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Suppose we have the moving average process: $X(t)=U(t)-aU(t-1)$ where $U(t)$ is a strictly stationary process. How can we prove that $X(t)$ is also strict and stationary in this case? For $X(t)$ to be strict sense stationery, all $N$-dimensional joint CDFs are invariant to arbitrary time-shifts $\tau$ : $P(X(t_{1})\le x_{1},...,X(t_{N})\le x_{N})=P(X(t_{1}+\tau)\le x_{1},...,X(t_{N}+\tau)\le x_{N})$

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    $\begingroup$ What is the definition of strictly stationary process that you are using? What conditions must $X(t)$ satisfy to be called a strictly stationary process? Can you prove that the simplest of such conditions must hold for $X(t)$ if $U(t)$ is assumed to be a strictly stationary process? Please do not answer in comments. Instead, edit your question to include the requested information; you can edit your question by clicking on the edit link below it. $\endgroup$ – Dilip Sarwate Jan 2 '18 at 16:04
  • $\begingroup$ Very good! Now, consider the case $N=1$. The distribution of $X(t_1)=U(t_1)-aU(t_1-1)$ depends on the joint distribution of $U(t_1)$ and $U(t_1-1)$. The distribution of $X(t_1+\tau))=U(t_1+\tau)-aU(t_1+\tau-1)$ depends on the joint distribution of $U(t_1+\tau)$ and $U(t_1+\tau-1)$. Since $U(t)$ is stationary, what, if anything, can you say about the joint distribution of $U(t_1)$ and $U(t_1-1)$ as compared to the joint distribution of $U(t_1+\tau)$ and $U(t_1+\tau-1)$? Are the two joint distributions the same? If so, why? If not, why not? Now what about $X(t_1)$ and $X(t_1+\tau)$? $\endgroup$ – Dilip Sarwate Jan 4 '18 at 3:56

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