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Mar 8, 2023 at 15:57 vote accept user123234
Mar 8, 2023 at 15:57 history undeleted user123234
Mar 8, 2023 at 15:37 history deleted user123234 via Vote
Mar 8, 2023 at 14:48 answer added user173262 timeline score: 4
Mar 8, 2023 at 14:11 comment added Kurt G. As far as I remember the $\mathcal{F}\otimes \mathcal{B}([0,\infty))$-measurability of $X$ (that maps $\Omega\times [0,\infty)$ to $\mathbb R$ (typically) or $\mathbb R^d$) is only a mild and technical condition. Much more important than that is the ${\cal F}_t$-measurability of every $X_t$ because that reflects the gain in information.
Mar 8, 2023 at 13:44 history asked user123234 CC BY-SA 4.0