If X and Y are stationary and ergodic processes, then, is XY a stationary and ergodic process? I think the answer is true, but I do not know how to find the mean (we do not know Y and X are independent of each other). Thank you.
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The answer is no to both questions, here are counterexamples. Take for $X$ a sequence of i.i.d. Bernoulli ($\pm 1$) random variables and take for $Z$ an independent sequence. To construct a counterexample to the first claim ($XY$ is stationary), set $Y_n=X_n$ for $n$ even and $Y_n=Z_n$ for $n$ odd. Then $Y$ by itself is again just i.i.d. Bernoulli, but $XY$ is equal to an i.i.d. Bernoulli sequence on odd indices and identically equal to $1$ on even indices, so that it is not stationary.
Even if $XY$ is stationary, it need not be ergodic. For that, take $X$ as before and let $\xi$ be another Bernoulli random variable that is independent of $X$ and set $Y = \xi X$. Then obviously $Y$ equal to $X$ in law, but $XY$ is identically equal to $1$ with probability $1/2$ and identically equal to $-1$ with probability $1/2$, so that it is stationary but not ergodic.
answered Mar 30, 2014 at 16:13
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$\begingroup$ but please look at the book (Asymptotic theory for econometricians by Halbert white proposition 3.35 page 45) it says : If $\{(Z_t,X_t,Y_t)\}$ is a stationary ergodic sequence, then $\{X_tX'_t\}$, $\{X_tY_t\}$, $\{Z_tX_t\}$, $\{Y_tY'_t\}$, $\{Z_tY_t\}$, and $\{Z_tZ'_t\}$ are stationary ergodic sequences. $\endgroup$– TPArrowCommented Mar 30, 2014 at 19:29
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2$\begingroup$ There is a difference between the pair $(X,Y)$ being a stationary ergodic sequence and $X$ and $Y$ each separately being a stationary ergodic sequence. Your question appeared to be for the latter but the reference you quote is for the former. $\endgroup$ Commented Mar 30, 2014 at 19:31
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$\begingroup$ But what is the difference, we can decompose a joint process to separate processes, is not it? $\endgroup$– TPArrowCommented Mar 30, 2014 at 19:33
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2$\begingroup$ Look at my first example: $X$ and $Y$ taken separately are stationary, but the pair $(X,Y)$ is not. $\endgroup$ Commented Mar 30, 2014 at 19:44
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$\begingroup$ Thank you. I have a question, it might seem easy but for me is difficult. If ${x_t,y_t}$ are joint stationary. then is $x_{t-a}y_t$ stationary for all $a$? $\endgroup$– TPArrowCommented Apr 11, 2014 at 16:03