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Consider two cumulative distribution functions $F(x)$ and $G(x)$ for $x\in[a,b]$ where $G(x)$ has the first-order stochastic dominance over $F(x)$. That is, $F(x)>G(x)$ for all $x\in(a,b)$. We assume $a<0$ and $b>0$. Let $f(x)$ and $g(x)$ be the probability density function of $F(x)$ and $G(x)$ respectively.

Suppose the expected value of $x$ under $F(x)$ is positive: $$ \int_{a}^{b}xf(x)dx=\int_{a}^{0}xf(x)dx+\int_{0}^{b}xf(x)dx>0. $$

Under this condition, does $f(x)-g(x)>0$ always hold in any interval of $0<x<b$?

Graphical Expression of the Question is Here.

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  • $\begingroup$ Thank you for your comments. Could you tell me how I can edit my question? Should I delete this question and post a new question? $\endgroup$
    – Tom M.
    Jul 25 '18 at 16:01
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    $\begingroup$ Please don't delete the question. There is a link that allows you to edit the question just below it. (It is just above and to the left of the box that shows your name.) $\endgroup$ Jul 25 '18 at 16:02
  • $\begingroup$ Also, your definition of FOSD seems much stronger than the usual definition. Is that intentional? $\endgroup$ Jul 25 '18 at 16:04
  • $\begingroup$ Yes, it is intentional. Thank you for pointing it out. $\endgroup$
    – Tom M.
    Jul 25 '18 at 16:05
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Counterexample:

Suppose $a=-1$ and $b=2$, that $F$ is the uniform distribution in the interval $[-1,2]$, and that $G$ is the uniform distribution in the interval $[0,2]$. Clearly the expected value of $x$ under $F$ is positive but $$g(x)=\frac{1}{2}>\frac{1}{3}=f(x)\quad\text{for all}\;x\in[0,b].$$

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