Let $X=\{X_t, t \ge 0\}$ and $Y=\{Y_t, t \ge 0\}$ are two stochastic processes defined on common probability space. Then two show X and Y are independent, does it suffice to show for any $t$ and $s$, $X_t$ and $Y_s$ are independent? Or do we need stronger conditions?
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You need to show that the sigma-algebras $\sigma(\{X_t \mid t \ge 0\})$ and $\sigma(\{Y_t \mid t \ge 0\})$ are independent, it isn't sufficient that all pairs $X_t, Y_t$ are independent.
This can easily be seen by the following example: Let $X, Y$ be two independent random variables with nontrivial distribution. Then the processes $$\begin{align*}X_t &= X \cdot I\{t = 0\} + Y \cdot I\{t = 1\} \\ Y_t &= X \cdot I\{t = 1\} + Y \cdot I\{t = 0\}\end{align*}$$ are not independent, although all pairs $X_t, Y_t$ are independent.
Edit: The question was changed since I've answered it. To answer the new question, consider the standard example of random variables $X_1, X_2, X_3$ that are pairwise independent, but not jointly independent. Now consider the processes $$\begin{align*}Y_t &= X_1 \cdot I\{t = 0\} + X_2 \cdot I\{t = 1\} \\ Z_t &= X_3 \cdot I\{t = 0\}\end{align*}.$$
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$\begingroup$ How about any $t$ and $s$ though? In above case $X_1$ and $Y_0$ are not independent for example. $\endgroup$ Sep 25, 2015 at 22:13
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1$\begingroup$ You can still easily produce counterexamples. Simply look up a typical example of three random variables that are pairwise independent, but not jointly independent. $\endgroup$– DominikSep 25, 2015 at 22:15