:51
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