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?

  • 4
    $\begingroup$ 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. $\endgroup$ – BKay Feb 9 '16 at 16:06
  • $\begingroup$ It would be helpful as a reference for this question (and in understanding the below answer) if you stated what Lemma 3.2 is. $\endgroup$ – Kitsune Cavalry Mar 15 '16 at 17:36

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, $u \circ z = 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.