<p>Notice, $z$ is in $Y$, and so it is a lottery. Lemma 3.1 uses the standard vNM techniques to identify $u: Y \to \mathbb{R}$. Now, $J$ is defined as a function from $L_0$ to $\mathbb{R}$, so the statement $J(z) = 0$ is via the natural identification between $Y$ and constant acts, $L_c$.</p>

<p>Condition (ii) of Lemma 3.2 says that $J(y) = u(y)$ for all $y \in L_c \cong Y$. So, $u(z) = J(z) = 0$, or, in the other notation, $u \circ z = 0$.</p>

CC BY-SA 3.0
2016-03-14T19:49:36.177Z
2016-03-14T19:49:36.177Z
201p
<p>I have a question on  the proof of <em>lemma 3.3</em>: Let $g\in L_0$ satisfy $u \circ g = b$, $z\in Y$ satisfy $J(z)=0$ where $J$ satisfy <em>lemma 3.2</em>, and let $f=\alpha g + (1-\alpha)z$. Then $u\circ f = \alpha u\circ g + (1-\alpha) u \circ z$. It is then stated that $\alpha u\circ g + (1-\alpha) u \circ z = \alpha b$. This implies that $u \circ z = 0$. Why is this the case?</p>

CC BY-SA 3.0
2016-02-09T16:06:10.323Z
This question could be improved to be easier to answer and be more of a reference to others by linking to the underlying paper and explaining the notation, particularly what that circular operator does.
BKay
CC BY-SA 3.0
2016-03-15T17:36:32.450Z
It would be helpful as a reference for this question (and in understanding the below answer) if you stated what Lemma 3.2 is.
Kitsune Cavalry
<mathematical-economics><decision-theory>
Question about Gilboa and Schmeidler "A Maxmin Expected Utility With Non-unique Prior" 1989
Jacob Morsbøl
# Question about Gilboa and Schmeidler "A Maxmin Expected Utility With Non-unique Prior" 1989

(Asked by: username_0 on 2016-02-09)

I have a question on the proof of *lemma 3.3*: Let $g\in L\_0$ satisfy $u \circ g = b$, $z\in Y$ satisfy $J(z)=0$ where $J$ satisfy *lemma 3.2*, and let $f=\alpha g + (1-\alpha)z$. Then $u\circ f = \alpha u\circ g + (1-\alpha) u \circ z$. It is then stated that $\alpha u\circ g + (1-\alpha) u \circ z = \alpha b$. This implies that $u \circ z = 0$. Why is this the case?

**Tags:** mathematical-economics, decision-theory

> **Comments:**
> **username_1**: This question could be improved to be easier to answer and be more of a reference to others by linking to the underlying paper and explaining the notation, particularly what that circular operator does.  
> **username_2**: It would be helpful as a reference for this question (and in understanding the below answer) if you stated what Lemma 3.2 is.  

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## 1 answers

### Answer by username_3 on 2016-03-14. Score: 1

Notice, $z$ is in $Y$, and so it is a lottery. Lemma 3.1 uses the standard vNM techniques to identify $u: Y \to \mathbb{R}$. Now, $J$ is defined as a function from $L\_0$ to $\mathbb{R}$, so the statement $J(z) = 0$ is via the natural identification between $Y$ and constant acts, $L\_c$.

Condition (ii) of Lemma 3.2 says that $J(y) = u(y)$ for all $y \in L\_c \cong Y$. So, $u(z) = J(z) = 0$, or, in the other notation, $u \circ z = 0$.


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