<p>We are given two CDFs $F$ and $G$, such that $F$ FOSD $G$ i.e. $F(x) \leq G(x)$ $\forall x$. Consider the random variables $X\sim F$ and $Y\sim G$. Also, suppose $X$ and $Y$ take non-negative values.</p>

<p>We want to show that $\mathbb{E}(X) \geq \mathbb{E}(Y)$. </p>

<p>Here is the intuition: $F(x) \leq G(x)$ $\forall x$ means that the probability that the random variable $X$ take values smaller than $x$ is smaller than the probability that $Y$ take values smaller than $x$, and this is true for every $x$. Therefore, $X$ take higher values than $x$ more often than $Y$ takes indicating that $X$ will have the higher mean than $Y$. </p>

<p>Here is the proof:</p>

<p>\begin{eqnarray*}  &amp; F(x)  \leq  G(x) \ \ \ \forall x \\
\rightarrow  &amp; 1 - F(x)  \geq  1- G(x) \ \ \ \forall x \\  \rightarrow  &amp; \int_{0}^{\infty}1 - F(x) dx \geq  \int_{0}^{\infty} 1- G(x)dx \\  \rightarrow  &amp; \int_{0}^{\infty}\Pr(X&gt; x) dx \geq  \int_{0}^{\infty} \Pr(Y&gt; x)dx \\  \rightarrow  &amp; \int_{0}^{\infty} \int_{x}^{\infty}f_X(a) da dx \geq  \int_{0}^{\infty} \int_{x}^{\infty}f_Y(a) da dx \\  \rightarrow  &amp; \int_{0}^{\infty} \int_{0}^{a}f_X(a) dx da \geq  \int_{0}^{\infty} \int_{0}^{a}f_Y(a) dx da \\  \rightarrow  &amp; \int_{0}^{\infty}  \int_{0}^{a} dx \ f_X(a) \ da \geq  \int_{0}^{\infty} \int_{0}^{a} dx \ f_Y(a) \ da \\  \rightarrow  &amp; \int_{0}^{\infty}  a \ f_X(a) \ da \geq  \int_{0}^{\infty} a \ f_Y(a) \ da  \\  \rightarrow  &amp; \mathbb{E}(X) \geq \mathbb{E}(Y) \ \end{eqnarray*}</p>

CC BY-SA 3.0
2017-01-23T17:56:19.780Z
2017-01-23T17:56:19.780Z
Amit
<blockquote>
  <p>Use the utility index $U(x) = x$ to prove that if the distribution of $F$ first-order stochastically dominates distribution $G$, then the mean of $x$ under $G$ cannot exceed the mean of $x$ under $F$.</p>
</blockquote>

<p>Attempted proof - Suppose the $F$ is first-order stochastically dominates $G$ then $$F(x) \leq G(x) \ \ \forall x$$ Since the expectation preserves linearity then it follows that $$\mathbb{E}\left[F(x)\right] \leq \mathbb{E}\left[G(x)\right] \ \ \forall x$$</p>

<p>I am not sure if this is correct or rigorous enough. Any suggestions are greatly appreciated. </p>

CC BY-SA 3.0
2017-01-23T16:30:01.900Z
What does preservation of linearity has to do with any of these...?
Giskard
<utility><financial-economics>
Implication of first-order stochastic dominance
Wolfy
# Implication of first-order stochastic dominance

(Asked by: username_0 on 2017-01-23)

> Use the utility index $U(x) = x$ to prove that if the distribution of $F$ first-order stochastically dominates distribution $G$, then the mean of $x$ under $G$ cannot exceed the mean of $x$ under $F$.

Attempted proof - Suppose the $F$ is first-order stochastically dominates $G$ then $$F(x) \leq G(x) \ \ \forall x$$ Since the expectation preserves linearity then it follows that $$\mathbb{E}\left[F(x)\right] \leq \mathbb{E}\left[G(x)\right] \ \ \forall x$$

I am not sure if this is correct or rigorous enough. Any suggestions are greatly appreciated.

**Tags:** utility, financial-economics
> **Comments:**
> **username_1**: What does preservation of linearity has to do with any of these...?  

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## 1 answers

### Answer by username_2 on 2017-01-23. Score: 4 | Accepted answer

We are given two CDFs $F$ and $G$, such that $F$ FOSD $G$ i.e. $F(x) \leq G(x)$ $\forall x$. Consider the random variables $X\sim F$ and $Y\sim G$. Also, suppose $X$ and $Y$ take non-negative values.

We want to show that $\mathbb{E}(X) \geq \mathbb{E}(Y)$.

Here is the intuition: $F(x) \leq G(x)$ $\forall x$ means that the probability that the random variable $X$ take values smaller than $x$ is smaller than the probability that $Y$ take values smaller than $x$, and this is true for every $x$. Therefore, $X$ take higher values than $x$ more often than $Y$ takes indicating that $X$ will have the higher mean than $Y$.

Here is the proof:

\begin{eqnarray\*} & F(x) \leq G(x) \ \ \ \forall x \\
\rightarrow & 1 - F(x) \geq 1- G(x) \ \ \ \forall x \\ \rightarrow & \int\_{0}^{\infty}1 - F(x) dx \geq \int\_{0}^{\infty} 1- G(x)dx \\ \rightarrow & \int\_{0}^{\infty}\Pr(X> x) dx \geq \int\_{0}^{\infty} \Pr(Y> x)dx \\ \rightarrow & \int\_{0}^{\infty} \int\_{x}^{\infty}f\_X(a) da dx \geq \int\_{0}^{\infty} \int\_{x}^{\infty}f\_Y(a) da dx \\ \rightarrow & \int\_{0}^{\infty} \int\_{0}^{a}f\_X(a) dx da \geq \int\_{0}^{\infty} \int\_{0}^{a}f\_Y(a) dx da \\ \rightarrow & \int\_{0}^{\infty} \int\_{0}^{a} dx \ f\_X(a) \ da \geq \int\_{0}^{\infty} \int\_{0}^{a} dx \ f\_Y(a) \ da \\ \rightarrow & \int\_{0}^{\infty} a \ f\_X(a) \ da \geq \int\_{0}^{\infty} a \ f\_Y(a) \ da \\ \rightarrow & \mathbb{E}(X) \geq \mathbb{E}(Y) \ \end{eqnarray\*}


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