Given that a Borel probability measure $\mu$ on [0,1] is characterized by its moments, it seems natural to consider the following stochastic order: say that $\mu\le\mu'$ if $$\forall k\in\mathbb N,\qquad\int_0^1 x^k \mu(dx)\le \int_0^1 x^k \mu'(dx).$$ Has this notion been investigated? Is it equivalent to the (seemingly stronger) increasing convex order?
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$\begingroup$ in en.wikipedia.org/wiki/Stochastic_dominance#Second-order they go over at least the case of nondecreasing and concave utility functions $\endgroup$– Thomas KojarCommented Sep 23, 2022 at 17:35
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1$\begingroup$ Probability measures on [0,1] are characterized by their moments, so we do get an order relation. $\endgroup$– Christophe LeuridanCommented Sep 23, 2022 at 18:14
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$\begingroup$ Sure, hence the title of my question. But what I am asking is whether this order is known, and equivalent to the classical increasing convex order. $\endgroup$– DRJCommented Sep 24, 2022 at 7:16
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