Consider a 0-1 random process $X(t)$, that takes values only in $\{0,1\}$, such that $\lim_{T \rightarrow \infty} \frac{1}{T}\sum_{t=1}^{T}X(t) = \overline{X}$ almost surely. If $Y(t) \in \{0, 1\}$ is a i.i.d. process that is also independent of $X(t)$ then can we say $\lim_{T \rightarrow \infty} \frac{1}{T}\sum_{t=1}^{T}X(t)Y(t) = \overline{X}\overline{Y}$, where $\overline{Y} = \mathbb{E}[Y(t)]$?
1 Answer
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There is a well-known characterization of weakly mixing dynamical systems as the ones whose product by any ergodic system is also ergodic. In your case $Y(t)$ is weakly mixing, because it is Bernoulli, $X(t)$ is ergodic by your assumption on convergence of its Cesaro averages to a constant, and $(X(t),Y(t))$ is a product process by the assumption on the indepenence of the processes $X$ and $Y$. Therefore, the sequence of pairs $(X(t),Y(t))$ is ergodic, whence the sequence of products $X(t)Y(t)$ is also ergodic. Thus, its Cesaro averages converge to $\mathbf E X_1(t)Y_1(t)=\mathbf E X_1(t) \mathbf E Y_1(t)=\overline X\overline Y$.
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1$\begingroup$ I think you’re assuming that $X(t)$ is stationary here (while the OP didn’t assume that). The result holds even for non-stationary $X$. Also it’s certainly not true that convergence of the Cesaro average of $X_i$’s implies that $X$ is ergodic. $\endgroup$ Commented Dec 3, 2017 at 3:18
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$\begingroup$ Assuming they converge to a constant it does :) $\endgroup$– R WCommented Dec 3, 2017 at 3:49