<p>Let me show that for the EU model, FOSD and monotonicity of the utility function implies preference dominance.</p>
<p>The reasoning is easier if we have a finite number of outcomes, say:
<span class="math-container">$$
x_1 &gt; x_2 \ldots &gt; x_n,
$$</span>
Consider associated probabilities <span class="math-container">$p_1,\ldots, p_n$</span> and <span class="math-container">$q_1,\ldots, q_n$</span> over these outcomes.</p>
<p>Let <span class="math-container">$P_i = \sum_{j \le i} p_i$</span> be the probability that the outcome is larger or equal to <span class="math-container">$x_i$</span> and similarly, define <span class="math-container">$Q_i = \sum_{j \le i} q_i$</span>.</p>
<p>We have that <span class="math-container">$p$</span> FOSD <span class="math-container">$q$</span> if <span class="math-container">$P_i \ge Q_i$</span> for all <span class="math-container">$i$</span> with at least one strict inequality. Note that <span class="math-container">$P_n = Q_n = 1$</span>.</p>
<p>Let <span class="math-container">$u(x_1) &gt; u(x_2) &gt; \ldots &gt; u(x_n)$</span>. We would like to show that <span class="math-container">$\sum_{i = 1}^n p_i u(x_i) &gt; \sum_{i = 1}^n q_i u(x_i)$</span>.</p>
<p>Indeed:
<span class="math-container">$$
\begin{align*}
\sum_{i = 1}^n (p_i - q_i) u(x_i) &amp;= \sum_{i = 1}^n \left(\sum_{j \le i} p_j - \sum_{j \le i} q_j \right) u(x_i) - \sum_{i = 2}^n \left(\sum_{j &lt; i} p_i - \sum_{j &lt; i} q_i\right) u(x_i),\\
&amp;= \sum_{i = 1}^n (P_i - Q_i) u(x_i) - \sum_{i = 2}^n (P_{i-1} - Q_{i-1}) u(x_i),\\
&amp;= \sum_{i = 1}^n (P_i - Q_i) u(x_i) - \sum_{j = 1}^{n-1} (P_j - Q_j)u(x_{j+1}),\\
&amp;= \sum_{i = 1}^{n-1}\underbrace{(P_i - Q_i)}_{\ge 0} \underbrace{(u(x_i) - u(x_{j+1}))}_{&gt; 0} + \underbrace{(P_n - Q_n)}_{ = 0} u(x_n),\\
&amp;&gt; 0
\end{align*}
$$</span></p>
<p>Consider now the case of a continuous outcome <span class="math-container">$x \in [0,1]$</span>. Let <span class="math-container">$P$</span> and <span class="math-container">$Q$</span> be two cdf, i.e. <span class="math-container">$P(x)$</span> is the probability that the outcome is smaller than or equal to <span class="math-container">$x$</span>. We have that <span class="math-container">$P$</span> FOSD <span class="math-container">$Q$</span> if <span class="math-container">$P(x) \le Q(x)$</span> for all <span class="math-container">$x$</span> and <span class="math-container">$P(x) &lt; Q(x)$</span> for at least one <span class="math-container">$x$</span>. Assume that <span class="math-container">$P, Q$</span> are <span class="math-container">$C^1$</span> (and hence also continuous).</p>
<p>This implies that <span class="math-container">$P$</span> and <span class="math-container">$Q$</span> have have densities, say <span class="math-container">$p$</span> and <span class="math-container">$q$</span> and that there is an open neighbourhood of <span class="math-container">$[0,1]$</span> such that <span class="math-container">$P(x) &lt; Q(x)$</span> for all <span class="math-container">$x$</span> in an open subset of <span class="math-container">$[0,1]$</span>.</p>
<p>Finally, we assume that <span class="math-container">$u: [0,1] \to \mathbb{R}$</span> is also <span class="math-container">$C^1$</span>. Monotonicity of <span class="math-container">$u$</span> then requires <span class="math-container">$u'(x) &gt; 0$</span>.</p>
<p>Note that:
<span class="math-container">$$
(u(x)P(x))' = u'(x) P(x) + u(x) p(x)
$$</span>
So, we have the integration by parts identity.
<span class="math-container">$$
\underbrace{u(1)P(1) - u(0)P(0)}_{u(1)} = \int_0^1 u'(x) P(x) dx + \int_0^1 u(x) p(x) dx.
$$</span></p>
<p>We want to show that <span class="math-container">$\int_0^1 p(x) u(x) dx \ge \int_0^1 q(x) u(x) dx$</span>.</p>
<p>Note:
<span class="math-container">$$
\begin{align*}
\int_0^1 p(x) u(x) dx &amp;= u(1) - \int_0^1 \underbrace{u'(x)}_{&gt; 0} \underbrace{P(x)}_{\le Q(x)} dx,\\
&amp;&gt; u(1) - \int_0^1 u'(x) Q(x) dx,\\
&amp;= \int_0^1 q(x) u(x) dx.
\end{align*}
$$</span>
The strict inequality follows from the fact that <span class="math-container">$P(x) \le Q(x)$</span> with a strict inequality for all <span class="math-container">$x$</span> in an open (non-empty) subset of <span class="math-container">$[0,1]$</span>.</p>

CC BY-SA 4.0
2024-03-19T09:26:00.823Z
Very detailed answer but is this what the OP was asking?
Giskard
CC BY-SA 4.0
2024-03-19T09:02:29.867Z
2024-03-19T09:02:29.867Z
tdm
<p>I think this result is very simple and very useful. However, I find no paper cover it.</p>
<p>Is it on some textbook like MWG?</p>
<p>Strict monotonicity means: <span class="math-container">$f(s)\succ g(s)$</span> for all <span class="math-container">$s$</span> implies <span class="math-container">$f\succ g$</span> and <span class="math-container">$f(s)\succsim g(s)$</span> for all <span class="math-container">$s$</span> implies <span class="math-container">$f\succsim g$</span>.</p>

CC BY-SA 4.0
2024-03-18T22:56:09.297Z
First-order stochastic dominance is about lotteries over numbers ("money"). You seem to discuss acts in an Anscombe-Aumann-type framework. The monotonicity assumption you mention is always satisfied under (subjective) expected utility.
Michael Greinecker
<microeconomics><decision-theory>
Strict monotonicity implies FOSD under EU?
dodo
# Strict monotonicity implies FOSD under EU?

(Asked by: username_0 on 2024-03-18)

I think this result is very simple and very useful. However, I find no paper cover it.

Is it on some textbook like MWG?

Strict monotonicity means: $f(s)\succ g(s)$ for all $s$ implies $f\succ g$ and $f(s)\succsim g(s)$ for all $s$ implies $f\succsim g$.

**Tags:** microeconomics, decision-theory
> **Comments:**
> **username_1**: First-order stochastic dominance is about lotteries over numbers ("money"). You seem to discuss acts in an Anscombe-Aumann-type framework. The monotonicity assumption you mention is always satisfied under (subjective) expected utility.  

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## 1 answers

### Answer by username_2 on 2024-03-19. Score: 1

Let me show that for the EU model, FOSD and monotonicity of the utility function implies preference dominance.

The reasoning is easier if we have a finite number of outcomes, say:
$$
x\_1 > x\_2 \ldots > x\_n,
$$
Consider associated probabilities $p\_1,\ldots, p\_n$ and $q\_1,\ldots, q\_n$ over these outcomes.

Let $P\_i = \sum\_{j \le i} p\_i$ be the probability that the outcome is larger or equal to $x\_i$ and similarly, define $Q\_i = \sum\_{j \le i} q\_i$.

We have that $p$ FOSD $q$ if $P\_i \ge Q\_i$ for all $i$ with at least one strict inequality. Note that $P\_n = Q\_n = 1$.

Let $u(x\_1) > u(x\_2) > \ldots > u(x\_n)$. We would like to show that $\sum\_{i = 1}^n p\_i u(x\_i) > \sum\_{i = 1}^n q\_i u(x\_i)$.

Indeed:
$$
\begin{align\*}
\sum\_{i = 1}^n (p\_i - q\_i) u(x\_i) &= \sum\_{i = 1}^n \left(\sum\_{j \le i} p\_j - \sum\_{j \le i} q\_j \right) u(x\_i) - \sum\_{i = 2}^n \left(\sum\_{j < i} p\_i - \sum\_{j < i} q\_i\right) u(x\_i),\\
&= \sum\_{i = 1}^n (P\_i - Q\_i) u(x\_i) - \sum\_{i = 2}^n (P\_{i-1} - Q\_{i-1}) u(x\_i),\\
&= \sum\_{i = 1}^n (P\_i - Q\_i) u(x\_i) - \sum\_{j = 1}^{n-1} (P\_j - Q\_j)u(x\_{j+1}),\\
&= \sum\_{i = 1}^{n-1}\underbrace{(P\_i - Q\_i)}\_{\ge 0} \underbrace{(u(x\_i) - u(x\_{j+1}))}\_{> 0} + \underbrace{(P\_n - Q\_n)}\_{ = 0} u(x\_n),\\
&> 0
\end{align\*}
$$

Consider now the case of a continuous outcome $x \in [0,1]$. Let $P$ and $Q$ be two cdf, i.e. $P(x)$ is the probability that the outcome is smaller than or equal to $x$. We have that $P$ FOSD $Q$ if $P(x) \le Q(x)$ for all $x$ and $P(x) < Q(x)$ for at least one $x$. Assume that $P, Q$ are $C^1$ (and hence also continuous).

This implies that $P$ and $Q$ have have densities, say $p$ and $q$ and that there is an open neighbourhood of $[0,1]$ such that $P(x) < Q(x)$ for all $x$ in an open subset of $[0,1]$.

Finally, we assume that $u: [0,1] \to \mathbb{R}$ is also $C^1$. Monotonicity of $u$ then requires $u'(x) > 0$.

Note that:
$$
(u(x)P(x))' = u'(x) P(x) + u(x) p(x)
$$
So, we have the integration by parts identity.
$$
\underbrace{u(1)P(1) - u(0)P(0)}\_{u(1)} = \int\_0^1 u'(x) P(x) dx + \int\_0^1 u(x) p(x) dx.
$$

We want to show that $\int\_0^1 p(x) u(x) dx \ge \int\_0^1 q(x) u(x) dx$.

Note:
$$
\begin{align\*}
\int\_0^1 p(x) u(x) dx &= u(1) - \int\_0^1 \underbrace{u'(x)}\_{> 0} \underbrace{P(x)}\_{\le Q(x)} dx,\\
&> u(1) - \int\_0^1 u'(x) Q(x) dx,\\
&= \int\_0^1 q(x) u(x) dx.
\end{align\*}
$$
The strict inequality follows from the fact that $P(x) \le Q(x)$ with a strict inequality for all $x$ in an open (non-empty) subset of $[0,1]$.

> **Comments:**
> **username_3**: Very detailed answer but is this what the OP was asking?  

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