1
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Assume I have a particular function $u(x)= \sqrt x $ (note: $u'(x)\geq0, u'(x) \leq 0, \forall x$)

Assume that I have two continuous, probability distributions over $x$, $F(x)$ and $G(x)$ but that they are unknown (we know they exist but we dont know what they look like)

Assume the following about the expectations given the two distributions: $\int u(x)dF(x) \geq \int u(x)dG(x)$

(1)

Given the information above, based on one specific functional form of $u(x)$, can I conclude: $F(x)\leq G(x), \forall x$?

(2)

If I assume $u(x)$ such that $\quad u'(x)\geq0, \quad u'(x) \leq 0 \quad \forall x$ (but do not specify what functional form $u(x)$ takes)

and I know $\int u(x)dF(x) \geq \int u(x)dG(x)$

can I conclude: $F(x)\leq G(x), \forall x$ ?

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For (1), the answer is clearly no. For example, let $F(x)$ be the uniform distribution over $[0,1]$ and $G(x)$ be the cdf of the degenerate distribution $$\mathbf P(X=1/2)=1.$$ Then it is easy to see that $$F(x)=\begin{cases}0, &\text{if }x<0\\x,&\text{if }0\le x<1\\1,&\text{if } x\ge 1\end{cases},\,\,\,G(x)=\begin{cases}0,&\text{if }x<1/2\\1,&\text{if }x\ge 1/2\end{cases}.$$ Clearly, (Note that although $G$ is not continuous here, it can be arbitrarily approximated by a distribution function that is continuous) $$\int_{-\infty}^\infty u(x)\,dF(x)=\frac 23< \int_{-\infty}^\infty u(x)\,dG(x)=\sqrt{\frac 12}.$$ But clearly, neither "$F(x)\ge G(x)$ for all $x$" nor "$G(x)\ge F(x)$ for all $x$" holds.

For (2), if you mean "for SOME $u(x)$ satisfying $u'\ge 0$ and $u''\le 0$, the inequality holds", then the answer is clearly the same as (1), since $u(x)$ could rightly be the function you specified in (1). But if you mean "for ALL $u(x)$ satisfying $u'\ge 0$ and $u''\le 0$, the inequality holds", then the answer is yes, because $F$ first-order stochastically dominates $G$ if and only if $$\int u(x)\,dF(x)\ge\int u(x)\,dG(x)$$ for EACH concave and increasing function $u(x)$, while each concave and increasing function can be approximated by a function which is twice differentiable, increasing, and concave.

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  • $\begingroup$ Hi Onol, Thank you for responding. Well then I am not sure what to say any more. If I know $F(x)\leq G(x), \forall x$ I can always conclude $\int u(x)dF(x) \geq \int u(x)dG(x)$, for all nondecreasing $u$, I dont need to check what possible functions satify it as long as I know they are nondecreasing. but based on what your saying and what I suspected I cannot say if $\int u(x)dF(x) \geq \int u(x)dG(x)$ then $F(x)\leq G(x), \forall x$ unless I check every possible $u$ in the universe. Just because I have $\int u(x)dF(x) \geq \int u(x)dG(x)$ for 1 specific function $u(x)$ I cannot assume FOSD. $\endgroup$
    – jessica
    Jun 13 '17 at 19:36
  • $\begingroup$ For case (1) mentioned above. Could I at least conclude SOSD? Adding the assumption that $u$ is concave. For some that is under consideration? ie I dont have to check for all $u$'s. $\endgroup$
    – jessica
    Jun 13 '17 at 19:53
  • $\begingroup$ @jessica For your first comment, yes, you cannot conclude $F(x)\le G(x)$ for all $x$ just based on testing the integral inequality for SOME function $u$. You know, that property is often useful in the converse way: if you know that $F$ FOSD $G$ then for any function nondecreasing you can obtain the integral inequality. $\endgroup$
    – OnoL
    Jun 13 '17 at 21:54
  • $\begingroup$ @jessica For your second comment, unfortunately, the answer is still no. To illustrate, let us shift the degenerate distribution a little bit to $\mathbf P(X=1/2-\epsilon)=1$ for some sufficiently small $\epsilon>0$. Then you can still get the integral inequality for $u(x)=\sqrt x$, but it is easy to see that $\int_0^1 xdF(x)=1/2>\int_0^1xdG(x)=1/2-\epsilon$, which means that $G$ fails to second order stochastically dominate $F$. $\endgroup$
    – OnoL
    Jun 13 '17 at 22:03
  • $\begingroup$ Onol, I really appreciate your help. The reason I am asking these questions is because lets say you have a "big U" $U(L)$ expected utility function that is VNM, with a concave bernuli utility, and it satisfies all the necessary axioms of EUH. Since $U(L)$ is a utility representing preferences of lotteries, we can say: $U(L_1)>U(L_2)$ IFF agent prefers lottery $L_1$ to $L_2$. From what I gather from our discussion is even if an agent, with a VNM function that satifies EUH, and has concave bernulli utility, a preference lottery $L_1$ to $L_2$ does not imply either $L_1$ FOSD or SOSD $L_2$ $\endgroup$
    – jessica
    Jun 13 '17 at 22:28

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