Preferences over rich sets of random variables: on the incompatibility of convexity and semicontinuity in measure

Abstract

This paper considers a decision maker whose preferences are locally upper- or/and lower-semicontinuous in measure. We introduce the notion of a rich set which encompasses any standard vector space of random variables but also much smaller sets containing only random variables with at most two different outcomes in their support. Whenever preferences are complete on a rich set of random variables, lower- (resp. upper-) semicontinuity in measure becomes incompatible with convexity of strictly better (resp. worse) sets. We discuss implications for utility representations and risk-measures. In particular, we show that the value-at-risk criterion violates convexity exactly because it is lower-semicontinuous in measure.

Appendix: Formal proofs

Proof of Proposition 1

The if-part We show that $\lim _{n\rightarrow \infty }d_{k}\left( X,X_{n}\right) =0$ implies $X_{n}\rightarrow _{\mu }X$. For an arbitrary $\epsilon$ with $0<\epsilon <k$ define the event

$$\begin{aligned} A_{n}=\left\{ \omega \in \Omega \mid \mu \left( \left| X_{n}\left( \omega \right) -X\left( \omega \right) \right| >\epsilon \right) \right\} \text {.} \end{aligned}$$

Because of

$$\begin{aligned} d_{k}\left( X,X_{n}\right)= & {} \int _{\Omega }\left( \left| X-Y\right| \wedge k\right) 1_{A_{n}}+\left| X-X_{n}\right| 1_{A_{n}^{c}}d\mu \\\ge & {} \int _{\Omega }\epsilon \cdot 1_{A_{n}}+0\cdot 1_{A_{n}^{c}}d\mu \\= & {} \epsilon \mu \left( A_{n}\right) \text {,} \end{aligned}$$

$\lim _{n\rightarrow \infty }d_{k}\left( X,X_{n}\right) =0$ implies $\lim _{n\rightarrow \infty }\mu \left( A_{n}\right) =0$.

The only-if part We show that $X_{n}\rightarrow _{\mu }X$ implies $\lim _{n\rightarrow \infty }d_{k}\left( X,X_{n}\right) =0$. For an arbitrary $\epsilon$ with $0<\epsilon <k$ pick some sufficiently fine partition such that $\mu \left( A_{n}\right) <\epsilon$. Note that

$$\begin{aligned} d_{k}\left( X,X_{n}\right)= & {} \int _{\Omega }\left( \left( \left| X-Y\right| \wedge k\right) 1_{A_{n}}+\left( \left| X-Y\right| \right) 1_{A_{n}^{c}}\right) d\mu \\\le & {} \int _{\Omega }\left( k\cdot 1_{A_{n}}+\epsilon \cdot 1_{A_{n}^{c}}\right) d\mu \\= & {} k\mu \left( A_{n}\right) +\epsilon \left[ 1-\mu \left( A_{n}\right) \right] \text {.} \end{aligned}$$

For every $\epsilon >0$ we have that $X_{n}\rightarrow _{\mu }X$ implies $\lim _{n\rightarrow \infty }\mu \left( A_{n}\right) =0$. Consequently, $\lim _{n\rightarrow \infty }d_{k}\left( X,X_{n}\right) \le \epsilon$ for all $\epsilon >0$ which proves the only-if part. $\square$

Proof of Lemma 1

Consider a rich set $R\left( \mathcal {F}\right) \subseteq L$ such that $Y\in \mathcal {F}$ as well as $Y\in A$ for some open and convex $A\subseteq \left( L,d_{k}\right)$. To prove the Lemma, we have to show that $\mathcal {F}\subseteq A$.

Step 1 Fix some $k>0$. Next, fix $Y\in \mathcal {F}$ and consider an arbitrary $X\in \mathcal {F}$. Since $R\left( \mathcal {F}\right)$ is rich, we have

$$\begin{aligned} Y_{i_{n}}=Y+n\left( X-Y\right) 1_{\Omega _{i_{n}}}\in R\left( \mathcal {F} \right) \end{aligned}$$

for $n\ge 1$. Next note that

$$\begin{aligned} d_{k}\left( Y,Y_{i_{n}}\right)= & {} \int _{\Omega }\left( \left| Y-Y_{i_{n}}\right| \wedge k\right) d\mu \\= & {} \int _{\Omega }\left( \left| n(X-Y)\right| \wedge k\right) 1_{\Omega _{i_{n}}}d\mu \\< & {} \int _{\Omega }k1_{\Omega _{i_{n}}}d\mu =\frac{k}{n}\text {.} \end{aligned}$$

Consequently, we have for all $Y_{i_{n}}$, $i_{n}\in \left\{ 1,\ldots ,n\right\}$, that

$$\begin{aligned} Y_{i_{n}}\in B_{\varepsilon }\left( Y;d_{k}\right) \text { for }\varepsilon \ge \frac{k}{n} \end{aligned}$$

where $B_{\varepsilon }\left( Y;d_{k}\right)$ denotes an open ball around Y in $\left( L,d_{k}\right)$ with radius $\varepsilon .$

Step 2 Fix any open set $A\subseteq \left( L,d_{k}\right)$ such that $Y\in A$. By definition, there must exist some sufficiently small $\varepsilon >0$ such that

$$\begin{aligned} B_{\varepsilon }\left( Y;d_{k}\right) \subseteq A \end{aligned}$$

so that, by Step 1, $Y_{i_{n}}\in A$ for all $i_{n}\in \left\{ 1,\ldots ,n\right\}$ with $n\ge \frac{k}{\varepsilon }$.

Step 3 Finally, note that convexity of A implies

$$\begin{aligned} \frac{1}{n}\sum _{i_{n}=1}^{n}Y_{i_{n}}= & {} \frac{1}{n}\sum _{i_{n}=1}^{n}Y+n \left( X-Y\right) 1_{\Omega _{i_{n}}} \\= & {} Y+\sum _{i_{n}=1}^{n}(X-Y)1_{\Omega _{i_{n}}} \\= & {} Y+X-Y=X\text {,} \end{aligned}$$

which gives the desired result $X\in A$ for any $X\in \mathcal {F}$. $\square$

Proof of Theorem 1

Ad (i) Assume that $S^{*}\left( X\right)$ is convex. Because of $X\prec Y$, we have $Y\in S^{*}\left( X\right)$. If $S^{*}\left( X\right)$ was open, Lemma 1 implies that $X\in S^{*}\left( X\right)$, a contradiction. Consequently, $\preceq$ cannot be lower-semicontinuous in measure above X. $\square$

Ad (ii) Assume now that $s^{*}\left( Y\right)$ is convex. Because of $X\prec Y$, we have $X\in s^{*}\left( Y\right)$. An open $s^{*}\left( Y\right)$ would, by Lemma 1, imply the contradiction $Y\in s^{*}\left( Y\right)$. Consequently, $\preceq$ cannot be upper-semicontinuous in measure below Y. $\square$

Proof of Observation 2

If $\preceq$ is not lower-semicontinuous above X, the strictly better set $S^{*}\left( X\right)$ is not open. That is, there must exist some $Y\in S^{*}\left( X\right)$ which is not an interior point of $S^{*}\left( X\right)$, i.e., for all $\delta >0$, there are Z such that

$$\begin{aligned} Z\in B_{\delta }\left( Y;d_{k}\right) \text { but }Z\notin S^{*}\left( X\right) \text {.} \end{aligned}$$

(14)

Let $\delta _{n}=\frac{1}{n}$ and pick $Y_{n}\in B_{\delta _{n}}\left( Y;d_{k}\right)$ such that $Y_{n}\notin S^{*}\left( X\right)$. By (14), such $Y_{n}$ exist for all $n\ge 1$. This constructs a converging sequence $\left\{ Y_{n}\right\}$, $d_{k}\left( Y_{n},Y\right) \rightarrow 0$, such that $Y_{n}\notin S^{*}\left( X\right)$ for all n, implying

$$\begin{aligned} U\left( Y_{n}\right) \le U\left( X\right) \text { for all }n\text { whereas } U\left( X\right) <U\left( Y\right) \text {.} \end{aligned}$$

Let

$$\begin{aligned} \varepsilon =U\left( Y\right) -U\left( X\right) >0 \end{aligned}$$

to see that, for all n,

$$\begin{aligned} U\left( Y_{n}\right) \le U\left( Y\right) -\varepsilon \text {.} \end{aligned}$$

But this violates (3) so that U is not lower-semicontinuous at Y. The argument for upper-semicontinuity proceeds analogously. $\square$

Proof of Observation 3(i)

If $S^{*}\left( X\right)$ is empty, the result obtains trivially. Suppose therefore that $Y,Z\in S^{*}\left( X\right)$ so that $U\left( X\right) <U\left( Y\right)$ as well as $U\left( X\right) <U\left( Z\right)$. If U is concave on a convex L, we have for any $\lambda \in \left( 0,1\right)$

$$\begin{aligned} U\left( \lambda Y+\left( 1-\lambda \right) Z\right)\ge & {} \lambda U\left( Y\right) +\left( 1-\lambda \right) U\left( Z\right) \\> & {} U\left( X\right) \text {,} \end{aligned}$$

implying

$$\begin{aligned} \lambda Y+\left( 1-\lambda \right) Z\in S^{*}\left( X\right) \text {.} \end{aligned}$$

$\square$

Proof of Observation 4

If u is concave we have, for all $\omega$ ,

$$\begin{aligned} u\left( \lambda X\left( \omega \right) +\left( 1-\lambda \right) Y\left( \omega \right) \right) \ge \lambda u\left( X\left( \omega \right) \right) +\left( 1-\lambda \right) u\left( Y\left( \omega \right) \right) \text {. } \end{aligned}$$

Next observe that

$$\begin{aligned} U\left( \lambda X+\left( 1-\lambda \right) Y\right)= & {} I\left( u\left( \lambda X+\left( 1-\lambda \right) Y\right) \right) \\\ge & {} I\left( \lambda u\left( X\right) +\left( 1-\lambda \right) u\left( Y\right) \right) \text {, by monotonicity} \\\ge & {} \lambda I\left( u\left( X\right) \right) +\left( 1-\lambda \right) I\left( u\left( Y\right) \right) \text {, by concavity} \\= & {} \lambda U\left( X\right) +\left( 1-\lambda \right) U\left( Y\right) \text { ,} \end{aligned}$$

which proves part (i). Part (ii) is proved analogously. $\square$