<p>Suppose <span class="math-container">$L \succ_{FOSD} L'$</span> and let <span class="math-container">$L$</span> have CDF <span class="math-container">$F$</span> and <span class="math-container">$L'$</span> have CDF <span class="math-container">$G$</span>.
That is, let 
<span class="math-container">$$F(x) \leq G(x) \quad \forall x \quad \mbox{or} \quad 1-F(x) \geq 1-G(x) \quad \forall x. $$</span>
Now consider two lotteries, one is <span class="math-container">$L'$</span> and the other one is <span class="math-container">$L'$</span> but instead of paying out <span class="math-container">$x$</span> it pays out
<span class="math-container">$$y(x) = F^{-1}(G(x)) \geq x,$$</span>
where the inequality follows by FOSD. That is, the lottery always pays out more. So everyone with an increasing utility function (more is better) prefers the second lottery. How are the payoffs from this lottery distributed?
<span class="math-container">$$Pr(y(X) \leq y) = Pr ( F^{-1}(G(X)) \leq y ) = Pr ( X \leq G^{-1}(F(y)))= 
G(G^{-1}(F(y)))= F(y).$$</span>
Thus,
<span class="math-container">$$\mathbb E[u(L)]= \int u (y(x)) d F(y(x)) = \int u (y(x)) d G(x) 
\geq \int u (x) d G(x) = \mathbb E[u(L')]$$</span>
for any increasing <span class="math-container">$u$</span> by the <span class="math-container">$y(x)\geq x$</span>. </p>

<p>The other direction would be that <span class="math-container">$\mathbb E[u(L)]\geq \mathbb E[u(L')]$</span> for all increasing <span class="math-container">$u$</span> implies 
<span class="math-container">$L \succ_{FOSD} L'$</span>. Let's show the equivalent statement that if <span class="math-container">$L$</span> does not FOSD <span class="math-container">$L'$</span>, then there is at least one increasing <span class="math-container">$\widetilde u$</span> such that  <span class="math-container">$\mathbb E[\widetilde u(L)]&lt; \mathbb E[\widetilde u(L')]$</span>.
If there exists some <span class="math-container">$z$</span> such that <span class="math-container">$1-F(z) &lt; 1-G(z)$</span>, then there is the increasing utility function
<span class="math-container">$$ \widetilde u (x) =\begin{cases}
1 \mbox{ if } x&gt; z\\
0 \mbox{ if } x \leq z \end{cases}$$</span>
such that
<span class="math-container">$$\mathbb E[\widetilde u(L)]= 1-F(z)  &lt; 1-G(z) = \mathbb E[\widetilde u(L')].$$</span>
That's it.</p>

CC BY-SA 4.0
2019-05-23T22:39:28.410Z
2019-05-23T22:39:28.410Z
Bayesian
<p>Let <span class="math-container">$L$</span>,<span class="math-container">$L'$</span> be two lotteries over the real numbers. Let <span class="math-container">$u$</span> be an increasing Bernoulli utility function. Let <span class="math-container">$F_L$</span>, <span class="math-container">$F_{L'}$</span> be the CDFs of the two lotteries. We wish to show that 
<span class="math-container">$$L \succ_{FOSD} L' \iff \mathbb E(u(L))\geq \mathbb E(u(L')).$$</span>
I am able to prove the necessity. I am not able to prove sufficiency. I get an expression like <span class="math-container">$\int u'(x) f_L dx \leq \int u'(x) f_{L'} dx$</span> after integrating by parts, but am unable to proceed after that.
This is not a homework problem.</p>

CC BY-SA 4.0
2019-05-24T14:51:11.213Z
Note that I have edited your question. I think this is what you mean, but your original statement was wrong [L FOSD L' means F_L(x) < F_L' (x) not >]. However, I carried over your small abuse of notation regarding the utility function [first u is a function of a lottery, then it is a function of money].
Bayesian
<microeconomics><expected-utility>
First Order Stochastic Domination and lottery preferences
Student
# First Order Stochastic Domination and lottery preferences

(Asked by: username_0 on 2019-05-23)

Let $L$,$L'$ be two lotteries over the real numbers. Let $u$ be an increasing Bernoulli utility function. Let $F\_L$, $F\_{L'}$ be the CDFs of the two lotteries. We wish to show that
$$L \succ\_{FOSD} L' \iff \mathbb E(u(L))\geq \mathbb E(u(L')).$$
I am able to prove the necessity. I am not able to prove sufficiency. I get an expression like $\int u'(x) f\_L dx \leq \int u'(x) f\_{L'} dx$ after integrating by parts, but am unable to proceed after that.
This is not a homework problem.

**Tags:** microeconomics, expected-utility

> **Comments:**
> **username_1**: Note that I have edited your question. I think this is what you mean, but your original statement was wrong [L FOSD L' means F_L(x) < F_L' (x) not >]. However, I carried over your small abuse of notation regarding the utility function [first u is a function of a lottery, then it is a function of money].  

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## 1 answers

### Answer by username_1 on 2019-05-23. Score: 1 | Accepted answer

Suppose $L \succ\_{FOSD} L'$ and let $L$ have CDF $F$ and $L'$ have CDF $G$.
That is, let
$$F(x) \leq G(x) \quad \forall x \quad \mbox{or} \quad 1-F(x) \geq 1-G(x) \quad \forall x. $$
Now consider two lotteries, one is $L'$ and the other one is $L'$ but instead of paying out $x$ it pays out
$$y(x) = F^{-1}(G(x)) \geq x,$$
where the inequality follows by FOSD. That is, the lottery always pays out more. So everyone with an increasing utility function (more is better) prefers the second lottery. How are the payoffs from this lottery distributed?
$$Pr(y(X) \leq y) = Pr ( F^{-1}(G(X)) \leq y ) = Pr ( X \leq G^{-1}(F(y)))=
G(G^{-1}(F(y)))= F(y).$$
Thus,
$$\mathbb E[u(L)]= \int u (y(x)) d F(y(x)) = \int u (y(x)) d G(x)
\geq \int u (x) d G(x) = \mathbb E[u(L')]$$
for any increasing $u$ by the $y(x)\geq x$.

The other direction would be that $\mathbb E[u(L)]\geq \mathbb E[u(L')]$ for all increasing $u$ implies
$L \succ\_{FOSD} L'$. Let's show the equivalent statement that if $L$ does not FOSD $L'$, then there is at least one increasing $\widetilde u$ such that $\mathbb E[\widetilde u(L)]< \mathbb E[\widetilde u(L')]$.
If there exists some $z$ such that $1-F(z) < 1-G(z)$, then there is the increasing utility function
$$ \widetilde u (x) =\begin{cases}
1 \mbox{ if } x> z\\
0 \mbox{ if } x \leq z \end{cases}$$
such that
$$\mathbb E[\widetilde u(L)]= 1-F(z) < 1-G(z) = \mathbb E[\widetilde u(L')].$$
That's it.


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