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Given a sequence of random variables $\{X_{t}\}_{t=1}^{T}$ , and a filtration $\mathcal{F}_{T}$, what does it mean that $X_{t+1}$ is independent of $\mathcal{F}_{t}$? Is it simply that $X_{t+1}$ is independent of any $X_{j}$, when j $\leq$ t, or is it more sufisticated?

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  • $\begingroup$ If $\mathcal{F}_t=\sigma(X_j\mid j\leq t)$, then yes. Otherwise, no. $\endgroup$ – Stefan Hansen Jun 12 '14 at 11:28
  • $\begingroup$ @StefanHansen Sorry but $X_2$ can be independent of $X_0$ and independent of $X_1$ without being independent of $F_1=\sigma(X_0,X_1)$. $\endgroup$ – Did Jun 12 '14 at 11:41
  • $\begingroup$ @Did Sure. I (mis)read it as $X_{t+1}$ should be simultaneously independent of $X_j$ for $j\leq t$. I guess this is not what the OP meant. $\endgroup$ – Stefan Hansen Jun 12 '14 at 11:43
  • $\begingroup$ I don't know if this is implicit in the question but it needs to be stated that $\{X_t\}_{t=1}^T$ is an adapted process, right? $\endgroup$ – Calculon Aug 21 '14 at 9:50
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It just means that $X_{t+1}$ is independent of every ${\cal F}_t$-measurable random variable.

In the case ${\cal F}_t = \sigma(X_0, \dots, X_t)$, it means that $X_{t+1}$ is independent of $(X_0, \dots, X_t)$.

In practice, you often have ${\cal F}_t \supseteq \sigma(X_0, \dots, X_t)$ but not always equality.

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  • $\begingroup$ Regarding the second sentence of your post, note that $X_2$ can be independent of $X_0$ and independent of $X_1$ without being independent of $F_1=\sigma(X_0,X_1)$. $\endgroup$ – Did Jun 12 '14 at 11:42
  • $\begingroup$ yes. thanks for the review! $\endgroup$ – mookid Jun 12 '14 at 11:47

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