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Let $X, Y$ be random variables in $L^1$. We will write $X \preceq_c Y$ if for all convex functions $u$ on $\mathbb{R}$

$$ \mathbb{E} [u(X)] \leq \mathbb{E}[u(Y)].$$

Now let $q_X$ and $q_Y$ be the (left-continuous) quantile functions of $X$ and $Y$.

How to prove that

$$ \int_0^t q_X(s) d s \geq \int_0^t q_Y(s) ds$$

for all $0 \leq t \leq 1$ given $X \preceq_c Y$?

The case $t=0$ is trivial. For $t=1$ note that if $U$ is uniformly distributed on $(0,1)$, then $q_X (U)$ is distributed as $X$. Thus,

$$\int_0^1 q_X (s) ds = \mathbb{E}[q_X(U)] = \mathbb{E} [X] = \mathbb{E}[Y] = \int_0^1 q_Y (s) ds $$

since $x \mapsto x$ and $x \mapsto -x$ are both convex.

But I don't see how this can be generalized to prove the case where $t$ is not $0$ or $1$.

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I think you are trying to prove something like Theorem 5 in David Blackwell's paper Comparison of Experiments . Blackwell's functions $F_M$ and $F_m$ are, I think, the inverse functions of your $q_X$ and $q_Y$ (or maybe vice versa; the notations make my head spin). In general, your hypothesis $Eu(X)\le Eu(Y)$ for all convex $u$ is pretty strong, and has well-known necessary and sufficient equivalents, as in the famous paper of Cartier, et al. The exposition in Le Cam's Comparison of Experiments - A short review (see esp. the bottom of page 130) might serve as a roadmap.

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  • $\begingroup$ Thank you for your answer. I was hoping to find a direct proof somewhere. There is also a (convoluted) proof in Föllmer&Schieds "Stochastic Finance" (Chapter 2). Seems like one has to dissect that to arrive at a direct argument. $\endgroup$ – LittleDoe Sep 11 '18 at 9:13
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It is relatively well-known that $X \preceq_c Y$ if and only if $Y$ is a mean-preserving spread of $X$. (If this is not something you already know, I imagine it might be found in the links in kimchi lover's answer. Otherwise, it's probably in Shaked and Shanthikumar.)

The result you are looking for is stated as lemma $1$ of Brooks and Du (2018). The proof is in their appendix.

Apologies for the brief answer. I don't have time to write a detailed one at the moment, and I thought you'd rather a brief one sooner rather than a detailed one (possibly much) later. I'll try to find the time to develop this into a full answer later on.

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I found a source for a direct proof. Thanks to the two answers for telling me where to look.

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  • $\begingroup$ Nice to see you find something that worked. The source you cite is older, but also employs essentially the same argument found in Brooks and Du. I suppose B&D were unable to find your source when they tried to prove their lemma. $\endgroup$ – Theoretical Economist Sep 14 '18 at 11:08
  • $\begingroup$ This might make a good Comment (or edit) on your original Question, but it does not serve as a useful Answer in its present form. While links can be helpful, without a summary of the information given there they are flagged as "link-only" Answers. See Provide context for links, "Always quote the most relevant part of an important link, in case the target site is unreachable or goes permanently offline." $\endgroup$ – hardmath Sep 14 '18 at 15:06

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