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Apr 20, 2017 at 16:14 vote accept amakelov
Apr 14, 2017 at 8:55 comment added user541686 Should probably be on Math.SE rather than here... I also wouldn't really call this "convex analysis"...
Apr 14, 2017 at 8:36 answer added coudy timeline score: 5
Apr 14, 2017 at 8:20 answer added Fedor Petrov timeline score: 4
Apr 14, 2017 at 8:19 comment added Oscar Cunningham If we write $\mu$ as a weighted average of some other distributions $\nu_i$, then the distribution formed by the weighted average of $\tilde{\nu}_i$ has this property. One could probably state this as Jensen's inequality on $X$ where $X$ is some random variable in a joint distribution with some other random variable $Y$ which we are conditioning on. I guess $\mathbb E\left(f\left(X\right)\right)\geq\mathbb E\left(f\left(\mathbb E\left(X|Y\right)\right)\right)$.
Apr 14, 2017 at 8:15 history edited amakelov CC BY-SA 3.0
conveyed question better
Apr 14, 2017 at 8:05 history asked amakelov CC BY-SA 3.0