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.

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    $\begingroup$ 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]. $\endgroup$ – Bayesian May 24 '19 at 14:51

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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