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Let $\mathcal{F_t}$ denote the filtration (history) of the process $\{X_t\}_{t \in Z}$ up to time t. In some vague sense I know that the filtration of a process up to time t represents the history of the process $\{X_t\}_{t \in Z}$ up to time t, however any formal measure theoretic definition is beyond my current reach.

In analogy with conditional expectation w.r.t. a random variable or particular realisation of a random variable, can $E[X_t|\mathcal F_{t-1}] $ be regarded in a way similar to $E[X|Y]$ or as $E[X|Y=y]$ where X and Y are both random variables. Or can it mean both depending on the context?

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Definition of $E[X|Y=y]$. If $P(Y=y)$, there is the following definition: $$ E[X|Y=y]=\frac{E[X\cdot 1_{\{Y=y\}}]}{P(Y=y)}.$$ In general, one has to use measure theory: $$ E[X|Y=y]=\int_{\mathbb{R}} x\,P_{X|Y=y}(dx).$$

The random variable $E[X|Y]$ has two definitions (which coincide). One is the following: $E[X|Y](\omega)=E[X|Y=Y(\omega)]$. The other definition uses $\sigma$-algebras: if you consider the $\sigma$-algebra generated by $Y$, $\sigma(Y)=\{Y^{-1}(B):B\text{ Borel set}\}$ (that is, the set of events given by $Y$), then $E[X|Y]$ is defined as $E[X|\sigma(Y)]$.

For example, consider the following situation: let $X$ be the random variable describing the number after rolling a dice, with $P(X=i)=1/6$ for $i=1,\ldots 6$; and let $Y$ be one if the number is even ($2$, $4$ or $6$) and $0$ otherwise ($1$, $3$ or $5$). Then $E[X|Y=1]=4$ and $E[X|Y=0]=3$, so $E[X|Y]=3 \cdot 1_{\{Y=0\}}+4\cdot 1_{\{Y=1\}}$.

For general $\sigma$-algebras $\mathcal{F}$, $E[X|\mathcal{F}]$ is defined as the unique random variable which is $\mathcal{F}$-measurable and $E[X\cdot 1_A]=E[E[X|\mathcal{F}]\cdot 1_A]$ for all $A\in\mathcal{F}$. If $\mathcal{F}=\sigma(Y)$, as I said $E[X|\mathcal{F}]=E[X|Y]$.

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