<p>The first order stochastic dominance relation is convex. </p>

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

CC BY-SA 4.0
2019-04-02T17:18:06.817Z
2019-04-02T17:18:06.817Z
Herr K.
<p>A convex relation is that <span class="math-container">$x\succeq y$</span> implies <span class="math-container">$\alpha x+(1-\alpha)y\succeq y$</span>.</p>

<p>Let <span class="math-container">$&gt;_{FOSD}$</span> be <span class="math-container">$\succ$</span>, is the FOSD convex? Intuitively it seems convex.</p>

CC BY-SA 4.0
2019-04-02T01:23:47.807Z
By FOSD do you mean "first order stochastic dominance"? If so, the answer is yes.
Herr K.
CC BY-SA 4.0
2019-04-02T10:58:54.703Z
Yes, I can think of no other relation commonly referred to as FOSD.
Bayesian
<microeconomics><mathematical-economics><decision-theory><theory><probability>
Is First Order Stochastic Dominance (FOSD) relation convex?
High GPA
# Is First Order Stochastic Dominance (FOSD) relation convex?

(Asked by: username_0 on 2019-04-01)

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.

**Tags:** microeconomics, mathematical-economics, decision-theory, theory, probability

> **Comments:**
> **username_1**: By FOSD do you mean "first order stochastic dominance"? If so, the answer is yes.  
> **username_2**: Yes, I can think of no other relation commonly referred to as FOSD.  

---

## 1 answers

### Answer by username_1 on 2019-04-02. Score: 7 | Accepted answer

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 $G$.


---

