Information and dynamic trading with the Gambler’s fallacy

Abstract

A multi-period stock trading model is developed in which there are two types of traders—a “rational” type and a “gambler’s fallacy” type—both observing a public signal about the fundamental value in each period. The rational type holds correct beliefs on the signals, whereas the gambler’s fallacy type mistakenly believes that the sequence of the signals exhibits systematic reversals. We explore the dynamic equilibrium in which the two types trade with each other to speculate future price changes based on their inferences about the fundamental value. It is shown that the presence of the gambler’s fallacy type can generate both short-term momentum and long-term reversal. Furthermore, the pattern is robust even in a market with a large proportion of rational traders as the price is closer to the gambler’s fallacy type’s valuation. We also show that the gambler’s fallacy type becomes more influential in determining the price as the market switches from momentum to reversal. Interestingly, to an outside observer, it would appear as though the rational traders act as if they have “hot-hand” fallacy in prices, and that a trend following strategy is optimal in this model.

Appendices

Mathematical Appendix A

Proof of Proposition 1: (Kalman Filtering)

We first reformulate agents’ beliefs into state space models. The beliefs evolution therefore is described by Kalman Filtering²⁰. To be specific, the beliefs of the rational type described by (2) is already a standard state space model and is presented as below in (12), (13):

$$\begin{aligned} y^R_t= & {} V_t+\xi ^R_t \end{aligned}$$

(12)

$$\begin{aligned} V_t= & {} V_{t-1}+\omega ^R_t. \end{aligned}$$

(13)

where $V_1\sim {\mathscr {N}}(\mu ,\sigma ^2_V)$, $\xi _{t}^{R}\sim N(0,\sigma _{\xi }^{2})$, $\omega _{t}^{R}\sim N(0,\sigma _{\omega }^{2})$ and $\sigma _{\omega }^{2}=0$.

Instead, the beliefs of type G in (3), (4) can be reformulated into a state space model as in (14), (15) by letting $\epsilon _{t}\equiv \underset{k=0}{\overset{t}{\varSigma }}\delta ^{k}\kappa _{t-k}$:

$$\begin{aligned} y^G_t= & {} ZS_{t-1}+\eta _t \end{aligned}$$

(14)

$$\begin{aligned} S_t= & {} \varUpsilon S_{t-1}+\tau _t \end{aligned}$$

(15)

where

$$\begin{aligned} S_t= & {} [V_{t}\;,\epsilon _t]',\\ Z= & {} [1,-\beta ],\\ \varUpsilon= & {} \left[ \begin{array}{cc} 1 &{} 0\\ 0 &{} \delta -\beta \end{array}\right] , \\ \eta _t= & {} \omega ^G_t+\xi _t^G, \\ \tau _t= & {} \left[ \omega ^G_{t}\;,\xi _{t}^G\right] '. \end{aligned}$$

Similar to the rational case, we assume $V_1\sim {\mathscr {N}}(\mu ,\sigma ^2_V)$, $\xi _{t}^{G}\sim N(0,\sigma _{\xi }^{2}),\;\omega _{t}^{G}\sim N(0,\sigma _{\omega }^{2}),$ and $\sigma _{\omega }^{2}=0$. We further assume $\epsilon _0=\kappa _{0}=\xi _0^G$ and is randomly drawn from ${\mathscr {N}}(0, \sigma ^2_\xi )$ implying both types of investors have the same priors on V so that the assumption ${\mathscr {F}}_0^i={\mathscr {F}}_0$ above is satisfied. Compared to the state space model for the rational agents with $V_t$ as the state variable, the fallacy agents’ problem have two state variables, $V_t$ and $\epsilon _t$.

The rational investor’s recursive-filtering problem is standard. The results are directly from Tsay [40], Chapter 11.1. We only prove the problem for the fallacy investors.

The problem described in (14), (15) can be reformulated as below:

$$\begin{aligned} y_{t}= & {} Z\varUpsilon ^{-1}S_{t}-Z\varUpsilon ^{-1}\tau _{t}+\eta _{t} \\ S_{t}= & {} \varUpsilon S_{t-1}+\tau _{t} \end{aligned}$$

And we have

$$\begin{aligned} S^{G}_{t+1|t}= & {} E_{G}\left( \varUpsilon S_{t}+\tau _{t+1}|{\mathscr {F}}_{t}\right) =\varUpsilon S^{G}_{t|t}\\ v_{t}^{G}= & {} y_{t}-y_{t|t-1}^{G}=Z\varUpsilon ^{-1}\left( S_{t}-S^{G}_{t|t-1}\right) -Z\varUpsilon ^{-1}\tau _{t}+\eta _{t} \end{aligned}$$

Use the formula for the conditional expectation of the bivariate normal distribution, we have

$$\begin{aligned} S^{G}_{t|t}=E_{G}\left( S_{t}|{\mathscr {F}}_{t},\; v_{t}\right) =S^{G}_{t|t-1}+Cov_{G}(S_{t},\; v_{t}|{\mathscr {F}}_{t-1})\cdot Var_{G}(v_{t}|{\mathscr {F}}_{t-1})^{-1}v_{t} \end{aligned}$$

where

$$\begin{aligned} Cov_{G}(S_{t},\; v_{t}|{\mathscr {F}}_{t})= & {} Cov_{G}\left( S_{t},\; Z\varUpsilon ^{-1}(S_{t}-S^{G}_{t|t-1}-Z\varUpsilon ^{-1}\tau _{t}+\eta _{t})|{\mathscr {F}}_{t-1}\right) \\= & {} \varSigma _{S,t|t-1}^{G}\left( Z\varUpsilon ^{-1}\right) ^{\mathbf {T}}+Cov_{G}\left( S_{t},-Z\varUpsilon ^{-1}\tau _{t}+\eta _{t}|{\mathscr {F}}_{t-1}\right) \\= & {} \varSigma _{S,t|t-1}^{G}\left( Z\varUpsilon ^{-1}\right) ^{\mathbf {T}}+Cov_{G}\left( \tau _{t},-Z\varUpsilon ^{-1}\tau _{t}+\eta _{t}|{\mathscr {F}}_{t-1}\right) \\= & {} \varSigma _{S,t|t-1}^{G}\left( Z\varUpsilon ^{-1}\right) ^{\mathbf {T}}-L_{t}\left( Z\varUpsilon ^{-1}\right) ^{\mathbf {T}}+\varOmega _{t}\\ Var_{G}\left( v_{t}|{\mathscr {F}}_{t-1}\right)= & {} Var_{G}\left( Z\varUpsilon ^{-1}(S_{t}-S^{G}_{t|t-1})-Z\varUpsilon ^{-1}\tau _{t}+\eta _{t}|{\mathscr {F}}_{t-1}\right) \\= & {} Z\varUpsilon ^{-1}\varSigma _{S,t|t-1}^{G}\left( Z\varUpsilon ^{-1}\right) ^{\mathbf {T}}+Var_{G}\left( -Z\varUpsilon ^{-1}\tau _{t}+\eta _{t}|{\mathscr {F}}_{t-1}\right) \\&+2Cov_{G}\left( Z\varUpsilon ^{-1}S_{t},\;-Z\varUpsilon ^{-1}\tau _{t}+\eta _{t}|{\mathscr {F}}_{t-1}\right) . \end{aligned}$$

Furthermore, it is easily to show that

$$\begin{aligned}&Var_{G}\left( -Z\varUpsilon ^{-1}\tau _{t}+\eta _{t}|{\mathscr {F}}_{t-1}\right) =H_{t}+Z\varUpsilon ^{-1}L_{t}\left( Z\varUpsilon ^{-1}\right) ^{\mathbf {T}}-2Z\varUpsilon ^{-1}\varOmega _{t}\\&Cov_{G}\left( Z\varUpsilon ^{-1}S_{t},\;-Z\varUpsilon ^{-1}\tau _{t}+\eta _{t}|{\mathscr {F}}_{t-1}\right) =-Z\varUpsilon ^{-1}L_{t}\left( Z\varUpsilon ^{-1}\right) ^{\mathbf {T}}+Z\varUpsilon ^{-1}\varOmega _{t}. \end{aligned}$$

Let $K_{t}=\varUpsilon Cov_{G}\left( S_{t},\; v_{t}^{G}\right) Var_{G}\left( v_{t}\right) ^{-1}$, we proved that

$$\begin{aligned} S^{G}_{t+1|t}=\varUpsilon S^{G}_{t|t-1}+K_{t}v_{t}^{G} \end{aligned}$$

Next, we prove the recursive equation for the conditional variance.

$$\begin{aligned} \varSigma _{S,t+1|t}^{G}=\varUpsilon \varSigma _{S,t|t}^{G}\varUpsilon ^{\mathbf {T}}+L_{t+1}. \end{aligned}$$

It is easy to show that

$$\begin{aligned} \varSigma _{S,t|t}^{G}=\varSigma _{S,t|t-1}^{G}-Var_{G}\left( v_{t}^{G}\right) \varUpsilon ^{-1}K_{t}K_{t}^{\mathbf {T}}\left( \varUpsilon ^{\mathbf {T}}\right) ^{-1}. \end{aligned}$$

Bring it back to the original formula, we prove that

$$\begin{aligned} \varSigma _{S,t+1|t}^{G}=\varUpsilon \varSigma _{t|t-1}^{G}\varUpsilon ^{{\mathbf {T}}}-[Y\varSigma _{t|t-1}^{G}Y^{\mathbf {T}}+H-YLY^{\mathbf {T}}]K_{t}K_{t}^{\mathbf {T}}+L, \end{aligned}$$

The formula of $K_t$ can be easily derived from here.

The definitions of $\varOmega _t$, $L_t$, $H_t$ and Y are defined as below:

$$\begin{aligned} \varOmega _t= & {} E[\eta _t \tau _t]\\= & {} [0,\;\sigma ^2_\xi ]',\quad L_t=E[\tau _t \tau _t']=\left[ \begin{array}{cc} 0 &{} 0\\ 0 &{} \sigma _{\xi }^{2} \end{array}\right] , \quad H_t=E[\eta _t^2]=\sigma ^2_\xi , \quad Y=Z\varUpsilon ^{-1} \end{aligned}$$

$\square$

Proof of Proposition 2(Linear Equilibrium) and Proposition 4(Optimal Holdings )

Step 1: Conjectured Price

Assumption 1

Given the economy defined in Sect. 2.1 and two type of investors with beliefs defined in Equation (2)-(4), there exists a competitive equilibrium, in which the price of the risky asset is a linear function of the state variables of the economy:

$$\begin{aligned} P_t=\alpha _t\varPsi _t, \end{aligned}$$

where $\alpha _{t}=\left[ \alpha _{0,t},\;\alpha _{1,t},\;\alpha _{2,t},\;\alpha _{3,t}\right]$ is a $1\times 4$ vector.

We use $P_{t}$ and $Q_{t}$ to denote the stochastic process of real price and return. Because beliefs are different, we further use $P_{t}^i$ and $Q_{t}^i$ to denote the beliefs of type i on the stochastic process of $P_t$ and $Q_t$, and $\varPsi _t^i$ to denote the beliefs on the stochastic process $\varPsi _t$²¹. Given the assumption in Step 1, we have the following Lemma 3.

Lemma 3

(Price and Return Process) In a linear equilibrium where $P_t=\alpha _t\varPsi _t$, $Q_t$ and $\varPsi _t$ are both measurable with respect to ${\mathscr {F}}_t$ and take the following Gaussian processes under information $\{{\mathscr {F}}_{t}:1\le t\le T\}$ for type i, $i\in {R,G}$:

$$\begin{aligned} Q_{t+1}^i= & {} A_{Q,t+1}^{i}\varPsi ^i_{t}+B_{Q,t+1}^{i}v_{t+1}^i \end{aligned}$$

(16)

$$\begin{aligned} \varPsi _{t+1}^i= & {} A_{\varPsi ,t+1}^{i}\varPsi ^i_{t}+B_{\varPsi ,t+1}^{i}v_{t+1}^i \end{aligned}$$

(17)

where $v_{t+1}^i$ are signal shocks defined as before and are normally distributed conditional on ${\mathscr {F}}_t$; $A_{\varPsi ,t}^i$ are $4\times 4$ matrices, $A_{Q,t}^i$ are $1\times 4$ vectors, $B_{\varPsi ,t}^i$ are $4\times 1$ vectors and $B_{Q,t}^i$ are constants given below.

Assume $P_t=\alpha _t\varPsi _t$ holds, we first prove the Lemma 3. The way to derive $A_{\varPsi ,t}^i$ and $B_{\varPsi ,t}^i$ is from the recursive process presented in Proposition 1. We only give the proof for $A_{\varPsi ,t}^G$ and $B_{\varPsi ,t}^G$ and use similar method, we can derive $A_{\varPsi ,t}^R$ and $B_{\varPsi ,t}^G$.

From G’s perspective,

$$\begin{aligned} \begin{array}{clc} V_{t+1|t}^{R}&{}=V_{t|t-1}^{R}+K_{t}^{R}v_{t}^{R}\\ &{}=V_{t|t-1}^{R}+K_{t}^{R}\left( V_{t|t-1}^{G}-\dfrac{\beta }{\delta -\beta }\epsilon ^{G}_{t|t-1}+v_{t}^{G}-V_{t|t-1}^{R}\right) \\ &{}=\left( 1-K_{t}^{R}\right) V_{t|t-1}^{R}+K_{t}^{R}V_{t|t-1}^{G}-\dfrac{\beta }{\delta -\beta }K_{t}^{R}\epsilon ^{G}_{t|t-1}+K_{t}^{R}v_{t}^{G}; \end{array} \end{aligned}$$

similarly,

$$\begin{aligned} V_{t+1|t}^{G}= & {} V_{t|t-1}^{G}+K_{t}^{G}v_{t}^{G}\\ \epsilon ^{G}_{t+1|t}= & {} \left( \delta -\beta \right) \epsilon ^{G}_{t|t-1}+k_{t}^{G}v_{t}^{G}. \end{aligned}$$

These equations above give us the expression that

$$\begin{aligned} A_{\varPsi ,t}^{G}=\left( \begin{array}{cccc} 1 &{} 0 &{} 0 &{} 0\\ 0 &{} 1-K_{t}^{R} &{} K_{t}^{R} &{} -\frac{\beta }{\delta -\beta }K_{t}^{R}\\ 0 &{} 0 &{} 1 &{} 0\\ 0 &{} 0 &{} 0 &{} \delta -\beta \end{array}\right) ,\quad \quad B_{\varPsi ,t}^{G}=\left( \begin{array}{c} 0\\ K_{t}^{R}\\ K_{t}^{G}\\ k_{t} \end{array}\right) . \end{aligned}$$

Similarly, we have

$$\begin{aligned} A_{\varPsi ,t}^{R}=\left( \begin{array}{cccc} 1 &{} 0 &{} 0 &{} 0\\ 0 &{} 1 &{} 0 &{} 0\\ 0 &{} K_{t}^{G} &{} 1-K_{t}^{G} &{} \frac{\beta }{\delta -\beta }K_{t}^{G}\\ 0 &{} k_{t} &{} -k_{t} &{} \delta -\beta +\frac{\beta }{\delta -\beta }k_{t} \end{array}\right) ,\quad \quad B_{\varPsi ,t}^{R}=\left( \begin{array}{c} 0\\ K_{t}^{R}\\ K_{t}^{G}\\ k_{t} \end{array}\right) . \end{aligned}$$

Furthermore, given the assumption that $P_t=\alpha _t\varPsi _t$, we can show that from G’s perspective,

$$\begin{aligned} Q^i_{t+1}= & {} \alpha _{0,t+1}-\alpha _{0,t}\\&+\alpha _{1,t+1}\left[ V_{t+1|t}^{R}+K_{t+1}^{R}\left( V_{t+1|t}^{G}-\dfrac{\beta }{\delta -\beta }\epsilon ^{G}_{t+1|t}+v_{t+1}^{G}-V_{t+1|t}^{R}\right) \right] \\&\alpha _{2,t+1}\left[ V_{t+1|t}^{G}+K_{t+1}^{G}v_{t+1}^{G}\right] \\&\alpha _{3,t+1}\left[ \left( \delta -\beta \right) \epsilon ^{G}_{t+1|t}+k_{t+1}^{G}v_{t+1}^{G}\right] \\&-\alpha _{1,t}V_{t+1|t}^{R}-\alpha _{2,t}V_{t+1|t}^{G}-\alpha _{3,t}\epsilon ^{G}_{t+1|t} \end{aligned}$$

Reorganise the right hand according to $V_{t+1|t}^R$, $V_{t+1|t}^G$, $\epsilon ^{G}_{t+1|t}$ and $v_t^G$ will give give us the matrix $A_{Q,t+1}^G$ and $B_{Q,t+1}^G$ in the Proposition, that is: $A_{Q,t}^{G}=\alpha _{t}A_{\varPsi ,t}^{G}-\alpha _{t-1}$, $B_{Q,t}^{G}=\alpha _{t}B_{\varPsi ,t}^G$. Similarly, we have $A_{Q,t}^{R}=\alpha _{t}A_{\varPsi ,t}^{R}-\alpha _{t-1}$, $B_{Q,t}^{R}=\alpha _{t}B_{\varPsi ,t}^R.$ $\square$

Notice that based on Definition 1 and given the Proposition 2 holds, the real stochastic process of the state vector of the economy and excess return must coincide with the beliefs of the rational investors because their beliefs $y^i_t$ are correct. Therefore, the real process of $\varPsi _t$ and $Q_t$ can be written as

$$\begin{aligned} \varPsi _{t+1}= & {} A_{\varPsi ,t+1}^{R}\varPsi _{t}+B_{\varPsi ,t+1}^{R}v_{t+1}^R\\ Q_{t+1}= & {} A_{Q,t+1}^{R}Q_{t}+B_{Q,t+1}^{R}v_{t+1}^R. \end{aligned}$$

We further move on to the optimization problem which determines the optimal portfolios of two types of investors.

Step 2: optimization

With Gaussian assumptions on the price process in Step 1 and Lemma 3 and under the CARA utility function assumption, the optimization problem defined by Eqs. (7), (8) can be solved using dynamic programming.

Let $J(W_{t}^{i};\;\varPsi ^i_{t},\; t)$ be the value function of type i, we have the Bellman equation below:

$$\begin{aligned} J(W_{t}^{i};\;\varPsi ^i_{t},\; t)=\underset{X_{t}^{i}}{Max}E_{i}[J(W_{t+1}^{i};\;\varPsi ^i_{t+1},\; t+1)|{\mathscr {F}}_{t}]. \end{aligned}$$

The optimization problem (7), (8) is equivalent to the following problem:

$$\begin{aligned}&0=\underset{X_{t}^{i}}{Max}\left\{ E_{i}[J(W_{t+1}^{i};\;\varPsi ^i_{t+1},\; t+1)|{\mathscr {F}}_{t}]-J(W_{t}^{i};\;\varPsi ^i_{t},\; t)\right\} \\&\quad s.t. \;\; W^i_{t+1}=W^i_t+X^i_tQ^i_{t+1} \\&\quad J(W_{T}^{i};\;\varPsi ^i_{T},\; T)=-e^{-\lambda W_{T}^{i}}, \end{aligned}$$

Given the Lemma 3, we can derive Proposition 4. We prove the Proposition 4 by solving the optimization problem.

Assume the value function take the following format:

$$\begin{aligned} J(W_{t}^{i};\;\varPsi _{t},\; t)=-exp\left\{ -\lambda W_{t}^{i}-\frac{1}{2}\varPsi _{t}^{i\mathbf {T}}U_{t}^{i}\varPsi _{t}^{i}\right\} \end{aligned}$$

The first term represents the utility from the current investment, and the second term represents a risk adjusted utility from future investment.

We plug the Gaussian process of $Q_{t+1}^i$ and $\varPsi _t^i$ into the proposed value function and we have:

$$\begin{aligned} \begin{array}{clc} E_{i}[J(W_{t+1}^{i};\;\varPsi _{t+1}^{i},\; t+1)|{\mathscr {F}}_{t}]&{}=-\rho _{t+1}exp\left\{ -\lambda W_{t}^{i}-\frac{1}{2}\varPsi _{t}^{i\mathbf {T}}A_{\varPsi ,t+1}^{i\mathbf {T}}U_{t+1}^{i}A_{\varPsi ,t+1}^{i}\varPsi _{t}^{i}\right\} \\ &{}\times exp\left\{ -\lambda X_{t}^{i}A_{Q,t+1}^{i}\varPsi _{t}^{i}+\frac{1}{2}C_{t+1}^{i\mathbf {T}}\theta _{t+1}^{i}C_{t+1}^{i}\right\} \end{array}\nonumber \\ \end{aligned}$$

(18)

where

$$\begin{aligned} C_{t+1}^{i}= & {} \lambda B_{Q,t+1}^{i\mathbf {T}}X_{t}^{i}+B_{\varPsi ,t+1}^{\mathbf {T}}U_{t+1}^{i}A_{\varPsi ,t+1}^{i}\varPsi _{t}^{i},\\ \theta _{t+1}^{i}= & {} \left( Var(v_{t+1}^{i})^{-1}+B_{\varPsi ,t+1}^{i\mathbf {T}}U_{t+1}^{i}B_{\varPsi ,t+1}^{i}\right) ^{-1},\\ \rho ^i_{t+1}= & {} (1+Var(v_{t+1}^{i})B_{\varPsi ,t+1}^{i\mathbf {T}}U_{t+1}^{i}B_{\varPsi ,t+1}^{i})^{-\frac{1}{2}}. \end{aligned}$$

Differentiate the value function with respect to $X_t^i$, and from the F.O.C., we have

$$\begin{aligned} X_{t}^{i}=\dfrac{1}{\lambda }F_{t}^{i}\varPsi _{t}, \end{aligned}$$

where

$$\begin{aligned} F_{t}^{i}\equiv \left( B_{Q,t+1}^{i}\theta _{t+1}^{i}B_{Q,t+1}^{i\mathbf {T}}\right) ^{-1}\left( A_{Q,t+1}^{i}-B_{Q,t+1}^{i}\theta _{t+1}^{i}B_{\varPsi ,t+1}^{i\mathbf {T}}U_{t+1}^{i}A_{\varPsi ,t+1}^{i}\right) , \end{aligned}$$

The second order condition for optimality holds because $B_{Q,t+1}^{i}\theta _{t+1}^{i}B_{Q,t+1}^{i\mathbf {T}}>0$. Since $A_{Q,t+1}^{i}$ is defined by Eqs. (16) and (17), we have $A_{Q,t+1}^{i}\varPsi _t=E_i(Q_{t+1}^i|{\mathscr {F}}_t)$. Define $\varGamma _t^i\equiv \left( B_{Q,t+1}^{i}\theta _{t+1}^{i}B_{Q,t+1}^{i\mathbf {T}}\right) ^{-1}$ and $g_t^i\equiv B_{Q,t+1}^{i}\theta _{t+1}^{i}B_{\varPsi ,t+1}^{i\mathbf {T}}U_{t+1}^{i}A_{\varPsi ,t+1}^{i}$, we then have the conclusion in Proposition 4.

Furthermore, we need to show that

$$\begin{aligned} J(W_{t}^{i};\;\varPsi _{t},\; t)=-exp\left\{ -\lambda W_{t}^{i}-\frac{1}{2}\varPsi _{t}^{i\mathbf {T}}U_{t}^{i}\varPsi _{t}^{i}\right\} \end{aligned}$$

(19)

is the right format of the value function. This is the correct value function if we can find the right format of $U_t^i$. To do this, we substitute $X_{t}^{i}=\dfrac{1}{\lambda }F_{t}^{i}\varPsi _{t}$ into the Bellman equation, that is

$$\begin{aligned} J(W_{t}^{i};\;\varPsi _{t},\; t)=\underset{X_{t}^{i}}{Max}E_{i}[J(W_{t+1}^{i};\;\varPsi _{t+1}^{i},\; t+1)|{\mathscr {F}}_{t}]. \end{aligned}$$

²² Define

$$\begin{aligned} M_{t+1}^{i}\equiv & {} F_{t+1}^{i\mathbf {T}}\left( B_{Q,t+2}^{i}\theta _{t+2}^{i}B_{Q,t+2}^{i\mathbf {T}}\right) F_{t+1}^{i}+A_{\varPsi ,t+2}^{i\mathbf {T}}U_{t+2}^{i}A_{\varPsi ,t+2}^{i}\\&-\left( B_{\varPsi ,t+2}^{i\mathbf {T}}U_{t+2}^{i}A_{\varPsi ,t+2}^{i}\right) ^{\mathbf {T}}\theta _{t+2}^{i}\left( B_{\varPsi ,t+2}^{i\mathbf {T}}U_{t+2}^{i}A_{\varPsi ,t+2}^{i}\right) . \end{aligned}$$

We have

$$\begin{aligned} \begin{array}{clc} -\rho ^i_{t+1}exp\left\{ -\lambda W_{t}^{i}-\frac{1}{2}\varPsi _{t}^{i\mathbf {T}}M_{t}^{i}\varPsi _{t}^{i}\right\} =&-exp\left\{ -\lambda W_{t}^{i}-\frac{1}{2}\varPsi _{t}^{i\mathbf {T}}U_{t}^{i}\varPsi _{t}^{i}\right\} \end{array}. \end{aligned}$$

From here, we can derive the right form of $U_t^i$:

$$\begin{aligned} U_{t}^{i}=-2ln\rho _{t+1}^{i}I_{11}+M_{t}^i, \end{aligned}$$

where $I_{11}$ is a $4\times 4$ zero matrix with only the first row, first column element equal to 1. Therefore, the conjectured format of $-\rho ^i_{t+1}exp\left\{ -\lambda W_{t}^{i}-\frac{1}{2}\varPsi _{t}^{i\mathbf {T}}M_{t}^{i}\varPsi _{t}^{i}\right\}$ holds.

If the conjectured price is correct as in Proposition 2, we then complete the proof for Proposition 4. We prove this in Step 3.

Step 3. market-clearing and Equilibrium Price

The last step to prove the Proposition 2 and Proposition 4 is to show that the equilibrium price derived from the market-clearing condition is consistent with the proposed equilibrium price function in Step 1.

Since investors in this model are assumed to be non-myopic, given the optimal demands of both types solved in Step 2, the market-clearing conditions can be applied backwards recursively to derive the equilibrium price in each period.

To be more specific, the market-clearing condition is $\underset{i\in {\mathscr {I}}}{\sum }X_{t}^{i}=0$, for $t=1,\ldots ,T-1$. We first find the equilibrium price function at $T-1$. The dollar return at T is $Q_T=V-P_{T-1}$ and the equilibrium price can be solved explicitly. There is a continuum of investors on $[0,\; 1]$, with m proportion of rational investors and $n=1-m$ proportion of fallacy investors in the market, investor i’s optimal demand at $T-1$ has the linear form:

$$\begin{aligned} X_{T-1}^{i}=\dfrac{1}{\lambda }\dfrac{E^{i}[Q_{T}|{\mathscr {F}}_{T-1}]}{\varSigma _{V,T|T-1}^{i}}. \end{aligned}$$

From market-clearing condition at $T-1$:

$$\begin{aligned} mX_{T-1}^R+nX_{T-1}^G=0, \end{aligned}$$

we derive the explicit format for $P_{T-1}=\alpha _{T-1}\varPsi _t$, where

$$\begin{aligned} \alpha _{T-1}=[0,\;\dfrac{m\varSigma _{V,T|T-1}^{G}}{\left( m\varSigma _{V,T|T-1}^{G}+n\varSigma _{V,T|T-1}^{R}\right) },\;\dfrac{n\varSigma _{V,T|T-1}^{R}}{\left( m\varSigma _{V,T|T-1}^{G}+n\varSigma _{V,T|T-1}^{R}\right) },\;0] \end{aligned}$$

(20)

Proposition 10 below shows that from $\alpha _{T-1}$, the parameter vector $\alpha _{t}$ for any period t exist and is uniquely determined recursively.

Proposition 10

(Recursive Process) The equilibrium price $P_t$ is linear in $\varPsi _t$ with the parameter vector $\alpha _t$ uniquely determined by the following recursive process:

$$\begin{aligned} \alpha _{t}=\alpha _{t+1}{\overline{E}}_{t+1}, \end{aligned}$$

(21)

where ${\overline{E}}_{t+1}$ is a $4 \times 4$ matrix fully determined by m, $\lambda$, $A_{\varPsi ,t+1}^i$, $B_{\varPsi ,t_1}$, $A_{Q,t+1}^i$, $B_{Q,t_1}$, $\varSigma ^i_{V,t|t-1}$, $Var(v^i_{t+1})$.

The final step to complete the proof is to show that we can recursively solve the problem as shown in Proposition 10. Below, we present the proof of Proposition 10. By Definition 1, the proof for Proposition 2 can also be extended to derive a competitive equilibrium where the real signals have serial correlation. The equilibrium price process is still linear in the state variables of the economy but the real stochastic process will evolve according to the beliefs of the fallacy agents.

We first present the Proposition 10 completely and show the detailed format of related matrices involved.

Parameters in the equilibrium are determined by the following recursive process:

$$\begin{aligned} \alpha _{t}= & {} \alpha _{t+1}{\overline{E}}_{t+1} \end{aligned}$$

(22)

$$\begin{aligned} {\overline{E}}_{t+1}= & {} \left( \dfrac{m}{\theta _{t+1}^{R}}+\dfrac{n}{\theta _{t+1}^{G}}\right) ^{-1}\left( \dfrac{mE_{t+1}^{R}}{\theta _{t+1}^{R}}+\dfrac{nE_{t+1}^{G}}{\theta _{t+1}^{G}}\right) \end{aligned}$$

(23)

$$\begin{aligned} E_{t+1}^{i}= & {} A_{\varPsi ,t+1}^{i}-B_{\varPsi ,t+1}^{i}B_{\varPsi ,t+1}^{i\mathbf {T}}\theta _{t+1}^{i}U_{t+1}^{i}A_{\varPsi ,t+1}^{i},\quad for \; i=R,\;G \end{aligned}$$

(24)

$$\begin{aligned} U_{t+1}^{i}= & {} M_{t+1}^{i}-2log(\rho _{t+1}^{i})II, \end{aligned}$$

(25)

with II a $4\times 4$ zero matrix except the up-left corner set to 1.

We also have the following recursive equations:

$$\begin{aligned} \theta _{t+1}^{i}= & {} \left( Var(v_{t+1}^{i})^{-1}+B_{\varPsi ,t+1}^{i\mathbf {T}}U_{t+1}^{i}B_{\varPsi ,t+1}^{i}\right) ^{-1}, \quad for \; i=R,\;G \end{aligned}$$

(26)

$$\begin{aligned} \rho _{t+1}= & {} (1+Var(v_{t+1}^{i})B_{\varPsi ,t+1}^{i\mathbf {T}}U_{t+1}^{i}B_{\varPsi ,t+1}^{i})^{-\frac{1}{2}} \end{aligned}$$

(27)

$$\begin{aligned} M_{t+1}^{i}= & {} F_{t+1}^{\mathbf {T}}\left( B_{Q,t+2}^{i}\theta _{t+2}^{i}B_{Q,t+2}^{i\mathbf {T}}\right) F_{t+1}^23\nonumber \\&-\left( B_{\varPsi ,t+2}^{i\mathbf {T}}U_{t+2}^{i}A_{\varPsi ,t+2}^{i}\right) ^{\mathbf {T}}\theta _{t+2}\left( B_{\varPsi ,t+2}^{i\mathbf {T}}U_{t+2}^{i}A_{\varPsi ,t+2}^{i}\right) +A_{\varPsi ,t+2}^{i\mathbf {T}}U_{t+2}^{i}A_{\varPsi ,t+2}^{i} \end{aligned}$$

(28)

$$\begin{aligned} F_{t+1}^{i}= & {} \left( B_{Q,t+2}^{i}\theta _{t+2}^{i}B_{Q,t+2}^{i\mathbf {T}}\right) ^{-1}\left( A_{Q,t+2}-B_{Q,t+2}^{i}\theta _{t+2}^{i}B_{\varPsi ,t+2}^{i\mathbf {T}}U^i_{t+2}A_{\varPsi ,t+2}\right) ^24 \end{aligned}$$

(29)

²³²⁴

From Lemma 3, we have $A_{Q,t}^{i}=\alpha _{t}A_{\varPsi ,t}^{i}-\alpha _{t-1}$ and $B_{Q,t}^{i}=\alpha _{t}K_{t}$. Substitute these two formulas into $F_t^i$:

$$\begin{aligned} F_{t}^{i}=\dfrac{1}{\left( \alpha _{t+1}K_{t+1}\right) ^{2}\theta _{t+1}^{i}}\alpha _{t+1}E_{t+1}^{i}-\dfrac{1}{\left( \alpha _{t+1}K_{t+1}\right) ^{2}\theta _{t+1}^{i}}\alpha _{t} \end{aligned}$$

(30)

where $E_{t+1}^{i}=A_{\varPsi ,t+1}^{i}-K_{t+1}B_{\varPsi ,t+1}^{i\mathbf {T}}\theta _{t+1}^{i}U_{t+1}^{i}A_{\varPsi ,t+1}^{i}.$ Since $\varPsi _t^R=\varPsi _t^G$, substitute $X_{t}^{i}=\dfrac{1}{\lambda }F_{t}^{i}\varPsi _{t}$ into the market-clearing condition $mX_{t}^{R}+nX_{t}^{G}=0,$ we have

$$\begin{aligned} mF_t^R+nF_t^G=\mathbf {0}. \end{aligned}$$

(32)

Furthermore, substitute (31) into (32):

$$\begin{aligned} m\left( \dfrac{\alpha _{t+1}E_{t+1}^{R}}{\theta _{t+1}^{R}}-\dfrac{\alpha _{t}}{\theta _{t+1}^{R}}\right) +n\left( \dfrac{\alpha _{t+1}E_{t+1}^{G}}{\theta _{t+1}^{G}}-\dfrac{\alpha _{t}}{\theta _{t+1}^{G}}\right) =0 \end{aligned}$$

(33)

Define ${\overline{E}}_{t+1}\equiv \left( \dfrac{m}{\theta _{t+1}^{R}}+\dfrac{n}{\theta _{t+1}^{G}}\right) ^{-1}\left( \dfrac{mE_{t+1}^{R}}{\theta _{t+1}^{R}}+\dfrac{nE_{t+1}^{G}}{\theta _{t+1}^{G}}\right) ,$ we have $\alpha _t=\alpha _{t+1}{\overline{E}}_{t+1}$ and ${\overline{E}}_{t+1}$ is a $4\times 4$ matrix. Therefore, as long as we know the $\alpha _{T-1}$, we can solve the linear equilibrium price and the parameter matrices $A^i_{Q,t}$ and $B_{Q,t}^i$ at any date t recursively.

The last period problem is fairly easy to be solved. It is equivalent to the last period solution as in a 3 period model presented in Brunnermeier [10] text book. The last period $U_{T-1}^i$ is solved directly from the last period Bellman equation. The last period $\alpha _{T-1}$ is shown in equation 20 while the exact format of $U_{T-1}$ is given as below:

$$\begin{aligned} U_{T-1}^{i}=\left( \begin{array}{cccc} 0 &{} 0 &{} 0 &{} 0\\ 0 &{} u_{T-1}^{i} &{} -u_{T-1}^{i} &{} 0\\ 0 &{} -u_{T-1}^{i} &{} u_{T-1}^{i} &{} 0\\ 0 &{} 0 &{} 0 &{} 0 \end{array}\right) \end{aligned}$$

where

$$\begin{aligned} u_{T-1}^{R}=\dfrac{n^{2}\varSigma _{V,T|T-1}^{R}}{(m\varSigma _{V,T|T-1}^{G}+n\varSigma _{V,T|T-1}^{R})^{2}} ,\quad u_{T-1}^{G}=\dfrac{m^{2}\varSigma _{V,T|T-1}^{G}}{(m\varSigma _{V,T|T-1}^{G}+n\varSigma _{V,T|T-1}^{R})^{2}} \end{aligned}$$

A 3-period example of this proof is given in the Appendix B to demonstrate the recursive process.

Proof of Proposition 3: (Rational Benchmark)

In a single type market, at time t, because from the perspective of that type, the current equilibrium price must fully reflects all the informations up to t and the future expected returns must be 0, we have the $i\in {\mathscr {I}}$ type’s optimal demand as follows:

$$\begin{aligned} X_{t}^{i}= & {} \dfrac{E_{i}\left( P_{i,t+1}|{\mathscr {F}}_{t}\right) -P^i_{t}}{\lambda Var_{i}(P_{i,t+1}|{\mathscr {F}}_{t})}\\= & {} \dfrac{V^i_{t+1|t}-P_{i,t}}{\lambda Var_{i}(P_{i,t+1}|{\mathscr {F}}_{t})}=0. \end{aligned}$$

Therefore, we have

$$\begin{aligned} P_{i,t} = V^i_{t+1|t} \end{aligned}$$

We proved the part 1 of Proposition 3 and Lemma 1.

Furthermore, the result that the equilibrium price $P_{R,t}$ is a martingale sequence comes directly from the nature of Kalman filtering because the $P_{R,t}$ takes the expectation of the liquidation value, which is a martingale sequence under the rational investor’s belief setting. We also prove that $E_{R}\left[ Q_{R,t+k}|{\mathscr {F}}_{t}\right] =0$ because the returns $Q_{R,t}$ is a martingale difference sequence.

Finally, we point out that $Corr_{R}\left( Q_{t_1}^{R},\; Q_{t_2}^{R}\right) =0$ comes directly from the property of a martingale difference sequence. We proved the Proposition 3$\square$

Proof of Proposition 5 and Proposition 6: (Expectations in the Mixed Market)

We only give the proof for Proposition 5. Results in Proposition 6 can be derived by expressing the state variables of the economy recursively as in the proof for Proposition 5.

$$\begin{aligned} E_{R}\left[ P_{t}|{\mathscr {F}}_{0}\right]= & {} E_{R}\left[ \alpha _{t}\varPsi _{t}^{R}|{\mathscr {F}}_{0}\right] \\= & {} E_{R}\left[ \alpha _{t}\left( A_{\varPsi ,t}^{R}\varPsi _{t-1}^{R}+B_{\varPsi ,t}^{R}v_{t}^{R}\right) |{\mathscr {F}}_{0}\right] \\= & {} E_{R}\left[ \alpha _{t}A_{\varPsi ,t}^{R}\varPsi _{t-1}^{R}|{\mathscr {F}}_{0}\right] \\= & {} \alpha _{t}\overset{t-1}{\underset{l=0}{\prod }}A_{\varPsi ,t-l}^{R}\varPsi _{0}.\\ E_{R}\left[ Q_{t}|{\mathscr {F}}_{0}\right]= & {} E_{R}\left[ A_{Q,t}^{R}\varPsi _{t}^{R}+B_{Q,t}^{R}v_{t}^{R}|{\mathscr {F}}_{0}\right] \\= & {} E_{R}\left[ A_{Q,t}^{R}\varPsi _{t}^{R}|{\mathscr {F}}_{0}\right] \\= & {} A_{Q,t}\overset{t-2}{\underset{l=0}{\prod }}A_{\varPsi ,t-1-l}^{R}\varPsi _{0}. \end{aligned}$$

We finished out proof for Proposition 5. $\square$

Proof of Proposition 7: (Market Dominance)

We present the proof for the last two periods. The proof of other periods are similar to the last period and can be seen from the backwardation.

As shown before in the proof of Proposition 2, we first derive that at $t=T-1$,

$$\begin{aligned} \alpha _{T-1}= & {} \left( 0,\;\dfrac{m\varSigma _{V,T|T-1}^{G}}{m\varSigma _{V,T|T-1}^{G}+n\varSigma _{V,T|T-1}^{R}},\;\dfrac{n\varSigma _{V,T|T-1}^{R}}{m\varSigma _{V,T|T-1}^{G}+n\varSigma _{V,T|T-1}^{R}},\;0\right) \\ U_{T-1}^{R}= & {} \left( \begin{array}{cccc} 0 &{} 0 &{} 0 &{} 0\\ 0 &{} u_{T-1}^{R} &{} -u_{T-1}^{R} &{} 0\\ 0 &{} -u_{T-1}^{R} &{} u_{T-1}^{R} &{} 0\\ 0 &{} 0 &{} 0 &{} 0 \end{array}\right) ,\; U_{T-1}^{G}=\left( \begin{array}{cccc} 0 &{} 0 &{} 0 &{} 0\\ 0 &{} u_{T-1}^{G} &{} -u_{T-1}^{G} &{} 0\\ 0 &{} -u_{T-1}^{G} &{} u_{T-1}^{G} &{} 0\\ 0 &{} 0 &{} 0 &{} 0 \end{array}\right) , \end{aligned}$$

where

$$\begin{aligned} u_{T-1}^{R}=\dfrac{1}{\lambda }\dfrac{n^{2}\varSigma _{V,T|T-1}^{R}}{\left( m\varSigma _{V,T|T-1}^{G}+n\varSigma _{V,T|T-1}^{R}\right) ^{2}},\; u_{T-1}^{G}=\dfrac{1}{\lambda }\dfrac{m^{2}\varSigma _{V,T|T-1}^{G}}{\left( m\varSigma _{V,T|T-1}^{G}+n\varSigma _{V,T|T-1}^{R}\right) ^{2}}. \end{aligned}$$

Define $b_{d}\equiv \dfrac{\beta }{\delta -\beta }$, we have

$$\begin{aligned} A_{\varPsi ,T-1}^{R}= & {} \left( \begin{array}{cccc} 1 &{} 0 &{} 0 &{} 0\\ 0 &{} 1 &{} 0 &{} 0\\ 0 &{} K_{T-1}^{G} &{} 1-K_{T-1}^{G} &{} b_{d}K_{T-1}^{G}\\ 0 &{} k_{T-1} &{} -k_{T-1} &{} \delta -\beta +b_{d}k_{T-1} \end{array}\right) , \\ A_{\varPsi ,T-1}^{G}= & {} \left( \begin{array}{cccc} 1 &{} 0 &{} 0 &{} 0\\ 0 &{} 1-K_{T-1}^{R} &{} K_{T-1}^{R} &{} -b_{d}K_{T-1}^{R}\\ 0 &{} 0 &{} 1 &{} 0\\ 0 &{} 0 &{} 0 &{} \delta -\beta \end{array}\right) , \\ B_{\varPsi ,T-1}^{R}= & {} B_{\varPsi ,T-1}^{G}=\left( \begin{array}{c} 0\\ K_{T-1}^{R}\\ K_{T-1}^{G}\\ k_{T-1} \end{array}\right) \end{aligned}$$

Furthermore, we have

$$\begin{aligned} \theta _{T-1}^{i}= & {} \left( \left( \varSigma _{v,T-1|T-2}^{i}\right) ^{-1}+B_{\varPsi ,t+1}^{i\mathbf {T}}U_{t+1}^{i}B_{\varPsi ,t+1}^{i}\right) ^{-1}\\= & {} \left( \left( \varSigma _{v,T-1|T-2}^{i}\right) ^{-1}+\left( K_{T-1}^{R}-K_{T-1}^{G}\right) ^{2}u_{T-1}^{i}\right) ^{-1};\\ E_{T-1}^{i}= & {} A_{\varPsi ,T-1}^{i}-B_{\varPsi ,T-1}^{i}B_{\varPsi ,T-1}^{i\mathbf {T}}\theta _{T-1}^{i}U_{T-1}^{i}A_{\varPsi ,T-1}^{i}\\= & {} \left( I_{4}-B_{\varPsi ,T-1}^{i}B_{\varPsi ,T-1}^{i\mathbf {T}}\theta _{T-1}^{i}U_{T-1}^{i}\right) A_{\varPsi ,T-1}^{i} \end{aligned}$$

For simplicity, we omit the subscription $T-1$ and we have

$$\begin{aligned} E^{R}= & {} \left( \begin{array}{cccc} 1 &{} 0 &{} 0 &{} 0\\ 0 &{} 1-L^{R}K^{R}+L^{R}K^{R}K^{G} &{} L^{R}K^{R}-L^{R}K^{R}K^{G} &{} b_{d}L^{R}K^{R}K^{G}\\ 0 &{} -L^{R}K^{G}+K^{G}+L^{R}\left( K^{G}\right) ^{2} &{} 1+L^{R}K^{G}-K^{G}-L^{R}\left( K^{G}\right) ^{2} &{} b_{d}K^{G}+b_{d}L^{R}\left( K^{G}\right) ^{2}\\ 0 &{} -L^{R}k+L^{R}K^{G}k+k &{} L^{R}k-L^{R}K^{G}k-k &{} b_{d}L^{R}K^{G}k+\delta -\beta +b_{d}k \end{array}\right) \\ E^{G}= & {} \left( \begin{array}{cccc} 1 &{} 0 &{} 0 &{} 0\\ 0 &{} 1-K^{R}-L^{G}K^{R}+L^{G}\left( K^{R}\right) ^{2} &{} K^{R}-L^{G}\left( K^{R}\right) ^{2}+L^{G}K^{R} &{} -b_{d}K^{R}+b_{d}L^{G}\left( K^{R}\right) ^{2}\\ 0 &{} -L^{G}K^{G}+L^{G}K^{R}K^{G} &{} 1-L^{G}K^{R}K^{G}+L^{G}K^{G} &{} -b_{d}L^{G}K^{R}K^{G}\\ 0 &{} -L^{G}k+L^{G}K^{R}k &{} -L^{G}K^{R}k+L^{G}k &{} \delta -\beta +b_{d}L^{G}K^{R}k \end{array}\right) \end{aligned}$$

where $L^{i}=(K^{R}-K^{G})u^{i}\theta ^{i}$. We have ${\overline{E}}=\dfrac{\frac{m}{\theta ^{R}}E^{R}+\frac{n}{\theta ^{G}}E^{G}}{\frac{m}{\theta ^{R}}+\frac{n}{\theta ^{G}}}$.

We further define ${\bar{E}}\equiv \frac{m}{\theta ^{R}}E^{R}+\frac{n}{\theta ^{G}}E^{G}$ as the numerator of ${\overline{E}}$. Define $D^{R}\equiv \dfrac{L^{R}}{\theta ^{R}}=\left( K^{R}-K^{G}\right) u^{R}$, $D^{G}\equiv \dfrac{L^{G}}{\theta ^{G}}=\left( K^{R}-K^{G}\right) u^{G}$. Define the j’th column of ${\bar{E}}$ as ${\bar{E}}_{(:,j)}.$ We have ${\bar{E}}_{(:,1)}$ as below:

$$\begin{aligned} {\bar{E}}_{(:,1)}= & {} \left( \begin{array}{c} \frac{m}{\theta ^{R}}+\frac{n}{\theta ^{G}}\\ 0\\ 0\\ 0 \end{array}\right) ,\\ {\bar{E}}_{(:,2)}= & {} \left( \begin{array}{c} 0\\ \dfrac{m}{\theta ^{R}}-mD^{R}K^{R}+mD^{R}K^{R}K^{G}+\dfrac{n}{\theta ^{G}}-\dfrac{n}{\theta ^{G}}K^{R}-nD^{G}K^{R}+nD^{G}\left( K^{R}\right) ^{2}\\ -mD^{R}K^{G}+\dfrac{m}{\theta ^{R}}K^{G}+mD^{R}\left( K^{G}\right) ^{2}-nD^{G}K^{G}+nD^{G}K^{R}K^{G}\\ -mD^{R}k+mD^{R}K^{G}k+\dfrac{m}{\theta ^{R}}k-nD^{G}k+nD^{G}K^{R}k \end{array}\right) ,\\ {\bar{E}}_{(:,3)}= & {} \left( \begin{array}{c} 0\\ mD^{R}K^{R}-mD^{R}K^{R}K^{G}+\dfrac{n}{\theta ^{G}}K^{R}-nD^{G}\left( K^{R}\right) ^{2}+nD^{G}K^{R}\\ \dfrac{m}{\theta ^{R}}+mD^{R}K^{G}-\dfrac{m}{\theta ^{R}}K^{G}-mD^{R}\left( K^{G}\right) ^{2}-nD^{G}K^{R}K^{G}+\dfrac{n}{\theta ^{G}}+nD^{G}K^{G}\\ mD^{R}k-mD^{R}K^{G}k-\dfrac{m}{\theta ^{R}}k-nD^{G}K^{R}k+nD^{G}k \end{array}\right) ,\\ {\bar{E}}_{(:,4)}= & {} \left( \begin{array}{c} 0\\ b_{d}mD^{R}K^{R}K^{G}-\dfrac{nb_{d}K^{R}}{\theta ^{G}}+nb_{d}D^{G}\left( K^{R}\right) ^{2}\\ b_{d}\dfrac{m}{\theta ^{R}}K^{G}+b_{d}mD^{R}\left( K^{G}\right) ^{2}-nb_{d}D^{G}K^{R}K^{G}\\ b_{d}mD^{R}K^{G}k+\dfrac{m(\delta -\beta )}{\theta ^{R}}+\dfrac{mb_{d}}{\theta ^{R}}k+\dfrac{n(\delta -\beta )}{\theta ^{G}}+nb_{d}D^{G}K^{R}k \end{array}\right) . \end{aligned}$$

Since $\alpha _{t}=\alpha _{t+1}{\overline{E}}_{t+1}$ where ${\overline{E}}_{t+1}\equiv \dfrac{{\bar{E}}_{t+1}}{\frac{m}{\theta ^{R}}+\frac{n}{\theta ^{G}}}$, we have

$$\begin{aligned} \alpha _{T-2} = \alpha _{T-1}{\overline{E}}_{T-1}. \end{aligned}$$

Elements in $\alpha _{T-2}$ are the follows:

$$\begin{aligned}&\alpha _{0,T-2}=0;\nonumber \\&\quad \alpha _{1,T-2}\left( \frac{m}{\theta ^{R}}+\frac{n}{\theta ^{G}}\right) =\alpha _{T-1}{\bar{E}}_{(:,2)}\nonumber \\&\quad ={\alpha _{T-1} \cdot \left[ \begin{array}{c} 0\\ \dfrac{m}{\theta ^{R}}-mD^{R}K^{R}+mD^{R}K^{R}K^{G}+\dfrac{n}{\theta ^{G}}-\dfrac{n}{\theta ^{G}}K^{R}-nD^{G}K^{R}+nD^{G}\left( K^{R}\right) ^{2}\\ -mD^{R}K^{G}+\dfrac{m}{\theta ^{R}}K^{G}+mD^{R}\left( K^{G}\right) ^{2}-nD^{G}K^{G}+nD^{G}K^{R}K^{G}\\ -mD^{R}k+mD^{R}K^{G}k+\dfrac{m}{\theta ^{R}}k-nD^{G}k+nD^{G}K^{R}k \end{array}\right] } \end{aligned}$$

(34)

$$\begin{aligned}&\alpha _{2,T-2}\left( \frac{m}{\theta ^{R}}+\frac{n}{\theta ^{G}}\right) =\alpha _{T-1}{\bar{E}}_{(:,3)}\nonumber \\&\quad ={\alpha _{T-1} \cdot \left[ \begin{array}{c} 0\\ mD^{R}K^{R}-mD^{R}K^{R}K^{G}+\dfrac{n}{\theta ^{G}}K^{R}-nD^{G}\left( K^{R}\right) ^{2}+nD^{G}K^{R}\\ \dfrac{m}{\theta ^{R}}+mD^{R}K^{G}-\dfrac{m}{\theta ^{R}}K^{G}-mD^{R}\left( K^{G}\right) ^{2}-nD^{G}K^{R}K^{G}+\dfrac{n}{\theta ^{G}}+nD^{G}K^{G}\\ mD^{R}k-mD^{R}K^{G}k-\dfrac{m}{\theta ^{R}}k-nD^{G}K^{R}k+nD^{G}k \end{array}\right] }\nonumber \\ \end{aligned}$$

(35)

In all the elements above, on both left and right hand sides of equations, the variables without the time t notation are at $T-1$.

We first prove the part 1 in the proposition, that is $\alpha _{0,t}=0$ for all t.

Since $\alpha _{0,t} = \alpha _{t+1} {\bar{E}}_{(:,1)}$ and the ${\bar{E}}_{(:,1)}$ has the general form shown above, with $\alpha _{0,t+1}=0$, we have $\alpha _{0,t} = 0$.

Next, we prove part 2 of the proposition, that is, $\alpha _{1,t}+\alpha _{2,t} = 1$.

To do so, we sum the left and right hand sides of Eqs. 34 and 35 , and we have the followings,

$$\begin{aligned}&\alpha _{1,T-2}\left( \frac{m}{\theta ^{R}}+\frac{n}{\theta ^{G}}\right) +\alpha _{2,T-2}\left( \frac{m}{\theta ^{R}}+\frac{n}{\theta ^{G}}\right) \\&\quad =\alpha _{T-1}\cdot \left( \left[ \begin{array}{c} 0\\ \dfrac{m}{\theta ^{R}}-mD^{R}K^{R}+mD^{R}K^{R}K^{G}+\dfrac{n}{\theta ^{G}}-\dfrac{n}{\theta ^{G}}K^{R}-nD^{G}K^{R}+nD^{G}\left( K^{R}\right) ^{2}\\ -mD^{R}K^{G}+\dfrac{m}{\theta ^{R}}K^{G}+mD^{R}\left( K^{G}\right) ^{2}-nD^{G}K^{G}+nD^{G}K^{R}K^{G}\\ -mD^{R}k+mD^{R}K^{G}k+\dfrac{m}{\theta ^{R}}k-nD^{G}k+nD^{G}K^{R}k \end{array}\right] \right. \\&\left. \qquad +\left[ \begin{array}{c} 0\\ mD^{R}K^{R}-mD^{R}K^{R}K^{G}+\dfrac{n}{\theta ^{G}}K^{R}-nD^{G}\left( K^{R}\right) ^{2}+nD^{G}K^{R}\\ \dfrac{m}{\theta ^{R}}+mD^{R}K^{G}-\dfrac{m}{\theta ^{R}}K^{G}-mD^{R}\left( K^{G}\right) ^{2}-nD^{G}K^{R}K^{G}+\dfrac{n}{\theta ^{G}}+nD^{G}K^{G}\\ mD^{R}k-mD^{R}K^{G}k-\dfrac{m}{\theta ^{R}}k-nD^{G}K^{R}k+nD^{G}k \end{array}\right] \right) \\&\quad = \left( \alpha _{T-1,1}+\alpha _{T-1,2}\right) \left( \frac{m}{\theta ^{R}}+\frac{n}{\theta ^{G}}\right) \end{aligned}$$

The last step holds as the items in the two vectors cancel out each other.

Furthermore, on both sides of the above equation, the omitted notation t of $\frac{m}{\theta ^{R}}+\frac{n}{\theta ^{G}}$ are at $T-1$, and since they are scalars, we can cancel them out from both sides. Therefore, we have $\alpha _{T-2,1}+\alpha _{T-2,2}=\alpha _{T-1,1}+\alpha _{T-1,2}=1$

Since the proof above does not depend on t, the result can be generalised to any time t, that is $\alpha _{t,1}+\alpha _{t,2}=1$. We complete the proof of Proposition 7. $\square$

Proof of Proposition 9 and Lemma 2: (Volatility)

$$\begin{aligned} \varPsi _{t}^{R}= & {} A_{\varPsi ,t}^{R}\varPsi _{t-1}^{R}+B_{\varPsi ,t}^{R}\left( y_{t}-V_{t|t-1}^{R}\right) \nonumber \\= & {} G_{t}^{R}\varPsi _{t-1}^{R}+B_{\varPsi ,t}^{R}y_{t}\nonumber \\= & {} \overset{t}{\underset{j=1}{\varPi }}G_{j}^{R}\varPsi _{0}+\overset{t-1}{\underset{n=1}{\sum }}\left( \overset{t}{\underset{l=n+1}{\varPi }}G_{l}\right) B_{\varPsi ,n}^{R}y_{n}+B_{\varPsi ,t}^{R}y_{t}, \end{aligned}$$

(36)

with $G_{t}^{R}=A_{\varPsi ,t}^{R}-\left[ \begin{array}{cccc} 0 &{} 1 &{} 0 &{} 0\\ 0 &{} K_{t}^{R} &{} 0 &{} 0\\ 0 &{} K_{t}^{G} &{} 0 &{} 0\\ 0 &{} k_{t} &{} 0 &{} 0 \end{array}\right]$ for simplicity. Furthermore, we have

$$\begin{aligned} y_{n}=V+\xi _{n} \end{aligned}$$

(37)

Substitute equation  (37) into equation  (36) and reformat the equation we have

$$\begin{aligned} \varPsi _{t}^{R}=\overset{t}{\underset{j=1}{\varPi }}G_{j}^{R}\varPsi _{0}+\overset{t-1}{\underset{n=1}{\sum }}\left( \overset{t}{\underset{l=n+1}{\varPi }}G_{l}\right) B_{\varPsi ,n}^{R}V+\overset{t-1}{\underset{n=1}{\sum }}\left[ \left( \overset{t}{\underset{l=n+1}{\varPi }}G_{l}\right) B_{\varPsi ,n}^{R}\xi _{n}\right] . \end{aligned}$$

Therefore, we have

$$\begin{aligned} Var\left( \varPsi _{t}^{R}\right) =\left( \overset{t-1}{\underset{n=1}{\sum }}D_{n}\right) \sigma _{V}^{2}\left( \overset{t-1}{\underset{n=1}{\sum }}D_{n}\right) ^{\mathbf {T}}+\overset{t-1}{\underset{n=1}{\sum }}\left( D_{n}D_{n}^{\mathbf {T}}\right) \sigma _{\xi }^{2} \end{aligned}$$

where $D_{n}=\left( \overset{t-1}{\underset{l=n+1}{\varPi }}G_{n}\right) B_{\varPsi ,n}^{R}$.

Expand the last equation, we can easily derive the conclusion in the Proposition.

To derive the result in Lemma 2, write

$$\begin{aligned} \begin{array}{lcl} Var(P_{t+1}|{\mathscr {F}}_{t})&{}=&{}\alpha _{t+1}Var(\varPsi _{t+1}^{R}|{\mathscr {F}}_{t})\alpha _{t+1}^{\mathbf {T}}\\ &{}=&{}\alpha _{t+1}Var(A_{\varPsi ,t+1}^{R}\varPsi _{t}^{R}+B_{\varPsi ,t+1}^{R}v_{t+1}^{R}|{\mathscr {F}}_{t})\alpha _{t+1}^{\mathbf {T}}\\ &{}=&{}\left( \alpha _{1,t+1}K_{t+1}^{R}+\alpha _{2,t+1}K_{t+1}^{G}+\alpha _{3,t+1}k_{t+1}\right) ^{2}Var_{R}(v_{t+1}|{\mathscr {F}}_{t}). \end{array} \end{aligned}$$

We complete the proof for Lemma 2. $\square$

Proof of Proposition 8: (Autocorrelation)

Expand $Q_i$ recursively and use the results in Proposition 3 on the $Cov^R(v_i,\;v_j)$ we have

$$\begin{aligned} \begin{array}{lcl} Cov^{R}\left( Q_{i},\; Q_{j}\right) &{}=&{}A_{Q,i}^{R}Cov\left( \varPsi _{i},\ \varPsi _{j}\right) A_{Q,j}^{R\mathbf {T}}+A_{Q,i}E_{R}\left( \varPsi _{i}\cdot v_{j}^{\mathbf {T}}\right) B_{Q,j}^{\mathbf {T}}\\ &{}=&{} A_{Q,i}^{R}\overset{i-j-1}{\underset{k=0}{\prod }}A_{\varPsi ,i-k}^{R}\left[ Var\left( \varPsi _{j}\right) A_{Q,j}^{R\mathbf {T}}+B_{\varPsi ,j}^{R}Var\left( v_{j}\right) B_{Q,j}^{R\mathbf {T}}\right] \end{array} \end{aligned}$$

where the format of the $Var^{R}\left( \varPsi _{t}\right)$ is shown in the proof of Proposition 9.

Mathematical Appendix B

A 3-period Proof of the Equilibrium:

Consider a simple 3-period problem where the traders’ expected utility function is given by:

$$\begin{aligned} E[-exp(-\lambda W_3^i|{\mathscr {F}}_t)]. \end{aligned}$$

Using backward induction, at period $t=2$, the optimization problem is

$$\begin{aligned}&\underset{X_{2}^{i}}{Max}E_{i,2}\left[ -exp\left\{ -\lambda W_{2}^{i}-X_{2}^{i}(V-P_{2})\right\} |{\mathscr {F}}_{2}\right] \\&\quad =\underset{X^i_{2}}{Max}-exp\left\{ -\lambda W_{2}^{i}-\lambda X_{2}^{i}\left( V_{3|2}^{i}-P_{2}\right) +\dfrac{1}{2}\lambda ^{2}X_{2}^{i2}\varSigma _{V,3|2}^{i}\right\} \end{aligned}$$

From the first-order condition, we have $X_{2}^{i}=\dfrac{V_{3|2}^{i}-P_{2}}{\lambda \varSigma _{V,3|2}^{i}}$. Together with $P_{2}=\alpha _{2}\varPsi _{2}$, we have $X_{2}^{i}=\dfrac{(I_{1}-\alpha _{2})\varPsi _{2}}{\lambda \varSigma _{V,3|2}^{i}}$, where $I_{1}=[0,1,0,0]$, and $F_{2}^{i}=\dfrac{I_{1}-\alpha _{2}}{\varSigma _{V,3|2}^{i}}$.

Plug $X_{2}^{i}=\dfrac{(I_{1}-\alpha _{2})\varPsi _{2}}{\lambda \varSigma _{V,3|2}^{i}}$ into the original utility function, we can derive the value function at $t=2$:

$$\begin{aligned} J(W_{2};X_{2};2)= & {} -exp\left\{ -\lambda W_{2}^{i}-\dfrac{1}{2}\dfrac{\left( V_{3|2}^{i}-P_{2}\right) ^{2}}{\varSigma _{V,3|2}^{i}}\right\} \\= & {} -exp\left\{ -\lambda W_{2}^{i}-\dfrac{1}{2}\dfrac{\varPsi _{2}^{\mathbf {T}}(I_{1}-\alpha _{2})^{\mathbf {T}}(I_{1}-\alpha _{2})\varPsi _{2}}{\varSigma _{V,3|2}^{i}}\right\} \\= & {} -exp\left\{ -\lambda W_{2}^{i}-\dfrac{1}{2}\varPsi _{2}^{\mathbf {T}}U_{2}^{i}\varPsi _{2}\right\} \end{aligned}$$

where $U_{2}^{i}=\dfrac{(I_{1}-\alpha _{2})^{\mathbf {T}}(I_{1}-\alpha _{2})}{\varSigma _{V,3|2}^{i}}$.

Use the market-clearing condition: $mX_{2}^{R}+nX_{2}^{G}=0$, we have

$$\begin{aligned} m\dfrac{V_{3|2}^{R}-P_{2}}{\lambda \varSigma _{V,3|2}^{R}}+n\dfrac{V_{3|2}^{G}-P_{2}}{\lambda \varSigma _{V,3|2}^{G}}=0, \end{aligned}$$

thus

$$\begin{aligned} P_{2}=\dfrac{mV_{3|2}^{R}\varSigma _{V,3|2}^{G}+nV_{3|2}^{G}\varSigma _{V,3|2}^{R}}{m\varSigma _{V,3|2}^{G}+n\varSigma _{V,3|2}^{R}} \end{aligned}$$

Therefore, $\alpha _{2}=[0,\dfrac{m\varSigma _{V,3|2}^{G}}{m\varSigma _{V,3|2}^{G}+n\varSigma _{V,3|2}^{R}},\dfrac{n\varSigma _{V,3|2}^{R}}{m\varSigma _{V,3|2}^{G}+n\varSigma _{V,3|2}^{R}},0]$. Plug it into $U_{2}^{i}$, we have

$$\begin{aligned} \begin{array}{lcl} U_{2}^{R}&{}=&{}\left[ \begin{array}{c} 0\\ \dfrac{n\varSigma _{V,3|2}^{R}}{m\varSigma _{V,3|2}^{G}+n\varSigma _{V,3|2}^{R}}\\ \dfrac{-n\varSigma _{V,3|2}^{R}}{m\varSigma _{V,3|2}^{G}+n\varSigma _{V,3|2}^{R}}\\ 0 \end{array}\right] \cdot [0,\dfrac{n\varSigma _{V,3|2}^{R}}{m\varSigma _{V,3|2}^{G}+n\varSigma _{V,3|2}^{R}},\dfrac{-n\varSigma _{V,3|2}^{R}}{m\varSigma _{V,3|2}^{G}+n\varSigma _{V,3|2}^{R}},0]/\varSigma _{V,3|2}^{R}\\ &{}=&{}\left[ \begin{array}{cccc} 0 &{} 0 &{} 0 &{} 0\\ 0 &{} \left( \dfrac{n}{m\varSigma _{V,3|2}^{G}+n\varSigma _{V,3|2}^{R}}\right) ^{2}\varSigma _{V,3|2}^{R} &{} -\left( \dfrac{n}{m\varSigma _{V,3|2}^{G}+n\varSigma _{V,3|2}^{R}}\right) ^{2}\varSigma _{V,3|2}^{R} &{} 0\\ 0 &{} -\left( \dfrac{n}{m\varSigma _{V,3|2}^{G}+n\varSigma _{V,3|2}^{R}}\right) ^{2}\varSigma _{V,3|2}^{R} &{} \left( \dfrac{n}{m\varSigma _{V,3|2}^{G}+n\varSigma _{V,3|2}^{R}}\right) ^{2}\varSigma _{V,3|2}^{R} &{} 0\\ 0 &{} 0 &{} 0 &{} 0 \end{array}\right] \end{array} \end{aligned}$$

This is exactly what we have in the general problem. Similar for $U_{2}^{G}$.

Use backward induction, the problem we want to solve at $t=1$ is

$$\begin{aligned} \underset{X_{1}^{i}}{Max}E_{i}\left[ J(W_{2};X_{2};2)|{\mathscr {F}}_{1}\right]= & {} \underset{X_{1}^{i}}{Max}E_{i}\left[ -exp\left\{ -\lambda W_{2}^{i}-\dfrac{1}{2}\varPsi _{2}^{\mathbf {T}}U_{2}^{i}\varPsi _{2}\right\} |{\mathscr {F}}_{1}\right] \\= & {} \underset{X_{1}^{i}}{Max}E_{i}\left[ -exp\left\{ -\lambda X_{1}^{i}Q_{2}-\dfrac{1}{2}\varPsi _{2}^{\mathbf {T}}U_{2}^{i}\varPsi _{2}\right\} |{\mathscr {F}}_{1}\right] \end{aligned}$$

In a linear equilibrium, we have

$Q_{2}^{i}=A_{Q,2}^{i}\varPsi _{1}+B_{Q,2}^{i}v_{2}^{i}$, and $\varPsi _{2}^{i}=A_{\varPsi ,2}^{i}\varPsi _{1}+B_{\varPsi ,2}^{i}v_{2}^{i}$. Plug these two processes into the optimization problem, the expectation can be removed and replaced by a linear function of conditional expectations and variances. That is equation  (18).

$$\begin{aligned}&\underset{X_{1}^{i}}{Max}E_{i}\left[ J(W_{2};X_{2};2)|{\mathscr {F}}_{1}\right] \\&\quad =\underset{X_{1}^{i}}{Max}-\rho _{2}exp\left\{ -\lambda W_{1}^{i}-\frac{1}{2}\varPsi _{1}^{i\mathbf {T}}A_{\varPsi ,2}^{i\mathbf {T}}U_{2}^{i}A_{\varPsi ,2}^{i}\varPsi _{1}^{i}\right\} \times exp\left\{ -\lambda X_{1}^{i}A_{Q,2}^{i}\varPsi _{1}^{i}+\frac{1}{2}C_{2}^{i\mathbf {T}}\theta _{2}^{i}C_{2}^{i}\right\} \end{aligned}$$

where $C_{2}^{i}$, $\theta _{2}^{i}$ and $\rho _{2}$ are simply functions in $U_{2}$ as expressed in the equation  (18).

The first-order condition of the maximization problem gives that

$$\begin{aligned} X_{1}^{i}=\dfrac{1}{\lambda }F_{1}^{i}\varPsi _{1}\hbox {, where }F_{1}^{i}\equiv \left( B_{Q,2}^{i}\theta _{2}^{i}B_{Q,2}^{i\mathbf {T}}\right) ^{-1}\left( A_{Q,2}^{i}-B_{Q,2}^{i}\theta _{2}^{i}B_{\varPsi ,2}^{i\mathbf {T}}U_{2}^{i}A_{\varPsi ,2}^{i}\right) . \end{aligned}$$

Substitute $X_{1}^{i}=\dfrac{1}{\lambda }F_{1}^{i}\varPsi _{1}$ into

$$\begin{aligned} -\rho _{2}exp\left\{ -\lambda W_{1}^{i}-\frac{1}{2}\varPsi _{1}^{i\mathbf {T}}A_{\varPsi ,2}^{i\mathbf {T}}U_{2}^{i}A_{\varPsi ,2}^{i}\varPsi _{1}^{i}\right\} \times exp\left\{ -\lambda X_{1}^{i}A_{Q,2}^{i}\varPsi _{1}^{i}+\frac{1}{2}C_{2}^{i\mathbf {T}}\theta _{2}^{i}C_{2}^{i}\right\} , \end{aligned}$$

we have

$$\begin{aligned} J(W_{1};X_{1};1)=-\rho _{2}exp\left\{ -\lambda W_{1}^{i}-\frac{1}{2}\varPsi _{1}^{i\mathbf {T}}A_{\varPsi ,2}^{i\mathbf {T}}U_{2}^{i}A_{\varPsi ,2}^{i}\varPsi _{1}^{i}-F_{1}^{i}\varPsi _{1}^{i}A_{Q,2}^{i}\varPsi _{1}^{i}+\frac{1}{2}C_{2}^{i\mathbf {T}}\theta _{2}^{i}C_{2}^{i}\right\} . \end{aligned}$$

By the assumption that $J(W_{1};X_{1};1)=-exp\{-\lambda W_{1}^{i}-\dfrac{1}{2}\varPsi _{1}^{\mathbf {T}}U_{1}^{i}\varPsi _{1}\}$, we let

$$\begin{aligned}&-exp\{-\lambda W_{1}^{i}-\dfrac{1}{2}\varPsi _{1}^{\mathbf {T}}U_{1}^{i}\varPsi _{1}\}\\&\quad =-\rho _{2}exp\left\{ -\lambda W_{1}^{i}-\frac{1}{2}\varPsi _{1}^{i\mathbf {T}}A_{\varPsi ,2}^{i\mathbf {T}}U_{2}^{i}A_{\varPsi ,2}^{i}\varPsi _{1}^{i}-F_{1}^{i}\varPsi _{1}^{i}A_{Q,2}^{i}\varPsi _{1}^{i}+\frac{1}{2}C_{2}^{i\mathbf {T}}\theta _{2}^{i}C_{2}^{i}\right\} \end{aligned}$$

which give us that $U_{1}^{i}=-2ln\rho _{2}^{i}I_{11}+M_{1}^{i}$, where $M_{1}^{i}$ is a function of $U_{2}^{i}$ and is known. The expression of $M_{1}^{i}$ is given in the general proof.

Similar as in the process of deriving $\alpha _{2}$, the only undetermined parameter at this stage is $A_{Q,2}^{i}$, which contains $\alpha _{1}$. $U_{1}^{i}$ is still a function of $\alpha _{1}$. To derive the $\alpha _{1}$, we further apply the market-clearing condition: $mX_{1}^{R}+nX_{1}^{G}=0$.

Substitute $X_{1}^{i}=\dfrac{1}{\lambda }F_{1}^{i}\varPsi _{1}$ into the market-clearing condition, we have

$$\begin{aligned} \left( mF_{1}^{R}+nF_{1}^{G}\right) \varPsi _{1}=0. \end{aligned}$$

Since it holds for any value of $\varPsi _{1}$, we have

$$\begin{aligned} mF_{1}^{R}+nF_{1}^{G}=0_{1\times 4}. \end{aligned}$$

Substitute $A_{Q,2}^{i}=\alpha _{2}A_{\varPsi ,2}^{i}-\alpha _{1}$ and $B_{Q,2}^{i}=\alpha _{2}K_{2}$ into

$$\begin{aligned} F_{1}^{i}\equiv \left( B_{Q,2}^{i}\theta _{2}^{i}B_{Q,2}^{i\mathbf {T}}\right) ^{-1}\left( A_{Q,2}^{i}-B_{Q,2}^{i}\theta _{2}^{i}B_{\varPsi ,2}^{i\mathbf {T}}U_{2}^{i}A_{\varPsi ,2}^{i}\right) , \end{aligned}$$

we have $F_{1}^{i}=\dfrac{1}{\left( \alpha _{2}K_{2}\right) ^{2}\theta _{2}^{i}}\alpha _{2}E_{2}^{i}-\dfrac{1}{\left( \alpha _{2}K_{2}\right) ^{2}\theta _{2}^{i}}\alpha _{1}$, where $E_{2}^{i}=A_{\varPsi ,2}^{i}-K_{2}B_{\varPsi ,2}^{i\mathbf {T}}\theta _{2}^{i}U_{2}^{i}A_{\varPsi ,2}^{i}$ and is known. Plug the formula into the market-clearing condition, we have

$$\begin{aligned} m\left( \dfrac{\alpha _{2}E_{2}^{R}}{\theta _{2}^{R}}-\dfrac{\alpha _{1}}{\theta _{2}^{R}}\right) +n\left( \dfrac{\alpha _{2}E_{2}^{G}}{\theta _{2}^{G}}-\dfrac{\alpha _{1}}{\theta _{2}^{G}}\right) =0, \end{aligned}$$

which gives us the expression of $\alpha _{1}$ as a function of $\alpha _{2}$, that is

$$\begin{aligned} \alpha _{1}=\alpha _{2}{\overline{E}}_{2}, \end{aligned}$$

where ${\overline{E}}_{2}\equiv \left( \dfrac{m}{\theta _{2}^{R}}+\dfrac{n}{\theta _{2}^{G}}\right) ^{-1}\left( \dfrac{mE_{2}^{R}}{\theta _{2}^{R}}+\dfrac{nE_{2}^{G}}{\theta _{2}^{G}}\right)$ and is known. $\square$

Appendix C: Non-zero net supply

Fig. 13

Information shock on expected price. This figure plots the rational expectation in price after a signal shock at $t=10$. Parameters are set at the following values: $m=0.5$, $T=50$, $\lambda =1$, $\delta =0.8$, $\beta =0.6$, $\sigma _\xi ^2=1$, $\sigma _V^2=1$

We assume $\varTheta =0$ for simplicity because a positive $\varTheta$ together with the learning process generates non-zero correlations in prices, making the analysis of a signal shock on the prices and returns less clear.

By changing the net supply of the stock $\varTheta$ into positive, the equilibrium price now contains a constant term, that is $\alpha _{0,t}\ne 0$.

It can be shown that $\alpha _{1,t}$, $\alpha _{2,t}$ and $\alpha _{3,t}$ still evolve as in the main paragraph while $\alpha _{0,t}$ can be derived as below:

$$\begin{aligned} \alpha _{0,t}=\alpha _{0,t+1}{\overline{E}}_{(:,1)}-\dfrac{\varTheta }{\frac{m}{\theta _{t+1}^{R}}+\frac{n}{\theta _{t+1}^{G}}} \end{aligned}$$

where ${\overline{E}}_{(:,1)}$ is the first column of matrix ${\overline{E}}$. Consistent with previous studies, $\alpha _{0,t}$ is negative, reflecting the risk-averse attitude. Simulation shows that $\alpha _{0,t}$ increases over time because more information reduces the risk in variance, thus strengthening demands and boosting the asset price.

Another change lies in the optimal demands of two types from the simulation results. Simulations show that $F^R_t$ and $F^G_t$ now also contain a constant term. That is

$$\begin{aligned} X_{t}^{i}=\dfrac{1}{\lambda }F_{1,t}^{i}+\dfrac{1}{\lambda }F_{d,t}^{i}\left( V_{t+1|t}^{R}-V_{t+1|t}^{G}\right) +\dfrac{1}{\lambda }F_{\epsilon ,t}^{i}\epsilon ^{G}_{t+1|t} \end{aligned}$$

The effect of $F_{1,t}^i$ is independent of $F_{d,t}$ and $F_{\epsilon ,t}$.

Signs of the constant term $F_{1,t}^i$ reflect that the rational investors turns from long to short positive. The constant term reflect the risk attitudes of the investors, indicating that the rational investors are comparatively more risk-averse in the market when the liquidation date is close. This result is consistent with the analysis in Proposition 7.

With a positive net supply, even the price in the pure rational market increases as the risk in variance decreases over time. Here we provide the simulation results for the prices and returns following a signal shock. Results are similar to what we have before except the price has a general increasing trend as shown in Fig. 13. Other simulation results are available on request.