Live fast, die young

“Passionate lived, reasonable lasted”, Chamfort.

Abstract

Irrational agents are driven out of the market. This should favor learning: Irrational agents observing that rational agents are being more successful should adopt rational beliefs. We show that the threat of elimination is not sufficient to push agents toward rationality: A shorter “life” might be more rewarding than a longer one. Even if they are eliminated in the long run, irrational agents might rationally stay irrational in the sense that their ex-ante and ex-post welfare levels over their whole life are higher than (1) the welfare level that they would reach if they adopted rational expectations, (2) the welfare level reached by the otherwise identical (same initial wealth and same risk aversion) rational agents, (3) the welfare level that they would have if they were given the optimal allocation of the rational agent. Threat of elimination is not sufficient to push irrational agents toward rationality, and rational and surviving agents’ performances are not sufficiently high to generate learning through an adaptive process based on imitation of successful behaviors. A numerical illustration is provided.

Appendix

Proof of Proposition 1

The first-order conditions for Agent 1 and Agent 2 are given by

$$\begin{aligned} M_{i}\exp \left( -\frac{c_{i}^{*}}{\theta _{i}}\right) =\lambda _{i}q^{*},\quad i=1,2, \end{aligned}$$

where $M_{2}=\frac{\hbox {d}Q}{\hbox {d}P}$ and $M_{1}=1$ is introduced for the symmetry in the formulas and where $\lambda _{1}$ and $\lambda _{2}$ are Lagrange multipliers and are such that the budget constraint is saturated for each agent, i.e., $E\left[ \int _{0}^{T}q_{t}^{*}c_{i,t}^{*}\hbox {d}t\right] =\frac{1}{2}E\left[ \int _{0}^{T}q_{t}^{*}e_{t}^{*}\hbox {d}t\right] ,\quad i=1,2$.

This gives

$$\begin{aligned} c_{i}^{*}=\theta _{i}\ln \frac{M_{i}}{\lambda _{i}q^{*}},\quad i=1,2, \end{aligned}$$

and since we have $c_{1}+c_{2}=e^{*}$ we obtain

$$\begin{aligned} q^{*}=\frac{1}{\lambda _{1}^{\frac{\theta _{1}}{\theta _{1}+\theta _{2}} }\lambda _{2}^{\frac{\theta _{2}}{\theta _{1}+\theta _{2}}}}M_{1}^{\frac{\theta _{1}}{\theta _{1}+\theta _{2}}}M_{2}^{\frac{\theta _{2}}{\theta _{1} +\theta _{2}}}\exp \left( -\frac{e^{*}}{\theta _{1}+\theta _{2}}\right) \end{aligned}$$

and from there

$$\begin{aligned} c_{i,t}^{*}&=\frac{\theta _{i}}{\theta _{1}+\theta _{2}}e^{*} +\frac{\theta _{1}\theta _{2}}{\theta _{1}+\theta _{2}}\ln \left( \frac{M_{i} }{M_{j}}\frac{\lambda _{j}}{\lambda _{i}}\right) ,\quad i=1,2, \,j\ne i,\\&=C_{i}t+D_{i}W_{t}+E_{i}\nonumber \end{aligned}$$

(1)

where $C_{i}$ and $D_{i}$ are as above and where $E_{i}=\frac{\theta _{i} }{\theta _{1}+\theta _{2}}e_{0}^{*}+\frac{\theta _{1}\theta _{2}}{\theta _{1}+\theta _{2}}\ln \frac{\lambda _{j}}{\lambda _{i}}.$

Since the agents have the same initial endowment, we have

$$\begin{aligned} E\left[ \int _{0}^{\infty }q_{t}^{*}c_{1,t}^{*}\hbox {d}t\right] =E\left[ \int _{0}^{\infty }q_{t}^{*}c_{2,t}^{*}\hbox {d}t\right] . \end{aligned}$$

(2)

Simple computations give

$$\begin{aligned}&\lambda _{1}^{\frac{\theta _{1}}{\theta _{1}+\theta _{2}}}\lambda _{2} ^{\frac{\theta _{2}}{\theta _{1}+\theta _{2}}}\exp \left( \frac{e_{0}^{*} }{\theta _{1}+\theta _{2}}\right) E\left[ \int _{0}^{\infty }q^{*} c_{i,t}^{*}\hbox {d}t\right] \\&=\int _{0}^{\infty }E\left[ M_{1}^{\frac{\theta _{1}}{\theta _{1}+\theta _{2}} }M_{2}^{\frac{\theta _{2}}{\theta _{1}+\theta _{2}}}\exp \left( -\frac{e_{t}^{*}-e_{0}^{*}}{\theta _{1}+\theta _{2}}\right) c_{i,t}^{*}\right] \hbox {d}t\\&=\int _{0}^{\infty }E\left[ \exp \left( -\frac{1}{2}v^{2}t+\delta W\right) \exp \left( -\frac{\mu t+\sigma W_{t}}{\theta _{1}+\theta _{2}}\right) \left( C_{i} t+D_{i}W_{t}+E_{i}\right) \right] \hbox {d}t\\&=\int _{0}^{\infty } \left( C_{i}t+D_{i}Bt+E_{i}\right) \exp \left( \left( -A+\frac{1}{2} B^{2}\right) t\right) \hbox {d}t \end{aligned}$$

with A and B as above. Condition (C) gives us $-A+\frac{1}{2}B^{2}<0$ and the integral above converges which gives

$$\begin{aligned} U_{\gamma ,t} \left( c_{i,\gamma }^{*}\right)= & {} U_{i,t} (C_{i}^{*}) \,\mathrm{exp}\, \left( -{\frac{q \cdot c_{i}^{*}}{\theta _i}}\right) \,\mathrm{exp}\, \left( {\frac{w_{i,\gamma }}{\theta _{i,\gamma }}}q\cdot c_{i}^{*} \right) \nonumber \\&\mathrm{and}\, \prod _{\gamma \in \Gamma _{{i}}}\, U_{\gamma ,t}\,\, \left( c_{i,\gamma }^{*}\right) ^{\frac{\theta _{i,\gamma }}{{\theta _i}}}\,\, = U_{i,t}\,\, \left( c_{i}^{*}\right) . \end{aligned}$$

Equation (2) gives then

$$\begin{aligned} E_{1}-E_{2}=\frac{F_{1}+G_{1}B}{\left( -A+\frac{1}{2}B^{2}\right) } \end{aligned}$$

with $F_{1}$ and $G_{1}$ as above. From the expressions of $E_{1}$ and $E_{2}$ we also have

$$\begin{aligned} E_{1}+E_{2}=e_{0}^{*} \end{aligned}$$

which leads to the desired formula for $E_{i}$.

Solving for $c_{1}^{*}=c_{2}^{*}$ leads to $\left( \delta _{2} ,\theta _{2}\right) =\left( \delta _{1},\theta _{1}\right)$ or to $\left( \delta _{2},\theta _{2}\right) =\left( -\frac{2\mu }{\sigma } ,\frac{\sigma ^{2}\theta _{1}}{\sigma ^{2}-4\mu \theta _{2}}\right) .$ The second solution is in contradiction with Condition (C). $\square$

Proof of Proposition 2

It is immediate that

$$\begin{aligned} \frac{E\left[ c_{2}\right] }{E\left[ c_{1}\right] }=\frac{C_{2}t+E_{2} }{C_{1}t+E_{1}}\rightarrow \frac{C_{2}}{C_{1}}=\frac{\theta _{2}}{\theta _{1} }\frac{\mu -\frac{1}{2}\theta _{1}\delta _{2}^{2}}{\mu +\frac{1}{2}\theta _{2}\delta _{2}^{2}} \end{aligned}$$

which is smaller than 1 for $\theta _{2}<\frac{\mu \theta _{1}}{\mu -\theta _{1}\delta _{2}^{2}}$.

Similarly, we have

$$\begin{aligned} \frac{E\left[ \exp (c_{2,t})\right] }{E\left[ \exp (c_{1,t})\right] }&=\frac{E\left[ \exp \left( C_{2}t+D_{2}W_{t}+E_{2}\right) \right] }{E\left[ \exp \left( C_{1}t+D_{1}W_{t}+E_{1}\right) \right] },\\&=\exp (E_{2}-E_{1})\frac{\exp \left( C_{2}+\frac{1}{2}D_{2}^{2}\right) t}{\exp \left( C_{1}+\frac{1}{2}D_{1}^{2}\right) t}. \end{aligned}$$

The asymptotic behavior is governed by the sign of $\left( C_{2}+\frac{1}{2}D_{2}^{2}\right) -\left( C_{1}+\frac{1}{2}D_{1}^{2}\right) .$ It is negative for $\theta _{2}<\frac{2\mu \theta _{1}+\sigma ^{2}\theta _{1}}{2\mu +\sigma ^{2}+2\delta _{2}\theta _{1}\left( \sigma -\delta _{2}\right) }.$

Similarly,

$$\begin{aligned} \frac{E\left[ u(c_{2,t})\right] }{E\left[ u(c_{1,t})\right] }&=\frac{E\left[ \exp \left( -\frac{c_{2,t}}{\theta _{2}}\right) \right] }{E\left[ \exp (-\frac{c_{1,t}}{\theta _{1}})\right] }=\frac{E\left[ \exp \left( C_{2}^{\prime }t+D_{2}^{\prime }W_{t}+E_{2}^{\prime }\right) \right] }{E\left[ \exp \left( C_{1}^{\prime }t+D_{1}^{\prime }W_{t}+E_{1}^{\prime }\right) \right] },\\&=\exp \left( E_{2}^{\prime }-E_{1}^{\prime }\right) \frac{\exp \left( C_{2}^{\prime } +\frac{1}{2}D_{2}^{\prime 2}\right) t}{\exp \left( C_{1}^{\prime }+\frac{1}{2}D_{1}^{\prime 2}\right) t}. \end{aligned}$$

where $C_{i}^{\prime }=-\frac{C_{i}}{\theta _{i}},$ $D_{i}^{\prime }=-\frac{D_{i}}{\theta _{i}}$ and $E_{i}^{\prime }=-\frac{E_{i}}{\theta _{i}}$ for $i=1,2.$ We have $\left( C_{2}^{\prime }+\frac{1}{2}D_{2}^{\prime 2}\right) -\left( C_{1}^{\prime }+\frac{1}{2}D_{1}^{\prime 2}\right) =\delta _{2} \frac{\sigma +\theta _{1}\delta _{2}}{\theta _{1}+\theta _{2}}$ which ends the proof. $\square$

Proof of Proposition 3

Let us index by $\gamma \in \Gamma _{i}$ the individual members of Group i and let us denote respectively by $c_{i,\gamma }^{*}$, $\theta _{i,\gamma }$ and $w_{i,\gamma }$ their optimal consumption, risk tolerance level and endowment share within Group i. By construction, we have $\sum _{\gamma \in \Gamma _{i}}c_{i,\gamma }^{*} =c_{\gamma }^{*},$ $\sum _{\gamma \in \Gamma _{i}}\theta _{i,\gamma }=\theta _{i}$ and $\sum _{\gamma \in \Gamma _{i}}w_{i,\gamma }=1.$ It is easy to check that, at the equilibrium, we have $c_{i,\gamma }^{*}=\frac{\theta _{i,\gamma }}{\theta _{i}}c_{i,\gamma }^{*}+(w_{i,\gamma }-\frac{\theta _{i,\gamma }}{\theta _{i}})q\cdot c_{i}^{*}$ where $q\cdot c_{i}^{*}=E\left[ \int _{0}^{\infty }q_{t}c_{i,t}^{*}\hbox {d}t\right] .$ We have then

$$\begin{aligned} U_{\gamma ,t}\left( c_{i,\gamma }^{*}\right)= & {} U_{i,t}\left( c_{i}^{*}\right) \exp \left( -\frac{q\cdot c_{i}^{*}}{\theta _{i}}\right) \exp \left( \frac{w_{i,\gamma }}{\theta _{i,\gamma }}q\cdot c_{i}^{*}\right) \text { and } {\displaystyle \prod \limits _{\gamma \in \Gamma _{i}}} U_{\gamma ,t}\left( c_{i,\gamma }^{*}\right) ^{\frac{\theta _{i,\gamma } }{\theta _{i}}}\\= & {} U_{i,t}\left( c_{i}^{*}\right) . \end{aligned}$$

$\square$

Proof of Claim 4

From the first-order conditions, we have

$$\begin{aligned} \frac{U_{2}\left( c_{2}^{*}\right) }{U_{1}\left( c_{1}^{*}\right) }=\frac{\lambda _{2}}{\lambda _{1}}. \end{aligned}$$

We know that $E_{i}=\frac{\theta _{i}}{\theta _{1}+\theta _{2}}e_{0}^{*} +\frac{\theta _{1}\theta _{2}}{\theta _{1}+\theta _{2}}\ln \frac{\lambda _{j} }{\lambda _{i}}$ which gives

$$\begin{aligned} \frac{U_{2}\left( c_{2}^{*}\right) }{U_{1}\left( c_{1}^{*}\right) }=\exp \left( \frac{\theta _{1}+\theta _{2}}{\theta _{1}\theta _{2}}E_{1}-\frac{1}{\theta _{2}}e_{0}^{*}\right) . \end{aligned}$$

A direct computation gives

$$\begin{aligned} U_{i}\left( c_{i}^{*}\right)&=-\exp \left( -\frac{E_{i}}{\theta _{i} }\right) \int _{0}^{\infty }\exp \left( -\left( \frac{C_{i}}{\theta _{i}} +\frac{1}{2}\delta _{i}^{2}\right) s+\frac{1}{2}\left( \delta _{i}-\frac{D_{i}}{\theta _{i}}\right) ^{2}s\right) dt\\&=-\frac{1}{a_{i}}\exp \left( -\frac{E_{i}}{\theta _{i}}\right) \end{aligned}$$

where the convergence of the integral is insured by Condition (C). $\square$

Proof of Claim 5

We have

$$\begin{aligned} U_{i}^{\text {ex-post}}(c_{i}^{*})&=-\exp \left( -\frac{E_{i}}{\theta _{i}}\right) \int _{0}^{\infty }\exp \left( -\frac{C_{i}}{\theta _{i} }s\right) E^{P}\left[ \exp \left( -\frac{D_{i}W_{s}}{\theta _{i}}\right) \right] \hbox {d}s\\&=-\exp \left( -\frac{E_{i}}{\theta _{i}}\right) \int _{0}^{\infty } \exp \left( -\left( \frac{C_{i}}{\theta _{i}}-\frac{1}{2}\frac{D_{i}^{2} }{\theta _{i}^{2}}\right) s\right) \hbox {d}s\\&=-\frac{1}{b_{i}}\exp \left( -\frac{E_{i}}{\theta _{i}}\right) \end{aligned}$$

as far as $b_{1}\$and $b_{2}$ are positive. These conditions, respectively, correspond to Conditions (C) and (C1).

Furthermore, $b_{1}-b_{2}=\delta _{2}\frac{\sigma +\theta _{1} \delta _{2}}{\theta _{1}+\theta _{2}}$ and we have then $b_{1}>b_{2}$ for $\delta _{2}>0$ which means that $U_{2}^{\text {ex-post}}(c_{2}^{*} )>U_{1}^{\text {ex-post}}(c_{1}^{*}).$ $\square$

Proof of Claim 6

Since both agents have the same initial endowment, $c_{1}^{*}$ satisfies then the budget constraint of Agent 2. By optimality, we have $U_{2}(c_{1}^{*})\le U_{2}(c_{2}^{*})$. We even have $U_{2}(c_{2}^{*})>U_{2}(c_{1}^{*})$ as far as $c_{2}^{*}\ne c_{1}^{*}$ which is the case as far as $\left( \delta _{1},\theta _{1}\right) \ne \left( \delta _{2},\theta _{2}\right)$.

Let us now derive explicit expressions for $U_{2,t}(c_{1}^{*})$ for all t. We have

$$\begin{aligned} U_{2}(c_{1}^{*})&\!=\!\int _{0}^{\infty }-\exp \left( -\frac{C_{1}s\!+\!E_{1} }{\theta _{2}}\right) \exp \left( -\frac{1}{2}\delta _{2}^{2}s\right) E^{P}\left[ \exp \left( \left( \delta _{2}-\frac{D_{1}}{\theta _{2}}\right) W_{s}\right) \right] \hbox {d}s\\&=-\exp \left( -\frac{E_{1}}{\theta _{2}}\right) \int _{0}^{\infty } \exp \left( \left( -\frac{C_{1}}{\theta _{2}}+\frac{1}{2}\frac{D_{1}^{2}}{\theta _{2}^{2}}-\delta _{2}\frac{D_{1}}{\theta _{2}}\right) s\right) \hbox {d}s \end{aligned}$$

which converges under Condition (C2) and leads to

$$\begin{aligned} U_{2}(c_{1}^{*})=-\frac{1}{d_{1}}\exp \left( -\frac{E_{1}}{\theta _{2} }\right) . \end{aligned}$$

Similarly, we have

$$\begin{aligned} U_{2}^{\text {ex-post}}(c_{1}^{*})=-\exp \left( -\frac{E_{1}}{\theta _{2} }\right) \int _{0}^{t}\exp \left( \left( -\frac{C_{1}}{\theta _{2}}+\frac{1}{2} \frac{D_{1}^{2}}{\theta _{2}^{2}}\right) s\right) \hbox {d}s \end{aligned}$$

which converges under Condition (C3) and leads to

$$\begin{aligned} U_{2}^{\text {ex-post}}(c_{1}^{*})=-\frac{1}{f_{1}}\exp \left( -\frac{E_{1} }{\theta _{2}}\right) . \end{aligned}$$

$\square$

Proof of Claim 7

Under Condition (C4) , $\widetilde{c} _{2}^{*}$ is obtained through the formulas above for $\delta _{2}=0.$ We have then

$$\begin{aligned} \widetilde{c}_{2,t}^{*}=C_{0}t+D_{0}W_{t}+E_{0}. \end{aligned}$$

The corresponding ex-ante and ex-post welfare levels for Agent 2 are then given, as above, by

$$\begin{aligned} U_{2}(\widetilde{c}_{2}^{*})&=-\frac{1}{g_{0}}\exp \left( -\frac{E_{0} }{\theta _{2}}\right) ,\\ U_{2}^{\text {ex-post}}(\widetilde{c}_{2}^{*})&=-\frac{1}{h_{0}} \exp \left( -\frac{E_{0}}{\theta _{2}}\right) . \end{aligned}$$

Note that $U_{2}^{\text {ex-post}}(\widetilde{c}_{2}^{*})$ is well defined by construction and $U_{2}(\widetilde{c}_{2}^{*})$ is well defined under Condition (C5). $\square$

Finite horizon In a finite horizon setting T, under the same conditions and with the same notations as above, the different formulas above become

$$\begin{aligned} U_{i}\left( c_{i}^{*}\right)&=-\frac{1}{a_{i}}\exp \left( -\frac{E_{i}}{\theta _{i}}\right) \left[ 1-\exp \left( -a_{i}T\right) \right] \\ U_{i}^{\text {ex-post}}\left( c_{i}^{*}\right)&=-\frac{1}{b_{i}}\exp \left( -\frac{E_{i}}{\theta _{i}}\right) \left[ 1-\exp \left( -b_{i}T\right) \right] \\ U_{2}\left( c_{1}^{*}\right)&=-\frac{1}{d_{1}}\exp \left( -\frac{E_{1}}{\theta _{2} }\right) \left[ 1-\exp \left( -d_{1}T\right) \right] \\ U_{2}^{\text {ex-post}}\left( c_{1}^{*}\right)&=-\frac{1}{f_{1}}\exp \left( -\frac{E_{1}}{\theta _{2}}\right) \left[ 1-\exp \left( -f_{1}T\right) \right] \\ U_{2}\left( \widetilde{c}_{2}^{*}\right)&=-\frac{1}{g_{0}}\exp \left( -\frac{E_{0} }{\theta _{2}}\right) \left[ 1-\exp \left( -g_{0}T\right) \right] \\ U_{2}^{\text {ex-post}}\left( \widetilde{c}_{2}^{*}\right)&=-\frac{1}{h_{0}} \exp \left( -\frac{E_{0}}{\theta _{2}}\right) \left[ 1-\exp \left( -h_{0}T\right) \right] \end{aligned}$$

and we have $a_{1}=a_{2}$ which means that comparing $U_{1}\left( c_{1} ^{*}\right)$ and $U_{2}\left( c_{2}^{*}\right)$ in a finite horizon setting amounts to comparing them in an infinite horizon setting. Furthermore, for $\delta _{2}>0,$ we have $b_{1}>b_{2}$ which means that if $U_{2}^{\text {ex-post}}(c_{2}^{*})>U_{1}^{\text {ex-post}}(c_{1}^{*})$ in an infinite horizon setting and if $\delta _{2}>0$ then the inequality pertains for all T.

Finally, it is immediate that all the inequalities obtained in the numerical examples above pertain for T large enough.