# Expectations, double integrals and Jensen's inequality

$$\def\anonfunc#1{#1(\cdot)}$$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 $$x, so $$[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 $$E(u(x))$$.

• 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. – Dieter Kadelka Oct 29 '19 at 11:02
• 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. – LSpice Oct 29 '19 at 14:15

This inequality is false is general. E.g., let $$u(t)\equiv\min(0,t)$$, $$x=-1$$, $$y=5$$, $$b=4$$, $$f(c)=\frac14\,I\{0, $$g(v)=\frac12\,I\{-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 $$Eu(a-C)I\{a If $$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 However, because $$u(a-c)$$ is decreasing in $$c$$ and $$I\{a is nondecreasing in $$c$$, the Chebyshev integral inequality implies the inequality in the direction opposite to your desired one: $$Eu(a-C)I\{a and this inequality will usually be strict.
• 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 $c$? – carlogambino Oct 29 '19 at 21:29
• @carlogambino : Even if you assume that $C$ and $V$ are iid, your inequality will still be false in general. – Iosif Pinelis Oct 30 '19 at 1:41
• 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. – carlogambino Oct 30 '19 at 18:40
• @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: $X,Y_1,Z_2,\dots$. – Iosif Pinelis Oct 31 '19 at 0:53