$\def\anonfunc#1{#1(\cdot)}$Consider two random variables distributed $v\sim \anonfunc G$ and $c \sim \anonfunc F$ with pdfs $\anonfunc g$ and $\anonfunc f$. Let the supports of $c$ and $v$ be $[x,y]$. Let $x<a=E(v)<b<y$, so $[a,b]\subset [x,y]$. Now consider a strictly concave (twice differentiable and continuous) function $\anonfunc u$, with $\anonfunc{u'}>0$, $\anonfunc{u''}<0$, and $u(0)=0$ (passes through the origin). Establish sufficient conditions such that the expression $$ \int_{a}^{b}u(E(v)-c)f(c)dc-\int_{a}^{b}\int_{x}^{y}% u(E(v)-v)g(v)f(c)dvdc\geq0\quad\forall v,c, $$ where $E(v)=\int_{x}^{y}vg(v)dv.$

Things I've tried:

  1. $\int_{0}^{\bar{v}}u(E(v)-v)g(v)dv\leq0$ by Jensen's inequality. To see this, let $E(v)-v=t$. But $E(t)=E_{v}[E(v)-v]=0$, and so $E(u(t))\leq u(E(t))=0$, since $u(0)=0$ by assumption.

  2. Clearly, $\int_{a}^{b}u(E(v)-c)f(c)dc\leq0$, since we are integrating the integrand $(E(v)-c)$ from $a=E(v)$ to $b$.

  3. Intuitively, a variant of Jensen's inequality should apply if $c$ and $v$ are i.i.d. Let $c$ and $v$ be i.i.d. with identical supports. Then the integrands are the same, and we have the expression $\int_{a}^{b} u(E(v)-v)f(c)dc-\int_{a}^{b}\int_{x}^{y}u(E(v)-v)g(v)f(c)dvdc$. However, we can't apply Jensen's inequality directly since $\int_{a}^{b}u(E(v)-v)f(c)dc$ is not $u(E(x))$, even if we "factor out" the outer integrals. $\int_{x}% ^{y}\anonfunc u g(v)dv$ seems to be a form of $E(u(x))$.

  • 1
    $\begingroup$ Your notation is irritating. You use the same symbol $v$ for the random variable $v$ and value $v$. Similarly for $c$. Do you assume that $v$ and $c$ are independent? Even if not, the double integral $\int_a^b \int_x^y u \ldots dvdc$ seems to be a product of simple integrals. $\endgroup$ – Dieter Kadelka Oct 29 '19 at 11:02
  • $\begingroup$ While we're at it, some of this TeX seems weirdly autogenerated (for example, $u^{^{\prime\prime}}$ $u^{^{\prime\prime}}$ for the second derivative, which should be $u''$ $u''$ or $u^{\prime\prime}$ $u^{\prime\prime}$, and an unexpected \lbrack where [ would work fine). I have edited to try to fix everything that seemed suspicious, TeX-wise. I also deleted the "Thank you", since the consensus is to omit such pleasantries. $\endgroup$ – LSpice Oct 29 '19 at 14:15

This inequality is false is general. E.g., let $u(t)\equiv\min(0,t)$, $x=-1$, $y=5$, $b=4$, $f(c)=\frac14\,I\{0<c<4\}$, $g(v)=\frac12\,I\{-1<v<1\}$, where $I$ denotes the indicator, so that $a=EV=0$, where $V$ is the random variable you denoted by $v$. Then your inequality becomes $-2-(-\frac14)\ge0$, which is false.

(If you insist that the function $u$ be smooth and strictly concave with a strictly positive derivative, to achieve that you can just tweak slightly the function given by $u(t)\equiv\min(0,t)$.)

Even if you assume that $C$ and $V$ are iid, as you suggest in your comment, your inequality will still be false in general. Indeed, your inequality can be written as $$Eu(a-C)I\{a<C<b\}\ge Eu(a-V)\,EI\{a<C<b\}. $$ If $C$ and $V$ are identically distributed and $C$ is always less than $b$, then your inequality can be rewritten as $$Eu(a-C)I\{a<C\}\ge Eu(a-C)\,EI\{a<C\}. $$ However, because $u(a-c)$ is decreasing in $c$ and $I\{a<c\}$ is nondecreasing in $c$, the Chebyshev integral inequality implies the inequality in the direction opposite to your desired one: $$Eu(a-C)I\{a<C\}\le Eu(a-C)\,EI\{a<C\}, $$ and this inequality will usually be strict.

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  • $\begingroup$ The inequality is not true in general, but the question asks for (the weakest) sufficient conditions such that the inequality holds. For example, what if $v$ and $c$ are i.i.d.? Or if $v$ FOSD/SOSD dominates $c$? $\endgroup$ – carlogambino Oct 29 '19 at 21:29
  • $\begingroup$ @carlogambino : Even if you assume that $C$ and $V$ are iid, your inequality will still be false in general. $\endgroup$ – Iosif Pinelis Oct 30 '19 at 1:41
  • $\begingroup$ Hi, I'm not sure that $Eu(a-C)I\{a<C<b\}$ is the same as $\int_{a}^{b}u(a-c)f(c)dc$. They seem to be different values, at least that's what Mathematica tells me. $\endgroup$ – carlogambino Oct 30 '19 at 18:40
  • $\begingroup$ @carlogambino : For $h(c):=u(a-c)I\{a<c<b\}$, we have $$Eu(a-C)I\{a<C<b\}=Eh(C)=\int_{-\infty}^\infty h(c)f(c)\,dc=\int_{-\infty}^\infty u(a-c)I\{a<c<b\}f(c)\,dc=\int_a^b u(a-c)f(c)\,dc,$$ as was claimed. As for Mathematica, it can only do what you tell it to do. Also, as pointed out by Dieter Kadelka, it is not a good idea to denote a random variable (r.v.) and its values by the same symbol; that way, you can only confuse other people and even yourself. The standard convention is to denote r.v.'s (which are maps and not numbers) by capital Roman letters: $X,Y_1,Z_2,\dots$. $\endgroup$ – Iosif Pinelis Oct 31 '19 at 0:53

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