# Is First Order Stochastic Dominance (FOSD) relation convex?

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.

• By FOSD do you mean "first order stochastic dominance"? If so, the answer is yes. – Herr K. Apr 2 '19 at 1:23
• Yes, I can think of no other relation commonly referred to as FOSD. – Bayesian Apr 2 '19 at 10:58

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 $$G$$.