<blockquote>
<p>[...] integration by parts [...]</p>
</blockquote>
<p>Recall the formula for integration by parts: <span class="math-container">$\int_a^b u\,\mathrm dv = uv\vert_a^b - \int_a^b v\,\mathrm du$</span>. Now let <span class="math-container">$u=U$</span> and <span class="math-container">$v = H$</span> (so that <span class="math-container">$\mathrm dv = \mathrm dH = h\,\mathrm dx$</span>. Then
<span class="math-container">\begin{align}
\int_0^\infty \underbrace{U(x)}_{u}\,\underbrace{h(x)\mathrm dx}_{\mathrm dv} &amp;= \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} \\
&amp;=U(x)H(x)\bigg\vert_0^\infty - \int_0^\infty H(x)\, U'(x) \mathrm dx
\end{align}</span>
This yields the middle line in your second picture.</p>
<hr />
<blockquote>
<p>why <span class="math-container">$U(x)H(x)\vert_0^\infty=0$</span>?</p>
</blockquote>
<p>Note that <span class="math-container">$H(x)=[F(x)-G(x)]$</span>, and that <span class="math-container">$F(0)=G(0)=0$</span> and <span class="math-container">$\lim_{x\to\infty}F(x)=\lim_{x\to\infty}G(x)=1$</span> because <span class="math-container">$F$</span> and <span class="math-container">$G$</span> are cumulative distribution functions with support on <span class="math-container">$[0,\infty)$</span>. Therefore <span class="math-container">$H(0)=\lim_{x\to\infty}H(x)=0$</span>.</p>
<hr />
<blockquote>
<p>to get to <span class="math-container">$[F(x)−G(x)]≤0$</span></p>
</blockquote>
<p>Given the assumption that utility is non-decreasing, i.e. <span class="math-container">$U'(x)\ge0$</span>, it follows that
<span class="math-container">$H(x)=F(x)-G(x)\le0$</span> for every <span class="math-container">$x$</span>
ensures <span class="math-container">$$-\int U'(x)H(x)\mathrm dx \ge 0$$</span>
thereby proving that statement (2) implies statement (1).</p>

CC BY-SA 4.0
2022-10-18T10:20:39.987Z
Does that last line completely follow? I can think of functions that are negative in places whose integrals are positive
dm63
CC BY-SA 4.0
2022-10-18T14:25:34.897Z
@dm63: Thanks for the catch. The direction of implication should have been reversed.
Herr K.
CC BY-SA 4.0
2022-10-18T00:42:19.613Z
2022-10-18T14:24:55.000Z
2022-10-18T14:24:55.000Z
Herr K.
<p>Here's the theorem, consisting of 2 statements:
<a href="https://i.sstatic.net/EZb6T.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/EZb6T.png" alt="enter image description here" /></a></p>
<p>The equivalence is proven with the aid of this:</p>
<p><a href="https://i.sstatic.net/cKLaG.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/cKLaG.png" alt="enter image description here" /></a></p>
<p>There are 2 things I don't understand about it. Firstly, why <span class="math-container">$ U(x).H(x)|\infty, 0 = 0 $</span>. And secondly, to get to <span class="math-container">$[F(x) - G(x)] \leq 0$</span> from that proof. I know the definition of integration from parts, but I'm no hero with it and I haven't been able to find a way to statement 2 on my own. I figure part of it must consist of replacing <span class="math-container">$\int U(x)h(x)dx$</span> with <span class="math-container">$-\int U'(x)h(x)dx$</span>, which would give me <span class="math-container">$\int U'(x)h(x)dx \leq 0$</span>, 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.</p>

<mathematical-economics>
I'm trying to understand a proof for First order stochastic dominance
Nasan
# I'm trying to understand a proof for First order stochastic dominance

(Asked by: username_0 on 2022-10-17)

Here's the theorem, consisting of 2 statements:
[![enter image description here](https://i.sstatic.net/EZb6T.png)](https://i.sstatic.net/EZb6T.png)

The equivalence is proven with the aid of this:

[![enter image description here](https://i.sstatic.net/cKLaG.png)](https://i.sstatic.net/cKLaG.png)

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.

**Tags:** mathematical-economics

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## 1 answers

### Answer by username_1 on 2022-10-18. Score: 5 | Accepted answer

> [...] 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.

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> 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 $H(0)=\lim\_{x\to\infty}H(x)=0$.

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> 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).

> **Comments:**
> **username_2**: Does that last line completely follow? I can think of functions that are negative in places whose integrals are positive  
> **username_1**: @dm63: Thanks for the catch. The direction of implication should have been reversed.  

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