Perception, utility, and evolution

Abstract

This paper presents a model of the evolution of the hedonic utility function in which perception is imperfect. Netzer (Am Econ Rev 99(3):937–955, 2009) considers a model with perfect perception and finds that the optimal utility function allocates marginal utility where decisions are made frequently. This paper shows that it is also beneficial to allocate marginal utility away from more perceptible events. The introduction of perceptual errors can lead to qualitatively different utility functions, such as discontinuous functions with flat regions rather than continuous and strictly increasing functions.

Appendices

Lemma 2

Lemma 2

For any $U\in \mathcal {U}$, exists a monotone, non-decreasing function $U^*\in \mathcal {U}$, such that $V[U]\le V[U^*]$.

Proof

This proof is not as trivial as it seems, since changing U at any point will change the whole integral. The strategy will be to consider simple functions first, and then apply the dominated convergence theorem to prove the general case.

The objective function can be written as:

$$\begin{aligned} V[U]= K + 2\int _0^1\int _y^1 (x-y)f(x)f(y)G\big (U(x) - U(y)\big ) {\text {d}}x{\text {d}}y, \end{aligned}$$

where $K=\mathbb {E}[X]+\int _0^1\int _y^1(y-x)f(x)f(y){\text {d}}x{\text {d}}y$.

Suppose that $U_n$ is a simple function that is constant on the intervals $\left( \frac{s-1}{n},\frac{s}{n}\right]$, $s = 1,\dots ,n.$ Each simple function of this form can be written as the step function $U_n(x)= \sum _{s=1}^n u_s \mathbb {1}\left\{ \left( \frac{s-1}{n},\frac{s}{n}\right] \right\} (x).$ Now, we can define the n-vector u as the vector with values $u_s$, which is the value that $U_n(x)$ takes on the sth interval. From this definition, we see that each n vector corresponds to a step function.

We will now define the sorting operator S acting on the vector u as follows. Let l be an integer in $1,\dots ,n$, such that $u_l>u_m$ for some $l <m$. If l exists, set S(u) to be the n-vector with $u_m$ on the lth position and $u_l$ on the mth position, and all other elements equal to the corresponding elements of u. If such an l does not exist, set $S(u) = u.$ Then, S sorts two elements of u in increasing order.¹⁴

Similarly, let $x_n$ and $f_n$ be two simple functions; the idea is that $x_n\rightarrow x$ and $f_n\rightarrow f$. With a slight abuse of notation, also let $x_n$ and $f_n$ stand for the n vectors with $x_s$ and $f_s$ in the sth position. Therefore, we can approximate $V[U_n]$ with

$$\begin{aligned} V_n[U_n]&\equiv K + 2\int _0^1\int _y^1 (x_n(x)-x_n(y))f_n(x)f_n(y)G\big (U_n(x) - U_n(y)\big ) {\text {d}}x{\text {d}}y\\&= K + \frac{2}{n^2}\sum _{j=1}^n \sum _{i=j}^n (x_i-x_j)f_i f_j G(u_i-u_j). \end{aligned}$$

In Appendix B, I prove the following lemma.

Lemma 3

$V_n[U_n]\le V_n[S(U_n)]$.

Therefore, if we apply the sorting operator S sufficiently many times to $U_n$, to a maximum of n times, we find the rearranged vector $U^*_n$: a vector completely sorted in ascending order. Every application of the sorting operator S improves $V_n$, and since we apply it a finite number of times, $V_n[U^*_n]\equiv V_n[S \circ S \circ \dots \circ S U_n ] \ge V_n[U_n].$

Now that this holds for simple functions, we go for the general case. Since f, U, and the identity are measurable functions, there are sequences of bounded simple functions $\{f_n\}$, $\{U_n\}$, and $\{x_n\}$ converging almost everywhere to f, U, and x. Define the increasing rearrangement of U as

$$\begin{aligned} U^*(x)\equiv \inf \left\{ q\in \mathbb {R} \ : \ x\le \int _0^1 \mathbb {1}\{U(z)\le q\}dz \right\} . \end{aligned}$$

This rearrangement always exists; it is like sorting the values of U in increasing order. This is a quantile function.

From this definition, $U_n\rightarrow U$ a.e. implies that the rearrangements also converge a.e. to the rearrangement: $U_n^*\rightarrow U^*$ a.e. (see Chernozhukov et al. (2009) and the references therein). Moreover, $V_n[U_n]\rightarrow V[U]$ and, since the inequality works for every n, by Lebesgue’s dominated convergence theorem, $V[U]\le V[U^*]$. $\square$

Proofs

Proof of Lemma 3

Consider the following expression:

$$\begin{aligned} \frac{V_n[U_n]- V_n[S(U_n)]}{2}&= \frac{1}{n^2}\sum _{j=1}^n \sum _{i=j}^n (x_i-x_j)f_i f_j\left[ G(u_i-u_j)-G(Su_i-Su_j)\right] . \end{aligned}$$

First, observe that $G(u_i-u_j)-G(Su_i-Su_j)=0$ for all $i,j\ne l,m.$ Next, for $i=l$, there are $j=1,\dots ,l$ summands and for $i=m$, there are $j=1,\dots ,m$ summands. Similarly, for $j=l$, there are $i=l,\dots ,n$ summands and for $j=m$, there are $i=m,\dots ,n$ summands. However, we are double-counting three summands, but two are zero: The non-zero summand corresponds to $j=l$ and $i=m$. Finally, note that $G(u_m-u_j)-G(Su_m-Su_j)=-[G(u_j-u_m)-G(Su_j-Su_m)]$. Thus,

$$\begin{aligned} \phantom {=} n^2\frac{V_n[U_n]- V_n[S(U_n)]}{2}&= \sum _{j=1}^n \sum _{i=j}^n (x_i-x_j)f_i f_j\left[ G(u_i-u_j)-G(Su_i-Su_j)\right] \\&= \sum _{i=l}^n(x_i-x_l)f_jf_l \left[ G(u_i-u_l)-G(Su_i-Su_l)\right] \\&\quad + \sum _{i=m}^n(x_i-x_m)f_jf_m \left[ G(u_i-u_m)-G(Su_i-Su_m)\right] \\&\quad + \sum _{j=1}^l(x_l-x_j)f_lf_j \left[ G(u_l-u_j)-G(Su_l-Su_j)\right] \\&\quad + \sum _{j=1}^m(x_m-x_j)f_mf_j \left[ G(u_m-u_j)-G(Su_m-Su_j)\right] \\&= \sum _{i=1}^n (x_i-x_l)f_i f_l\left[ G(u_i-u_l)-G(Su_i-Su_l)\right] \\&\quad + \sum _{i=1}^n (x_i-x_m)f_i f_m\left[ G(u_i-u_m)-G(Su_i-Su_m)\right] \\&\quad - (x_m-x_l)f_mf_l \left[ G(u_m-u_l)-G(u_l-u_m)\right] . \end{aligned}$$

At this point, we use the definition of S and conclude that the preceding sum equals:

$$\begin{aligned}&= \sum _{i=1,i\ne l,m}^n (x_i-x_l)f_i f_l\left[ G(u_i-u_l)-G(u_i-u_m)\right] \\&\quad + \sum _{i=1,i\ne l,m}^n (x_i-x_m)f_i f_m\left[ G(u_i-u_m)-G(u_i-u_l)\right] \\&\quad + (x_m-x_l)f_mf_l \left[ G(u_m-u_l)-G(u_l-u_m)\right] . \\&=\sum _{i=1}^n \left[ (x_i-x_l)f_l-(x_i-x_m)f_m\right] f_i \left[ G(u_i-u_l)-G(u_i-u_m)\right] . \end{aligned}$$

Now, we have to show that the preceding expression is non-positive.

By assumption $u_l> u_m$. Moreover, G is a non-decreasing cdf and $x_l<x_m$. Therefore, $G(u_i-u_l)-G(u_i-u_m)\le 0$.

On the other hand, $(x_i-x_l)f_l-(x_i-x_m)f_m\ge 0$: Suppose that $f_l>f_m$, then $(x_i-x_l)f_l>(x_i-x_m)f_l>(x_i-x_m)f_m$. Now suppose that $f_l\le f_m$, then $f_m(x_m-x_i)>f_m(x_l-x_i)\ge f_l(x_l-x_i)$, which implies that $(x_i-x_l)f_l-(x_i-x_m)f_m\ge 0$.

Therefore, we conclude that $V_n[U_n]- V_n[S(U_n)]\le 0$. $\square$

Proof of Lemma 1

Define $h_y(x)\equiv 2(y-x)f(x)f(y)g\left( U(y)-U(x)\right)$, so that $I(y)=\int _0^1 h_y(x){\text {d}}x$. Since $U\in \mathcal {M}$, $h_y$ is (Lebesgue) integrable $\forall y\in [0,1]$. Since f and g are bounded, there exists a function B, such that $|h_y(x)|\le B$ almost everywhere. Now, fix y and consider a sequence $\{y_n\}_{n=1}^\infty \in \mathbb {R}$ converging to y. Then $h_{y_n}\rightarrow h_y$ a.e. and $|h_{y_n}(x)|\le B$ a.e. Therefore

$$\begin{aligned} \lim _{y_n\rightarrow y} I(y_n)= \lim _{y_n\rightarrow y} \int _0^1 h_{y_n}(x){\text {d}}x = \int _0^1 \lim _{y_n\rightarrow y} h_{y_n}(x){\text {d}}x =\int _0^1 h_y(x){\text {d}}x= I(y). \end{aligned}$$

$\square$

Second-order conditions

For the sake of completeness, note that $\left. \dfrac{\partial ^2}{\partial ^2 \alpha }V[U+\alpha \delta U]\right| _{\alpha =0}$ is equal to

$$\begin{aligned} \int _0^1\int _0^1 (x-y)f(x)f(y)g'\big (U(x)-U(y)\big )[\delta U(x)-\delta U(y)]^2dx {\text {d}}y \le 0, \end{aligned}$$

since $U\in \mathcal {M}$.

Corollaries

Proof of Corollary 4

Consider $x\in [0,a]$ and recall that $I(x)\le 0$ and $U(x)=0$ for such x. In particular $I(a)=0$ and $U(a)=0$. In other words,

$$\begin{aligned} I(a)=\int _0^1 (a-y) f(y) e^{-\frac{U(y)^2}{2\sigma ^2}} {\text {d}}y=0. \end{aligned}$$

The function $w(y,\sigma )\equiv e^{-\frac{U(y)^2}{2\sigma ^2}}$ is a positive weight. Since U(y) is monotone, w is increasing in $\sigma$ and decreasing in y. In particular, $w(y,\sigma )=1$ for $y\in [0,a)$ and for any $\sigma$, and $w(y,\sigma )<1$ for $y\in [a,1]$. Moreover, $(a-y)>0$ for $y\in [0,a)$ and $(a-y)\le 0$ for $y\in [a,1]$. Since f(y) is fixed, it follows that increasing $\sigma$ increases the weight of the negative part of the integral, relative to the positive part. To maintain $I(a)=0$, the positive part must increase in size, and thus, the interval [0, a) must expand. Then, a must increase.

The case of b is analogous. $\square$

Proof of Corollary 5

As in Corollary 4, consider

$$\begin{aligned} I(a)=\int _0^1 (a-y) \frac{y^{\alpha -1}(1-y)^{\alpha -1}}{\text {B}(\alpha ,\alpha )} g\left( U(y)\right) {\text {d}}y=0, \end{aligned}$$

where $\text {B}(\alpha ,\alpha )$ is the beta function, which becomes irrelevant. As $\alpha$ increases, the weight shifts towards 1/2 and away from the interval [0, a), which is the positive part of the integral. Since the equality $I(a)=0$ must be maintained, the interval must expand. Then, a must increase.

The case of b is analogous. $\square$

Proof of Corollary 6

As in Corollary 5, consider

$$\begin{aligned} I(a)=\int _0^1 (a-y) \frac{y^{\alpha -1}(1-y)^{\beta -1}}{\text {B}(\alpha ,\beta )} g\left( U(y)\right) {\text {d}}y=0, \end{aligned}$$

where $\text {B}(\alpha ,\beta )$ is the beta function, which becomes irrelevant. As $\alpha$ increases, the weight shifts away from the interval [0, a), which is the positive part of the integral. Since the equality $I(a)=0$ must be maintained, the interval must expand. Then, a must increase.

For b, consider

$$\begin{aligned} I(b)=\int _0^1 (b-y) \frac{y^{\alpha -1}(1-y)^{\beta -1}}{\text {B}(\alpha ,\beta )} g\left( U(y)-1\right) {\text {d}}y=0. \end{aligned}$$

As $\alpha$ increases, the weight shifts towards the interval (b, 1], which is the negative part of the integral. Since $I(b)=0$ must hold, the interval (b, 1] must shrink. Then, b must increase. $\square$