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A convex relation is that $x\succeq y$ implies $\alpha x+(1-\alpha)y\succeq y$.

Let $>_{FOSD}$ be $\succ$, is the FOSD convex? Intuitively it seems convex.

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    $\begingroup$ By FOSD do you mean "first order stochastic dominance"? If so, the answer is yes. $\endgroup$ – Herr K. Apr 2 at 1:23
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    $\begingroup$ Yes, I can think of no other relation commonly referred to as FOSD. $\endgroup$ – Bayesian Apr 2 at 10:58
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The first order stochastic dominance relation is convex.

An easy way to prove this is to use the property that a cdf $F$ FOSD another cdf $G$ if and only if $F(x)\le G(x)$ for all $x$. That is, $F$ FOSD $G$ if and only if the graph of $F$ is never above the graph of $G$. It is then easy to show that $F$ is never above any convex combination $H(x)=\alpha F(x)+(1-\alpha)G(x)$, which in turn is never above $G$.

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