Here's the theorem, consisting of 2 statements: enter image description here

The equivalence is proven with the aid of this:

enter image description here

There are 2 things I don't understand about it. Firstly, why $ U(x).H(x)|\infty, 0 = 0 $. And secondly, to get to $[F(x) - G(x)] \leq 0$ from that proof. I know the definition of integration from parts, but I'm no hero with it and I haven't been able to find a way to statement 2 on my own. I figure part of it must consist of replacing $\int U(x)h(x)dx$ with $-\int U'(x)h(x)dx$, which would give me $\int U'(x)h(x)dx \leq 0$, but that's as far as I'm able to get. Please to try to plain in your answer, I don't find it easy to deal with jargon and deep abstraction.


1 Answer 1


[...] integration by parts [...]

Recall the formula for integration by parts: $\int_a^b u\,\mathrm dv = uv\vert_a^b - \int_a^b v\,\mathrm du$. Now let $u=U$ and $v = H$ (so that $\mathrm dv = \mathrm dH = h\,\mathrm dx$. Then \begin{align} \int_0^\infty \underbrace{U(x)}_{u}\,\underbrace{h(x)\mathrm dx}_{\mathrm dv} &= \underbrace{U(x)}_u\,\underbrace{H(x)}_{v}\bigg\vert_0^\infty - \int_0^\infty \underbrace{H(x)}_{v}\,\underbrace{\mathrm dU(x)}_{\mathrm du} \\ &=U(x)H(x)\bigg\vert_0^\infty - \int_0^\infty H(x)\, U'(x) \mathrm dx \end{align} This yields the middle line in your second picture.

why $U(x)H(x)\vert_0^\infty=0$?

Note that $H(x)=[F(x)-G(x)]$, and that $F(0)=G(0)=0$ and $\lim_{x\to\infty}F(x)=\lim_{x\to\infty}G(x)=1$ because $F$ and $G$ are cumulative distribution functions with support on $[0,\infty)$. Therefore $H(0)=\lim_{x\to\infty}H(x)=0$.

to get to $[F(x)−G(x)]≤0$

Given the assumption that utility is non-decreasing, i.e. $U'(x)\ge0$, it follows that $H(x)=F(x)-G(x)\le0$ for every $x$ ensures $$-\int U'(x)H(x)\mathrm dx \ge 0$$ thereby proving that statement (2) implies statement (1).

  • $\begingroup$ Does that last line completely follow? I can think of functions that are negative in places whose integrals are positive $\endgroup$
    – dm63
    Commented Oct 18, 2022 at 10:20
  • $\begingroup$ @dm63: Thanks for the catch. The direction of implication should have been reversed. $\endgroup$
    – Herr K.
    Commented Oct 18, 2022 at 14:25

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