Timeline for Conditions under which a linear functional on a space of measures must be integration of a function
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 28, 2017 at 3:26 | vote | accept | Alex Mennen | ||
| Jan 26, 2017 at 23:47 | comment | added | Alex Mennen | I should have given more context for this question. The VNM theorem provides conditions for when a binary relation $\preceq$ on P(X) is given by $x\preceq y\iff f(x)\leq f(y)$ for some affine function $f$. This is normally interpreted as conditions under which $x\preceq y$ iff $U$ has higher expected value on y than on x, for some function $U$ on $X$, because those are equivalent if $X$ is finite (or if we only pay attention to finitely-supported probability measures on $X$). I want to strengthen the theorem so the conclusion is what the theorem is usually interpreted as saying, for any $X$. | |
| Jan 26, 2017 at 23:41 | comment | added | Alex Mennen | Ah, thanks. That condition clearly implies the condition I gave (which I came up with independently), but it doesn't seem obvious whether the reverse implication holds. | |
| Jan 25, 2017 at 8:13 | comment | added | Robert Furber | @Alex Mennen: The natural measure-theoretic condition coming from the theory of the Giry monad ( link.springer.com/chapter/10.1007/BFb0092872 ) is that a bounded measurable function $P(X) \to \mathbb{R}$ arises from integration against a bounded measurable function $X \to \mathbb{R}$ iff $$ f(\mathrm{proj}(\mathbb{P})) = \int_{P(X)}f d\mathbb{P} $$ for all $\mathbb{P} \in P(X)$. Is the condition you give in the question intended to rephrase this condition, or did you come up with it independently? | |
| Jan 22, 2017 at 23:08 | history | edited | Alex Mennen | CC BY-SA 3.0 |
added 330 characters in body
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| Jan 22, 2017 at 10:07 | answer | added | Robert Furber | timeline score: 3 | |
| Jan 22, 2017 at 8:53 | comment | added | Pietro Majer | In general, if $M$ is a linear space (here $M=M(X)$) and $F$ a linear subspace of the algebraic dual $M'$ of $M$, then any $f\in M'$ is in $F$ if and only if it is continuous w.r.to the weak topology $\sigma(M,F)$. (that is almost a tautology yet maybe a starting point) | |
| Jan 22, 2017 at 4:43 | history | asked | Alex Mennen | CC BY-SA 3.0 |