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Jun 12, 2020 at 20:47 comment added user158968 Thanks. $\qquad$
Jun 12, 2020 at 20:47 vote accept user158968
Jun 12, 2020 at 20:37 comment added ofer zeitouni Because $F(t)$ is the expression I wrote, and not the one you did. If you take $F=\lambda F_1+(1-\lambda) F_2$, where $F_i$ are deterministic functions, the measure you will get is not the convex combination of $\mu_1$ and $\mu_2$.
Jun 12, 2020 at 20:35 comment added user158968 Why do you disagree?
Jun 12, 2020 at 20:34 comment added user158968 That is what I wanted. Thank you.
Jun 12, 2020 at 20:33 history edited ofer zeitouni CC BY-SA 4.0
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Jun 12, 2020 at 20:32 comment added ofer zeitouni I thought you wanted a drift $F$ that will generate the measure $\mu$. I constructed one for you. Wasn't it your question? BTW, I somewhat disagree with what you wrote in the "deterministic case".
Jun 12, 2020 at 16:33 comment added user158968 I'm sorry, I'm not sure exactly what you're showing here.
Jun 12, 2020 at 15:17 history answered ofer zeitouni CC BY-SA 4.0