<p>Let <span class="math-container">$A = \mathbb{R}$</span>. For <span class="math-container">$p,q\in \mathcal{L}(A)$</span>, <span class="math-container">$p$</span> first-order stochastically dominates (FOSD) <span class="math-container">$q$</span> if <span class="math-container">$F_p(a)-F_q(a) \leq 0, \forall a\in A$</span>.</p>
<p>Show that <span class="math-container">$p$</span> first-order stochastically dominates <span class="math-container">$q$</span> if and only if for every non-decreasing function <span class="math-container">$v: A\to \mathbb{R}$</span>,</p>
<p><span class="math-container">$$
\sum_{a \in A} p(a) v(a) \geq \sum_{a \in A} q(a) v(a).
$$</span></p>
<p>Using Riemann sums instead of integrals made me life quite difficult for this proof. I am aware that <span class="math-container">$F_p(a)=\sum_{x\leq a}p(x)$</span>. Maybe I should use expected values somewhere but I am really confused with this. I will appreciate any help at the moment.</p>
<p>Best</p>

CC BY-SA 4.0
2023-06-10T10:38:23.320Z
Consider also Mathematics Stack Exchange. Perhaps you will find a similar question already answered there.
Richard Hardy
CC BY-SA 4.0
2023-06-10T11:28:05.573Z
The problem, it seems to me, is the meaning of a sum on a uncountanble set as in $\sum_{a \in A} q(a) v(a)$, if $A$ is $\mathbb{R}$ or an uncountable subset of $\mathbb{R}$. Have you a definition of this kind of sum?
BakerStreet
CC BY-SA 4.0
2023-06-10T19:23:24.417Z
By the way, why you write $A=\mathbb{R}$, instead of writing just $\mathbb{R}$ in the formulas? May be you meant $A \subseteq \mathbb{R}$?
BakerStreet
CC BY-SA 4.0
2023-06-11T13:16:12.550Z
@BakerStreet As far as I know, such a sum is usually defined for positive summands as the supremum of the sum over all finite subsets. For general summands, you can then split it into positive and negative parts.  Defined as such, it is equivalent to an integral with respect to the counting measure, see https://math.stackexchange.com/a/4664145/697506.

Not sure this helps, though.
Wittgenstein's Poker
CC BY-SA 4.0
2023-06-11T13:22:56.250Z
@BakerStreet Also, to my knowledge it is not called a "Riemann sum", so maybe OP is referring to something else.  A clarification would be useful.
Wittgenstein's Poker
CC BY-SA 4.0
2023-06-11T13:26:19.943Z
@  Wittgenstein's Poker Thank you for the answer. Yes, I knew this definition, but I'm not sure that the sum  can be finite for an uncountable set. I read about it in other posts on MathSE, where they  said  that a necessary condition for the sum to be finite is that the set is countable. But I'm not an expert of this, I was just asking.
BakerStreet
CC BY-SA 4.0
2023-06-11T13:27:59.297Z
@ Wittgenstein's PokerYes, I too was doubtful about the name 'Riemann sums', that for me is a different thing. I think that it could be clarified by the book the OP is referring to, there should be there these definitions.
BakerStreet
<utility><decision-theory>
Proof of First-Order Stochastic Dominance with Riemann Sums
homo-economitux
# Proof of First-Order Stochastic Dominance with Riemann Sums

(Asked by: username_0 on 2023-06-09)

Let $A = \mathbb{R}$. For $p,q\in \mathcal{L}(A)$, $p$ first-order stochastically dominates (FOSD) $q$ if $F\_p(a)-F\_q(a) \leq 0, \forall a\in A$.

Show that $p$ first-order stochastically dominates $q$ if and only if for every non-decreasing function $v: A\to \mathbb{R}$,

$$
\sum\_{a \in A} p(a) v(a) \geq \sum\_{a \in A} q(a) v(a).
$$

Using Riemann sums instead of integrals made me life quite difficult for this proof. I am aware that $F\_p(a)=\sum\_{x\leq a}p(x)$. Maybe I should use expected values somewhere but I am really confused with this. I will appreciate any help at the moment.

Best

**Tags:** utility, decision-theory

> **Comments:**
> **username_1**: Consider also Mathematics Stack Exchange. Perhaps you will find a similar question already answered there.  
> **username_2**: The problem, it seems to me, is the meaning of a sum on a uncountanble set as in $\sum_{a \in A} q(a) v(a)$, if $A$ is $\mathbb{R}$ or an uncountable subset of $\mathbb{R}$. Have you a definition of this kind of sum?  
> **username_2**: By the way, why you write $A=\mathbb{R}$, instead of writing just $\mathbb{R}$ in the formulas? May be you meant $A \subseteq \mathbb{R}$?  
> **username_3**: @username_2 As far as I know, such a sum is usually defined for positive summands as the supremum of the sum over all finite subsets. For general summands, you can then split it into positive and negative parts.  Defined as such, it is equivalent to an integral with respect to the counting measure, see https://math.stackexchange.com/a/4664145/697506.  Not sure this helps, though.  
> **username_3**: @username_2 Also, to my knowledge it is not called a "Riemann sum", so maybe OP is referring to something else.  A clarification would be useful.  
> **username_2**: @  username_3 username_3 Thank you for the answer. Yes, I knew this definition, but I'm not sure that the sum  can be finite for an uncountable set. I read about it in other posts on MathSE, where they  said  that a necessary condition for the sum to be finite is that the set is countable. But I'm not an expert of this, I was just asking.  
> **username_2**: @ username_3 PokerYes, I too was doubtful about the name 'Riemann sums', that for me is a different thing. I think that it could be clarified by the book the OP is referring to, there should be there these definitions.  

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## 0 answers

