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Let $Q_0$ and $Q_1$ be distributions on $\mathbb{R}$ with distribution function $G_0$ and $G_1$ respectively. I am trying to show that

  1. $G_0(y) \geq G_1(y), y \in \mathbb{R}$
  2. $G_0(y-) \geq G_1(y-), y \in \mathbb{R}$
  3. $\int h dQ_0 \leq \int h dQ_1 \: \: \: \text{for any non decreasing function} \: \: \: h: \mathbb{R} \to [0,\infty)$

these three are equivalent.

My idea was to use: $Z \geq 0$

$$\mathbb{E}(Z)=\int_{0}^{\infty}\mathbb{P}(Z \geq t)dt=\int_{0}^{\infty}\mathbb{P}(Z > t)dt$$

Any idea to proceed would be appreciated.

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closed as off-topic by Did, user284331, Saad, HK Lee, user223391 Apr 18 '18 at 14:10

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1) implies 2) follows by taking limits. 2) implies 1) follows by noting that distribution functions are continuous except at countable many points, so we get $G_0(y) \geq G_1 (y)$ at all but countable many points; but distribution functions are also right-continuous at all points so we get 1). 3) implies 1) follows by taking $h=I_{(y,\infty)}$. The tricky part is 1) implies 3). There are several approaches but I think the following is an elegant one. Consider $(0,1)$ as a probability space by providing it with Borel sigma algebra and Lebesgue measure. Let $X(\omega)=\inf \{t:G_0(\omega) \geq t\}$. This is the so-called generalized inverse of $G_0$. One check easily that $X(\omega) \leq t$ if and only if $G_0(\omega) \geq t$ and this implies that $G_0$ is the distribution function of $X$. Similarly let $Y(\omega)=\inf \{t:G_1(\omega) \geq t\}$ so $G_1$ is the distribution function of $Y$. By 1) it follows that $X(\omega) \leq Y(\omega)$ for all $\omega$. Hence $h(X(\omega)) \leq h(Y(\omega))$ for all $\omega$. Taking expectation we get 3)..

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