Strategic Exploitation of a Common-Property Resource Under Rational Learning About its Reproduction

Abstract

We build a workable game of common-property resource extraction under rational Bayesian learning about the reproduction prospects of a resource. We focus on Markov-perfect strategies under truthful revelation of beliefs. For reasonable initial conditions, exogenously shifting the prior beliefs of one player toward more pessimism about the potential of natural resources to reproduce can create anti-conservation incentives. The single player whose beliefs have been shifted toward more pessimism exhibits higher exploitation rate than before. In response, all other players reduce their exploitation rates in order to conserve the resource. However, the overall conservation incentive is weak, making the aggregate exploitation rate higher than before the pessimistic shift in beliefs of that single player. Due to this weakness in strategic conservation responses, if the number of players is relatively small, then in cases with common priors, jointly shifting all players’ beliefs toward more pessimism exacerbates the commons problem.

Appendix: Proofs

The proof of Proposition 1 relies on Lemmata 1 through 3, which is separate results. We state and prove Lemmata 1 and 3 below, while the proof of Lemma 2 appears in the literature. So, we state Lemma 2 and provide an appropriate citation for its proof.

Lemma 1

For any function $h:\mathbb {R}\rightarrow \mathbb {R}$ and any given prior distribution $\xi _{0}$ , for all $t\in \left\{ 1,2,\ldots \right\}$ and $\left\{ \xi _{s}\right\} _{s=0}^{t-1}$ generated through

$$\begin{aligned} \xi _{\tau +1}\left( \theta |\eta \right) =\frac{\phi \left( \eta |\theta \right) \xi _{\tau }\left( \theta \right) }{\int \limits _{\Theta }\phi \left( \eta |x\right) \xi _{\tau }\left( x\right) \text {d}x},\quad \tau =0,1,\ldots , \end{aligned}$$

(23)

from any sequence $\left\{ \eta _{s}\right\} _{s=0}^{t-1}$ of independent draws of the shock $\eta$, the conditional expectation

$$\begin{aligned}&E_{0}\left( \left. \underset{s=0}{\overset{t-1}{\prod }}h\left( \eta _{s}\right) \right| \xi _{0}\text {, update governed by (23)} \right) \equiv \\&\equiv \int \limits _{\mathcal {H}}\int \limits _{\Theta }\cdots \int \limits _{\mathcal {H} }\int \limits _{\Theta }\int \limits _{\mathcal {H}}\int \limits _{\Theta }\underset{s=0}{\overset{t-1}{ \prod }}h\left( \eta _{s}\right) \phi \left( \eta _{t-1}|\theta _{t-1}\right) \xi _{t-1}\left( \theta _{t-1}\right) \text {d}\theta _{t-1} \text {d}\eta _{t-1}\\ \quad \quad \quad&\times \phi \left( \eta _{t-2}|\theta _{t-2}\right) \xi _{t-2}\left( \theta _{t-2}\right) \text {d}\theta _{t-2}\text {d}\eta _{t-2}\times \cdots \times \phi \left( \eta _{0}|\theta _{0}\right) \xi _{0}\left( \theta _{0}\right) \text {d}\theta _{0}\text {d}\eta _{0} \end{aligned}$$

$$\begin{aligned} =\int \limits _{\Theta }\left[ \int \limits _{\mathcal {H}}h\left( \eta \right) \phi \left( \eta |\theta \right) \text {d}\eta \right] ^{t}\xi _{0}\left( \theta \right) \text {d}\theta , \end{aligned}$$

(24)

assuming that appropriate integrability conditions hold in (24) for all $t\in \left\{ 1,2,\ldots \right\}$.

Proof of Lemma 1

We express $\xi _{t-1}$ as a function of $\xi _{t-2}$, according to the Bayesian update of beliefs given by (23), and we substitute it into the LHS of (24),

$$\begin{aligned} \int \limits _{\mathcal {H}}\int \limits _{\Theta }\cdots \int \limits _{\mathcal {H}}\int \limits _{\Theta }\int \limits _{ \mathcal {H}}\int \limits _{\Theta }\underset{s=0}{\overset{t-1}{\prod }}h\left( \eta _{s}\right) \phi \left( \eta _{t-1}|\theta _{t-1}\right) \frac{\phi \left( \eta _{t-2}|\theta _{t-1}\right) \xi _{t-2}\left( \theta _{t-1}\right) }{ \int \nolimits _{\Theta }\phi \left( \eta _{t-2}|x\right) \xi _{t-2}\left( x\right) \text {d}x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\, \end{aligned}$$

$$\begin{aligned} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\times \,\text {d}\theta _{t-1}\text {d}\eta _{t-1}\phi \left( \eta _{t-2}|\theta _{t-2}\right) \xi _{t-2}\left( \theta _{t-2}\right) \text {d}\theta _{t-2}\text {d}\eta _{t-2}\times \cdots \times \phi \left( \eta _{0}|\theta _{0}\right) \xi _{0}\left( \theta _{0}\right) \text {d}\theta _{0}\text {d}\eta _{0}= \end{aligned}$$

$$\begin{aligned} =\int \limits _{\mathcal {H}}\int \limits _{\Theta }\cdots \int \limits _{\mathcal {H}}\int \limits _{\mathcal {H} }\int \limits _{\Theta }\underset{s=0}{\overset{t-1}{\prod }}h\left( \eta _{s}\right) \phi \left( \eta _{t-1}|\theta _{t-1}\right) \phi \left( \eta _{t-2}|\theta _{t-1}\right) \xi _{t-2}\left( \theta _{t-1}\right) \text {d}\theta _{t-1} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\, \end{aligned}$$

$$\begin{aligned} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\times \text {d}\eta _{t-1}\text {d}\eta _{t-2}\phi \left( \eta _{t-3}|\theta _{t-3}\right) \xi _{t-3}\left( \theta _{t-3}\right) \text {d}\theta _{t-3}\text {d}\eta _{t-3}\times \cdots \times \phi \left( \eta _{0}|\theta _{0}\right) \xi _{0}\left( \theta _{0}\right) \text {d}\theta _{0}\text {d}\eta _{0} , \end{aligned}$$

i.e., $\xi _{t-1}$ has been canceled from the expression. Continuing in this way up to period $0$, the LHS of (24) becomes

$$\begin{aligned} \int \limits _{\mathcal {H}}\cdots \int \limits _{\mathcal {H}}\int \limits _{\mathcal {H}}\int \limits _{\Theta } \underset{s=0}{\overset{t-1}{\prod }}h\left( \eta _{s}\right) \phi \left( \eta _{s}|\theta _{t-1}\right) \xi _{0}\left( \theta _{t-1}\right) \text {d} \theta _{t-1}\text {d}\eta _{t-1}\text {d}\eta _{t-2}\times \cdots \times \text {d}\eta _{0} , \end{aligned}$$

and because $\eta$’s are independent over time, this last expression is equal to the result given by (24). $\square$

Lemma 2

(matrix determinant lemma) Let $A$ be an $N\times N$ nonsingular matrix, and $x$,$~y$ be any $N\times 1$ vectors. Then,

$$\begin{aligned} \det \left( A+x\cdot y^{T}\right) =\left( 1+y^{T}\cdot A^{-1}\cdot x\right) \cdot \det (A). \end{aligned}$$

Proof of Lemma 2

See [9, p. 416, Theorem 18.1.1 and Corollaries 18.1.2 and 18.1.3]. $\square$

Lemma 3

Let the $N\times N$ linear system

$$\begin{aligned} \mathbf {A}\cdot \left[ \begin{array}{c} x_{1} \\ x_{2} \\ \vdots \\ x_{N} \end{array} \right] =\left[ \begin{array}{c} \underset{}{1} \\ \overset{}{1} \\ \vdots \\ \overset{}{1} \end{array} \right] , \end{aligned}$$

(25)

in which,

$$\begin{aligned} \mathbf {A}\equiv \left[ \begin{array}{l@{\quad }lll} a_{1} &{} 1 &{} \cdots &{} 1 \\ 1 &{} a_{2} &{} \cdots &{} 1 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 1 &{} 1 &{} \cdots &{} a_{N} \end{array} \right] , \end{aligned}$$

with $a_{i}\ne 1$ for all $i\in \left\{ 1,\ldots ,N\right\}$ . Then, the solution to this system (denoted by a star) is unique with,

$$\begin{aligned} x_{i}^{*}=\frac{\frac{1}{a_{i}-1}}{1+~\underset{j=1}{\overset{N}{\sum }} \frac{1}{a_{j}-1}},\quad \mathrm{for\,all }\quad \,\,i\in \left\{ 1,\ldots ,N\right\} . \end{aligned}$$

(26)

Proof of Lemma 3

Simple substitution of (26) into (25) proves the validity of the solution. The condition $a_{i}\ne 1$ for all $i\in \left\{ 1,\ldots ,N\right\}$ guarantees linear independence among all rows (and columns) of $A$, which implies the nonsingularity of $\mathbf {A}$, which, in turn, proves the uniqueness of (26), proving the Lemma. Yet, below we provide the steps that lead to identifying this solution, which may prove useful for characterizing similar results in other resource-extraction games with similar structure.¹⁸

Pick any $i\in \left\{ 1,\ldots ,N\right\}$, and notice that (25) implies,

$$\begin{aligned} x_{i}^{*}=\mathbf {0}_{\mathbf {1}_{i}}^{T}\cdot \mathbf {A}^{-1}\cdot \mathbf {1}_{N} , \end{aligned}$$

(27)

in which $\mathbf {1}_{N}$ is an $N\times 1$ vector of ones, and $\mathbf {0}_{ \mathbf {1}_{i}}$ is an $N\times 1$ vector of zeros, with the sole exception that its $i$-th element is equal to $1$. The matrix determinant lemma (Lemma 2) implies that

$$\begin{aligned} \det \left( \mathbf {A}-\mathbf {1}_{N}\cdot \mathbf {0}_{\mathbf {1} _{i}}^{T}\right) =\left( 1-\mathbf {0}_{\mathbf {1}_{i}}^{T}\cdot \mathbf {A} ^{-1}\cdot \mathbf {1}_{N}\right) \cdot \det \left( \mathbf {A}\right) . \end{aligned}$$

(28)

Setting,

$$\begin{aligned} \mathbf {A}_{i}\equiv \mathbf {A}-\mathbf {1}_{N}\cdot \mathbf {0}_{\mathbf {1} _{i}}^{T}, \end{aligned}$$

and combining (27) with (28), we obtain

$$\begin{aligned} x_{i}^{*}=1-\frac{\det \left( {\mathbf {A}}_{i}\right) }{\det ({\mathbf {A}})}. \end{aligned}$$

(29)

Equation (29) implies that, in order to characterize $x_{i}^{*}$ , we must first characterize $\det \left( \mathbf {A}_{i}\right)$ and $\det \left( \mathbf {A}\right)$. We start from characterizing $\det \left( \mathbf {A}\right)$. Let

$$\begin{aligned} \tilde{\mathbf {A}}\equiv \mathbf {A}-\mathbf {1}_{N}\cdot \mathbf {1}_{N}^{T} , \end{aligned}$$

which implies that $\tilde{\mathbf {A}}$ is a diagonal matrix. Denoting the $i$-th diagonal element of $\tilde{\mathbf {A}}$ by $\mathtt {diag}\left( \tilde{\mathbf {A}}\right) _{i}$, it is,

$$\begin{aligned} \mathtt {diag}\left( \tilde{\mathbf {A}}\right) _{i}=a_{i}-1 . \end{aligned}$$

(30)

Applying again the matrix determinant lemma (Lemma 2),

$$\begin{aligned} \det \left( \mathbf {A}\right) =\det \left( \tilde{\mathbf {A}}+\mathbf {1} _{N}\cdot \mathbf {1}_{N}^{T}\right) =\left( 1+\mathbf {1}_{N}^{T}\cdot \tilde{\mathbf {A}}^{-1}\cdot \mathbf {1}_{N}\right) \cdot \det (\tilde{\mathbf {A}}), \end{aligned}$$

which implies

$$\begin{aligned} \det \left( \mathbf {A}\right) =\left( 1~+~\overset{N}{\underset{i=1}{\sum }} \frac{1}{\mathtt {diag}\left( \tilde{\mathbf {A}}\right) _{i}}\right) \cdot \underset{i=1}{\overset{N}{\prod }}\mathtt {diag}\left( \tilde{\mathbf {A}} \right) _{i}. \end{aligned}$$

(31)

In order to characterize $\det \left( \mathbf {A}_{i}\right)$, we use the definitions of $\mathbf {A}_{i}$ and $\tilde{\mathbf {A}}$, noticing that $\mathbf {A}_{i}-\tilde{\mathbf {A}}=$ $\mathbf {1}_{N}\cdot \left( \mathbf {1} _{N}^{T}-\mathbf {0}_{\mathbf {1}_{i}}^{T}\right)$, which implies,

$$\begin{aligned} \mathbf {A}_{i}=\tilde{\mathbf {A}}+\mathbf {1}_{N}\cdot \mathbf {1}_{\mathbf {0} _{i}}^{T}, \end{aligned}$$

(32)

in which $\mathbf {1}_{\mathbf {0}_{i}}$ is an $N\times 1$ vector of ones, with the sole exception that its $i$-th element is equal to $0$. Combining (32) with the matrix determinant lemma (Lemma 2),

$$\begin{aligned} \det \left( \mathbf {A}_{i}\right) =\det \left( \tilde{\mathbf {A}}+\mathbf {1} _{N}\cdot \mathbf {1}_{\mathbf {0}_{i}}^{T}\right) =\left( 1+\mathbf {1}_{ \mathbf {0}_{i}}^{T}\cdot \tilde{\mathbf {A}}^{-1}\cdot \mathbf {1}_{N}\right) \cdot \det \left( \tilde{\mathbf {A}}\right) \!, \end{aligned}$$

which gives,

$$\begin{aligned} \det \left( \mathbf {A}_{i}\right) =\left( 1~+~\overset{N}{\underset{\underset{j\ne i}{j=1}}{\sum }}\frac{1}{\mathtt {diag}\left( \tilde{\mathbf {A}} \right) _{j}}\right) \cdot \underset{i=1}{\overset{N}{\prod }}\mathtt {diag} \left( \tilde{\mathbf {A}}\right) _{i} . \end{aligned}$$

(33)

Equation (26) is derived after combining (29) with (33), (31), and (30). The formula given by Eq. (26) holds for all $i\in \left\{ 1,\ldots ,N\right\}$, since the choice of $i$ was arbitrary. $\square$

Proof of Proposition 1

Our solution approach follows Levhari and Mirman (1980). We start from deriving RLMPNE in the finite-horizon setting. Then we use the finite-horizon RLMPNE results in order to generalize them to the infinite-horizon case. The approach of Levhari and Mirman (1980) for proving the result helps in exhibiting the informational structure of the problem.

The static problem (0-period-horizon problem)

The Nash-equilibrium solution is not unique in this case, but without loss of generality we can set,

$$\begin{aligned} c_{0}^{i}=\kappa _{i}^{\left( 0\right) }k, \end{aligned}$$

(34)

in which $\kappa _{i}^{\left( 0\right) }$ is some constant with $\kappa _{i}^{\left( 0\right) }\in \left[ 0,1\right]$ for all $i\in \left\{ 1,\ldots ,N\right\}$, with $\Sigma _{i=1}^{N}\kappa _{i}^{\left( 0\right) }=1$. (We are solving this problem recursively, so we denote the $n$-th iteration by a superscript “ $\left( n\right)$” wherever this is applicable.) In order to keep each player’s problem well defined in next iteration, so as to comply with the requirement that value functions are well defined (notice that logarithmic utility implies that zero consumption in one period implies a value function equal to minus–infinity, and zero-consumption choices prevail in a Markov-perfect solution) we focus on a solution with $\kappa _{i}^{\left( 0\right) }\in \left( 0,1\right)$ for all $i\in \left\{ 1,\ldots ,N\right\}$, and with $\Sigma _{i=1}^{N}\kappa _{i}^{\left( 0\right) }=1$. So, the value function of the agent in the static problem is,

$$\begin{aligned} V^{i,\left( 0\right) }\left( k;\Xi _{0}\right) =\ln \left( k\right) +\ln \left( \kappa _{i}^{\left( 0\right) }\right) , \end{aligned}$$

(35)

which is the only case with the value function not depending on $\Xi _{0}$.

The 1-period-horizon problem

The decision of player $i$ is determined by the Bellman equation,

$$\begin{aligned} V^{i,\left( 1\right) }\left( k;\Xi _{0}\right) \!=\!\underset{c_{i}\ge 0}{\max } \left\{ \ln \left( c_{i}\right) \underset{\underset{}{}}{\overset{\overset{}{ }}{}}\right. ~\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{aligned}$$

$$\begin{aligned} ~~\ \left. +\,\delta \int \limits _{\mathcal {H}}V^{i,\left( 0\right) }\left( B\cdot \left[ k-c_{i}\!-\!\underset{j\ne i}{\sum }C^{j,\left( 1\right) }\left( k;\Xi _{0}\right) \right] ^{\eta };\Xi _{1}\left( \cdot \left| \eta \right. \right) \right) \left[ \int \limits _{\Theta }\phi \left( \eta |\theta \right) \xi _{0}^{i}\left( \theta \right) \text {d}\theta \right] \text {d}\eta \right\} \end{aligned}$$

so, using (35),

$$\begin{aligned} V^{i,\left( 1\right) }\left( k;\Xi _{0}\right) =\underset{c_{i}\ge 0}{\max } \left\{ \ln \left( c_{i}\right) \underset{\underset{}{}}{\overset{\overset{}{ }}{}}\right. ~\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{aligned}$$

$$\begin{aligned} ~~\ \left. +\,\delta \int \limits _{\mathcal {H}}\eta \ln \left[ k-c_{i}\!-\!\underset{j\ne i}{\sum }C^{j,\left( 1\right) }\left( k;\Xi _{0}\right) \right] \left[ \int \limits _{\Theta }\!\phi \! \left( \eta |\theta \right) \xi _{0}^{i}\left( \theta \right) \text {d}\theta \right] \text {d}\eta \!+\!\delta \left[ \! \ln \left( B\right) \!+\!\ln \left( \! \kappa _{i}^{\left( 0\right) }\right) \right] \right\} \end{aligned}$$

(36)

with first-order condition,

$$\begin{aligned} \frac{1}{c_{i}}\!=\!\frac{\delta \int \nolimits _{\mathcal {H}}\eta \left[ \int \nolimits _{\Theta }\phi \left( \eta |\theta \right) \xi _{0}^{i}\left( \theta \right) \text {d} \theta \right] \text {d}\eta }{k-c_{i}-\underset{j\ne i}{\sum }C^{j,\left( 1\right) }\left( k;\Xi _{0}\right) } , \end{aligned}$$

which implies that, for all $i\in \left\{ 1,\ldots ,N\right\}$, $C^{i,\left( 1\right) }\left( k;\Xi _{0}\right)$ is of the multiplicatively separable form,

$$\begin{aligned} C^{i,\left( 1\right) }\left( k;\Xi _{0}\right) =c^{i,\left( 1\right) }\left( \Xi _{0}\right) \cdot k , \end{aligned}$$

(37)

in which $\left\{ c^{i,\left( 1\right) }\left( \Xi _{0}\right) \right\} _{i=1}^{N}$ is the unique solution to the linear system,

$$\begin{aligned} \left[ \begin{array}{llll} 1+\delta E_{0}\left( \eta |\xi _{0}^{1}\right) &{} 1 &{} \cdots &{} 1 \\ 1 &{} 1+\delta E_{0}\left( \eta |\xi _{0}^{2}\right) &{} \cdots &{} 1 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 1 &{} 1 &{} \cdots &{} 1+\delta E_{0}\left( \eta |\xi _{0}^{N}\right) \end{array} \right] \left[ \begin{array}{c} c^{1,\left( 1\right) }\left( \Xi _{0}\right) \\ c^{2,\left( 1\right) }\left( \Xi _{0}\right) \\ \vdots \\ c^{N,\left( 1\right) }\left( \Xi _{0}\right) \end{array} \right] =\left[ \begin{array}{c} 1 \\ 1 \\ \vdots \\ 1 \end{array} \right] , \end{aligned}$$

(38)

with

$$\begin{aligned} E_{0}\left( \eta |\xi _{0}^{i}\right) \equiv \int \limits _{\mathcal {H}}\eta \left[ \int \limits _{\Theta }\phi \left( \eta |\theta \right) \xi _{0}^{i}\left( \theta \right) \text {d}\theta \right] \text {d}\eta , \quad i=1,\ldots ,N. \end{aligned}$$

(39)

Lemma 3 applies to the linear system given by (38), and it implies that (38) has a unique solution with $c^{i,\left( 1\right) }\left( \Xi _{0}\right) \in \left( 0,1\right)$, and $\Sigma _{i}c^{i,\left( 1\right) }\left( \Xi _{0}\right) \in \left( 0,1\right)$. Substituting $\left\{ C^{i,\left( 1\right) }\left( \Xi _{0}\right) \right\} _{i=1}^{N}$ of the multiplicatively separable form given by (37) into (36 ), the Bellman equation, leads to a value function of the form,

$$\begin{aligned} V^{i,\left( 1\right) }\left( k;\Xi _{0}\right) =\left\{ 1+\delta \int \limits _{ \mathcal {H}}\eta \left[ \int \limits _{\Theta }\phi \left( \eta |\theta \right) \xi _{0}^{i}\left( \theta \right) \text {d}\theta \right] \text {d}\eta \right\} \ln \left( k\right) +\kappa ^{i,\left( 1\right) }\left( \Xi _{0}\right) , \end{aligned}$$

(40)

for all $i\in \left\{ 1,\ldots ,N\right\}$, in which $\kappa ^{i,\left( 1\right) }\left( \Xi _{0}\right)$ is a constant that does not affect optimization in future steps. Unlike $V^{i,\left( 0\right) }\left( k;\Xi _{0}\right)$, the value function $V^{i,\left( 1\right) }\left( k;\Xi _{0}\right)$ depends on $\Xi _{0}$. Yet, we have an explicit form regarding the way $V^{i,\left( 1\right) }\left( k;\Xi _{0}\right)$ depends on $\Xi _{0}$. Most interestingly, in Eq. (40) the coefficient of $\ln \left( k\right)$ depends on $\xi _{0}^{i}$ only and not on the beliefs of other individuals.

The 2-period-horizon problem

The decision of player $i$ is now determined by the Bellman equation,

$$\begin{aligned} V^{i,\left( 2\right) }\left( k;\Xi _{0}\right)&= \underset{c_{i}\ge 0}{\max } \left\{ \ln \left( c_{i}\right) \underset{\underset{}{}}{\overset{\overset{}{ }}{}} \!\!+\!\!\,\delta \int \limits _{\mathcal {H}}V^{i,\left( 1\right) }\left( B\cdot \left[ k-c_{i}\!-\!\underset{j\ne i}{\sum }C^{j,\left( 2\right) }\left( k;\Xi _{0}\right) \right] ^{\eta };\Xi _{1}\left( \cdot \left| \eta \right. \right) \right) \right. \\&\left. \times \left[ \int \limits _{\Theta }\phi \left( \eta |\theta \right) \xi _{0}^{i}\left( \theta \right) \text {d}\theta \right] \text {d}\eta \right\} \\ \end{aligned}$$

so, using (40),

$$\begin{aligned}&V^{i,\left( 2\right) }\left( k;\Xi _{0}\right) =\underset{c_{i}\ge 0}{\max } \left\{ \ln \left( c_{i}\right) \underset{\underset{}{}}{\overset{\overset{}{ }}{}} \!+\!\delta \int \limits _{\mathcal {H}}\eta _{0}\left\{ 1\!+\!\delta \int \limits _{\mathcal {H} }\eta _{1}\left[ \int \limits _{\Theta }\phi \left( \eta _{1}|\theta _{1}\right) \xi _{1}^{i}\left( \theta _{1}|\eta _{0}\right) \text {d}\theta _{1}\right] \text { d}\eta _{1}\right\} \right. \nonumber \\&\left. \quad \times \ln \left[ k\!-\!c_{i}-\underset{j\ne i}{\sum }C^{j,\left( 2\right) }\left( k;\Xi _{0}\right) \right] \times \left[ \int \limits _{\Theta }\phi \left( \eta _{0}|\theta \right) \! \xi _{0}^{i}\left( \theta \right) \text {d}\theta \right] \text {d}\eta _{0}\!+\!\delta \! \left\{ \ln \left( B\right) \right. \right. \nonumber \\&\left. \left. \quad \!+\!\int \limits _{\mathcal {H} }\kappa ^{i,\left( \!1\right) }\left( \!\Xi _{1}\left( \!\cdot \! \left| \eta \right. \right) \right) \left[ \! \int \limits _{\Theta }\phi \left( \! \eta |\theta \right) \xi _{0}^{i}\left( \!\theta \right) \text {d}\theta \right] \text {d} \eta \right\} \right\} \end{aligned}$$

(41)

subject to,

$$\begin{aligned} \xi _{1}^{i}\left( \theta |\eta \right) =\frac{\phi \left( \eta |\theta \right) \xi _{0}^{i}\left( \theta \right) }{\int \nolimits _{\Theta }\phi \left( \eta |x\right) \xi _{0}^{i}\left( x\right) \text {d}x},\quad i=1,\ldots ,N. \end{aligned}$$

(42)

What is crucial to observe here is the notation about the timing of shocks. In the problem expressed by (41), each player is deciding upon a strategy in period $0$, expecting both a shock $\eta _{0}$ in period $0$, after the decision has been made, and a shock $\eta _{1}$ in period 1. Yet, it is the shock $\eta _{0}$ which will determine how the prior distribution $\xi _{0}^{i}$ will evolve to $\xi _{1}^{i}$, which is an element that the analytic form of (41) allows us to see explicitly. So, with the time horizon being expanded, we can see how prior beliefs determine what type of information is expected to arrive and also how this information is expected to be exploited.

To simplify notation, we can re-write (41) as,

$$\begin{aligned} V^{i,\left( 2\right) }\left( k;\Xi _{0}\right) =\underset{c_{i}\ge 0}{\max } \left\{ \ln \left( c_{i}\right) \underset{\underset{}{}}{\overset{\overset{}{ }}{+}}\delta \left[ E_{0}\left( \eta _{0}|\xi _{0}^{i}\right) +\delta E_{0}\left( \eta _{1}\eta _{0}|\xi _{0}^{i}\right) \right] \right. ~\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{aligned}$$

$$\begin{aligned} \left. \times \ln \left[ k-c_{i}-\underset{j\ne i}{\sum }C^{j,\left( 2\right) }\left( k;\Xi _{0}\right) \right] +\delta \left[ \ln \left( B\right) +E_{0}\left[ \kappa ^{i,\left( 1\right) }\left( \Xi \left( \cdot \left| \eta \right. \right) \right) \right] \right] \right\} , \end{aligned}$$

(43)

with

$$\begin{aligned} E_{0}\left( \eta _{0}|\xi _{0}^{i}\right) \equiv \int \limits _{\mathcal {H}}\eta _{0} \left[ \int \limits _{\Theta }\phi \left( \eta _{0}|\theta \right) \xi _{0}^{i}\left( \theta \right) \text {d}\theta \right] \text {d}\eta _{0} , \end{aligned}$$

as in equation (39) above, and,

$$\begin{aligned} E_{0}\left( \eta _{1}\eta _{0}|\xi _{0}^{i}\right) \equiv \int \limits _{\mathcal {H} }\eta _{0}\int \limits _{\mathcal {H}}\eta _{1}\left[ \int \limits _{\Theta }\phi \left( \eta _{1}|\theta _{1}\right) \xi _{1}^{i}\left( \theta _{1}|\eta _{0}\right) \text {d}\theta _{1}\right] \text {d}\eta _{1}\left[ \int \limits _{\Theta }\phi \left( \eta _{0}|\theta \right) \xi _{0}^{i}\left( \theta \right) \text {d}\theta \right] \text {d}\eta _{0} , \end{aligned}$$

in which $\xi _{1}^{i}\left( \theta _{1}|\eta _{0}\right)$ is given from ( 42). The first-order conditions of (43) are given by,

$$\begin{aligned} \frac{1}{c_{i}}=\delta \left[ E_{0}\left( \eta _{0}|\xi _{0}^{i}\right) +\delta E_{0}\left( \eta _{1}\eta _{0}|\xi _{0}^{i}\right) \right] \frac{1}{ k-c_{i}-\underset{j\ne i}{\sum }C^{j,\left( 2\right) }\left( k;\Xi _{0}\right) } , \end{aligned}$$

which implies that $C^{i,\left( 1\right) }\left( k;\Xi _{0}\right)$ is of the multiplicatively separable form,

$$\begin{aligned} C^{i,\left( 2\right) }\left( k;\Xi _{0}\right) =c^{i,\left( 2\right) }\left( \Xi _{0}\right) \cdot k , \end{aligned}$$

(44)

for all $i\in \left\{ 1,\ldots ,N\right\}$, where $\left\{ c^{i,\left( 2\right) }\left( \Xi _{0}\right) \right\} _{i=1}^{N}$ is the unique solution to the linear system,

$$\begin{aligned} \left[ \begin{array}{cccc} A^{\left( 2\right) }\left( \xi _{0}^{1}\right) &{} 1 &{} \cdots &{} 1 \\ 1 &{} A^{\left( 2\right) }\left( \xi _{0}^{2}\right) &{} \cdots &{} 1 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 1 &{} 1 &{} \cdots &{} A^{\left( 2\right) }\left( \xi _{0}^{N}\right) \end{array} \right] \cdot \left[ \begin{array}{c} c^{1,\left( 2\right) }\left( \Xi _{0}\right) \\ c^{2,\left( 2\right) }\left( \Xi _{0}\right) \\ \vdots \\ c^{N,\left( 2\right) }\left( \Xi _{0}\right) \end{array} \right] =\left[ \begin{array}{c} 1 \\ 1 \\ \vdots \\ 1 \end{array} \right] , \end{aligned}$$

(45)

in which

$$\begin{aligned} A^{\left( 2\right) }\left( \xi _{0}^{i}\right) \equiv 1+\delta \left[ E_{0}\left( \eta _{0}|\xi _{0}^{i}\right) +\delta E_{0}\left( \eta _{1}\eta _{0}|\xi _{0}^{i}\right) \right] \!, \ i\in \left\{ 1,\ldots ,N\right\} . \end{aligned}$$

Again, Lemma 3 guarantees that (45) has a unique solution with $c^{i,\left( 2\right) }\left( \Xi _{0}\right) \in \left( 0,1\right)$, and $\Sigma _{i}c^{i,\left( 2\right) }\left( \Xi _{0}\right) \in \left( 0,1\right)$, while substitution of $\left\{ C^{i,\left( 2\right) }\left( \Xi _{0}\right) \right\} _{i=1}^{N}$ as given by (44) into the Bellman equation given by (43) gives a value function of the form,

$$\begin{aligned} V^{i,\left( 2\right) }\left( k;\Xi _{0}\right) =A^{\left( 2\right) }\left( \xi _{0}^{i}\right) \ln \left( k\right) +\kappa ^{i,\left( 2\right) }\left( \Xi _{0}\right) , \end{aligned}$$

where $\kappa ^{i,\left( 2\right) }\left( \Xi _{0}\right)$ is a constant that does not affect optimization in any future step. At this point we have seen enough of the problem’s structure to be able to deduce the formulas of the $n$-period horizon problem.

The $n$ -period-horizon problem

The strategy of player $i$ is determined by the Bellman equation,

$$\begin{aligned} V^{i,\left( n\right) }\left( k;\Xi _{0}\right) =\underset{c_{i}\ge 0}{\max } \left\{ \ln \left( c_{i}\right) \underset{\underset{}{}}{\overset{\overset{}{ }}{}}\right. ~\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{aligned}$$

$$\begin{aligned} ~~\ \left. +\delta \int \limits _{\mathcal {H}}V^{i,\left( n\!-\!1\right) }\left( B\cdot \left[ k-c_{i}-\underset{j\ne i}{\sum }C^{j,\left( n\right) }\left( k;\Xi _{0}\right) \right] ^{\eta };\Xi _{1}\left( \cdot \left| \eta \right. \right) \right) \left[ \!\int \limits _{\Theta }\!\phi \! \left( \!\eta \! |\theta \right) \xi _{0}^{i}\left( \theta \right) \text {d}\theta \right] \text {d}\eta \right\} \end{aligned}$$

with $V^{i,\left( n\right) }\left( k;\Xi _{0}\right)$ being of the form,

$$\begin{aligned} V^{i,\left( n\right) }\left( k;\Xi _{0}\right) =\left[ 1+\delta \underset{t=0 }{\overset{n-1}{\sum }}\delta ^{t}E_{0}\left( \eta _{t}|\xi _{0}^{i}\right) \right] \ln \left( k\right) +\kappa ^{i,\left( n\right) }\left( \Xi _{0}\right) , \end{aligned}$$

(46)

in which $\kappa ^{i,\left( n\right) }\left( \Xi _{0}\right)$ is a constant, and

$$\begin{aligned} E_{0}\left( \left. \underset{s=0}{\overset{t}{\prod }}\eta _{s}\right| ~\xi _{0}^{i}\right) \equiv \int \limits _{\mathcal {H}}\int \limits _{\Theta }\cdots \int \limits _{ \mathcal {H}}\int \limits _{\Theta }\int \limits _{\mathcal {H}}\int \limits _{\Theta }\underset{s=0}{ \overset{t}{\prod }}\eta _{s}\phi \left( \eta _{t}|\theta _{t}\right) \xi _{t}^{i}\left( \theta _{t}\right) \text {d}\theta _{t}\text {d}\eta _{t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\, \end{aligned}$$

$$\begin{aligned} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\times \phi \left( \eta _{t-1}|\theta _{t-1}\right) \xi _{t-1}^{i}\left( \theta _{t-1}\right) \text {d}\theta _{t-1} \text {d}\eta _{t-1}\times \cdots \times \phi \left( \eta _{0}|\theta _{0}\right) \xi _{0}^{i}\left( \theta _{0}\right) \text {d}\theta _{0}\text {d} \eta _{0} . \end{aligned}$$

Moreover, players’ strategies are of the form

$$\begin{aligned} C^{i,\left( n\right) }\left( \Xi _{0}\right) =c^{i,\left( n\right) }\left( \Xi _{0}\right) \cdot k,\quad \textit{i}=1,\ldots ,N , \end{aligned}$$

(47)

in which $\left\{ c^{i,\left( n\right) }\left( \Xi _{0}\right) \right\} _{i=1}^{N}$ is the unique solution to the linear system,

$$\begin{aligned} \left[ \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} A^{\left( n\right) }\left( \xi _{0}^{1}\right) &{} 1 &{} \cdots &{} 1 \\ 1 &{} A^{\left( n\right) }\left( \xi _{0}^{2}\right) &{} \cdots &{} 1 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 1 &{} 1 &{} \cdots &{} A^{\left( n\right) }\left( \xi _{0}^{N}\right) \end{array} \right] \cdot \left[ \begin{array}{c} c^{1,\left( n\right) }\left( \Xi _{0}\right) \\ c^{2,\left( n\right) }\left( \Xi _{0}\right) \\ \vdots \\ c^{N,\left( n\right) }\left( \Xi _{0}\right) \end{array} \right] =\left[ \begin{array}{c} 1 \\ 1 \\ \vdots \\ 1 \end{array} \right] , \end{aligned}$$

(48)

with

$$\begin{aligned} A^{\left( n\right) }\left( \xi _{0}^{i}\right) \equiv 1+\delta \underset{t=0}{\overset{n-1}{\sum }}\delta ^{t}E_{0}\left( \left. \underset{s=0}{\overset{t }{\prod }}\eta _{s}\right| ~\xi _{0}^{i}\right) ,\quad \ i\in \left\{ 1,\ldots ,N\right\} . \end{aligned}$$

To calculate $E_{0}\left( \eta _{t}|\xi _{0}^{i}\right)$ we rely on Lemma 1. From Eq. (24) of Lemma 1, after setting $h\left( \eta \right) =\eta$, the identity function, we obtain,

$$\begin{aligned} E_{0}\left( \left. \underset{s=0}{\overset{t}{\prod }}\eta _{s}\right| ~\xi _{0}^{i}\right) =\int \limits _{\Theta }\left[ \int \limits _{\mathcal {H}}\eta \phi \left( \eta |\theta \right) \text {d}\eta \right] ^{t+1}\xi _{0}^{i}\left( \theta \right) \text {d}\theta , \end{aligned}$$

and from (6) it is,

$$\begin{aligned} E_{0}\left( \left. \underset{s=0}{\overset{t}{\prod }}\eta _{s}\right| ~\xi _{0}^{i}\right) =\int \limits _{\Theta }\left[ \mu \left( \theta \right) \right] ^{t+1}\xi _{0}^{i}\left( \theta \right) \text {d}\theta . \end{aligned}$$

(49)

Substituting (49) into (46) we obtain,

$$\begin{aligned} V^{i,\left( n\right) }\left( k;\Xi _{0}\right) =\left[ 1+\delta \int \limits _{\Theta }\underset{t=0}{\overset{n-1}{\sum }}\delta ^{t}\left[ \mu \left( \theta \right) \right] ^{t+1}\xi _{0}^{i}\left( \theta \right) \text {d}\theta \right] \ln \left( k\right) +\kappa ^{i,\left( n\right) } (\Xi _{0}). \end{aligned}$$

(50)

The infinite-horizon problem

Notice that since $\mathcal {H}\subseteq \left( 0,1\right)$, $\mu \left( \theta \right) \in \left( 0,1\right)$ for all $\theta \in \Theta$. After taking the limit when $n\rightarrow \infty$, (50) gives,

$$\begin{aligned} V^{i,\left( \infty \right) }\left( k;\Xi _{0}\right) =V^{i}\left( k;\Xi _{0}\right) =\left[ 1+\delta \int \limits _{\Theta }\underset{t=0}{\overset{\infty }{ \sum }}\delta ^{t}\left[ \mu \left( \theta \right) \right] ^{t+1}\xi _{0}^{i}\left( \theta \right) \text {d}\theta \right] \ln \left( k\right) +\kappa ^{i,\left( \infty \right) }\left( \Xi _{0}\right) , \end{aligned}$$

so, $\mu \left( \theta \right) \in \left( 0,1\right)$ for all $\theta \in \Theta$ guarantees that $V^{i}\left( k;\Xi \right)$ is well defined. Since $\xi _{0}^{i}$ is a density function, $\int _{\Theta }\xi _{0}^{i}\left( \theta \right)$d$\theta =1$, which leads to,

$$\begin{aligned} V^{i}\left( k;\Xi \right) =\int \limits _{\Theta }\frac{1}{1-\delta \mu \left( \theta \right) }\xi ^{i}\left( \theta \right) \text {d}\theta \ln \left( k\right) +\kappa ^{i,\left( \infty \right) }\left( \Xi \right) . \end{aligned}$$

(Subscript “ $0$” of $\Xi _{0}$ has appeared in order to remind that $\Xi _{0}$ denotes prior beliefs in period $0$. In the infinite-horizon setup this timing does not matter any more, so subscript “ $0$” can be dropped. So, we drop it throughout the rest of the proof.) Moreover, the solution is again of the multiplicatively separable form

$$\begin{aligned} C^{i,\left( \infty \right) }\left( k;\Xi \right) =C^{i}\left( k;\Xi \right) =c^{i}\left( \Xi \right) \cdot k , \end{aligned}$$

and (48) is generalized to,

$$\begin{aligned} \mathbf {A}\cdot \left[ \begin{array}{c} c^{1}\left( \Xi \right) \\ c^{2}\left( \Xi \right) \\ \vdots \\ c^{N}\left( \Xi \right) \end{array} \right] =\left[ \begin{array}{c} \underset{}{1} \\ \overset{}{1} \\ \vdots \\ \overset{}{1} \end{array} \right] , \end{aligned}$$

(51)

in which

$$\begin{aligned} \mathbf {A}\equiv \left[ \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} \int \limits _{\Theta }\frac{1}{1-\delta \mu \left( \theta \right) }\xi ^{1}\left( \theta \right) \text {d}\theta &{} 1 &{} \cdots &{} 1 \\ 1 &{} \int \limits _{\Theta }\frac{1}{1-\delta \mu \left( \theta \right) }\xi ^{2}\left( \theta \right) \text {d}\theta &{} \cdots &{} 1 \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 1 &{} 1 &{} \cdots &{} \int \limits _{\Theta }\frac{1}{1-\delta \mu \left( \theta \right) } \xi ^{N}\left( \theta \right) \text {d}\theta \end{array} \right] . \end{aligned}$$

Since $\mu \left( \theta \right) \in \left( 0,1\right)$ for all $\theta \in \Theta$,

$$\begin{aligned} \int \limits _{\Theta }\frac{1}{1-\delta \mu \left( \theta \right) }\xi ^{i}\left( \theta \right) d\theta >1,\quad i\in \left\{ 1,\ldots ,N\right\} . \end{aligned}$$

(52)

Inequality (52) guarantees that Lemma 3 applies, which implies that the unique solution to (51) is given by Eq. (9).

Regarding the transversality condition given by (3), notice that according to (4) and (9),

$$\begin{aligned} \frac{\partial h^{i}\left( k_{t}^{*},k_{t+1}^{*};\Xi _{t}\right) }{ \partial k_{t}}k_{t}=\frac{1-\underset{j\ne i}{\sum }c^{j}\left( \Xi _{t}\right) }{c^{i}\left( \Xi _{t}\right) } . \end{aligned}$$

(53)

Combining the expression on the left-hand side of (3) with (53) gives,

$$\begin{aligned} \underset{t\rightarrow \infty }{\lim }\delta ^{t}E_{t}\left[ \frac{\partial h^{i}\left( k_{t}^{*},k_{t+1}^{*};\Xi _{t}\right) }{\partial k_{t}} k_{t}^{*}\right] =\underset{t\rightarrow \infty }{\lim }\delta ^{t}\frac{ 1-\underset{j\ne i}{\sum }c^{j}\left( \Xi _{t}\right) }{c^{i}\left( \Xi _{t}\right) }=0 , \end{aligned}$$

proving that the transversality condition given by (3) indeed holds.

In order to guarantee that $V^{i}\left( k;\Xi \right)$ is well defined, it remains to verify condition (11), which can prove important for identifying conditions on $\phi \left( \cdot \left| \eta \right. \right)$ and $\Xi _{0}$ that guarantee the boundedness of $\kappa ^{i,\left( \infty \right) }\left( \Xi _{0}\right)$ in applications using specific functional forms.¹⁹ Specifically, let us take a guess on the functional form of the value function of player $i\in \left\{ 1,\ldots ,N\right\}$,

$$\begin{aligned} V^{i}\left( k;\Xi \right) =\kappa ^{i}\left( \Xi \right) +\left[ \int \limits _{\Theta }f\left( \theta \right) \xi ^{i}\left( \theta \right) \text {d} \theta \right] \cdot \ln \left( k\right) , \end{aligned}$$

(54)

in which $\kappa ^{i}\left( \Xi \right) =\kappa ^{i,\left( \infty \right) }\left( \Xi _{0}\right)$ is an unknown functional and $f$ is given by

$$\begin{aligned} f\left( \theta \right) =\frac{1}{1-\delta \mu \left( \theta \right) }, \end{aligned}$$

(55)

which is the same for all players $i\in \left\{ 1,\ldots ,N\right\}$, since it is independent from specific beliefs $\xi ^{i}$. Substituting Eq. (54) into the Bellman equation (5) gives,

$$\begin{aligned} \kappa ^{i}\left( \Xi \right) \,+\,\left[ \int \limits _{\Theta }f\left( \theta \right) \xi ^{i}\left( \theta \right) \text {d}\theta \right] \,\cdot \, \ln \left( k\right) =\underset{c_{i}\ge 0}{\max } \left\{ \ln \left( c_{i}\right) \underset{\underset{}{}}{\overset{\overset{}{}}{+}}\,\delta \int \limits _{\mathcal {H} }\kappa ^{i}\left( \Xi \left( \cdot ~|\,\,\,\eta \right) \right) \left[ \int \limits _{\Theta }\phi \left( \eta |\theta \right) \xi ^{i}\left( \theta \right) \text {d}\theta \right] \text {d}\eta \right. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{aligned}$$

$$\begin{aligned} \left. +\,\delta \ln \left( B\right) +\delta \int \limits _{\mathcal {H}}\eta \ln \left[ k-c_{i}-\underset{j\ne i}{\sum }C^{j}\left( k;\Xi \right) \right] \int \limits _{\Theta }\frac{f\left( \theta \right) \phi \left( \eta |\theta \right) \xi ^{i}\left( \theta \right) }{\int \limits _{\Theta }\phi \left( \eta |x\right) \xi ^{i}\left( x\right) \text {d}x}\text {d}\theta \left[ \int \limits _{\Theta }\phi \left( \eta |\theta \right) \xi ^{i}\left( \theta \right) \text {d}\theta \right] \text {d}\eta \right\} \end{aligned}$$

which becomes,

$$\begin{aligned} \kappa ^{i}\left( \Xi \right) +\left[ \int \limits _{\Theta }f\left( \theta \right) \xi ^{i}\left( \theta \right) \text {d}\theta \right] \cdot \ln \left( k\right) =\underset{c_{i}\ge 0}{\max }\left\{ \ln \left( c_{i}\right) \underset{\underset{}{}}{\overset{\overset{}{}}{+}}\,\delta \int \limits _{\mathcal {H} }\kappa ^{i}\left( \Xi \left( \cdot ~|\,\,\,\eta \right) \right) \left[ \int \limits _{\Theta }\phi \left( \eta |\theta \right) \xi ^{i}\left( \theta \right) \text {d}\theta \right] \text {d}\eta \right. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{aligned}$$

$$\begin{aligned} ~\ \ \ \ \ \ \left. +\,\delta \ln \left( B\right) +\delta \int \limits _{\Theta }\left[ \int \limits _{\mathcal {H}}\eta \phi \left( \eta |\theta \right) \text {d}\eta \right] f\left( \theta \right) \xi ^{i}\left( \theta \right) \text {d}\theta \cdot \ln \left[ k-c_{i}-\underset{j\ne i}{\sum }C^{j}\left( k;\Xi \right) \right] \right\} . \end{aligned}$$

(56)

First-order conditions based on (56) give,

$$\begin{aligned} \frac{1}{c_{i}}=\frac{\delta \int \limits _{\Theta }\mu \left( \theta \right) f\left( \theta \right) \xi ^{i}\left( \theta \right) \text {d}\theta }{k-c_{i}- \underset{j\ne i}{\sum }C^{j}\left( k;\Xi \right) } , \end{aligned}$$

so,

$$\begin{aligned} \left[ 1+\delta \int \limits _{\Theta }\mu \left( \theta \right) f\left( \theta \right) \xi ^{i}\left( \theta \right) \text {d}\theta \right] c_{i}+\underset{ j\ne i}{\sum }C^{j}\left( k;\Xi \right) =k , \end{aligned}$$

(57)

and if we make the additional guess that the strategies of all other players are of the multiplicatively separable form $C^{j}\left( k;\Xi \right) =c^{j}\left( \Xi \right) \cdot k$, then (57) reconfirms that player $i$ is also of multiplicatively separable form, namely,

$$\begin{aligned} c_{i}=\frac{1-\underset{j\ne i}{\sum }c^{j}\left( \Xi \right) }{1+\delta \int \nolimits _{\Theta }\mu \left( \theta \right) f\left( \theta \right) \xi ^{i}\left( \theta \right) \text {d}\theta }\cdot k. \end{aligned}$$

(58)

Substituting (58) into (56) we identify terms that are multiplicatively separable expressions of $\ln \left( k\right)$ and terms that do not depend on the state variable $k$. Starting from the terms that do not depend on $k$, and which do not affect optimization, these should satisfy the recursion,

$$\begin{aligned} \kappa ^{i}\left( \Xi \right) =\ln \left[ \frac{1-\underset{j\ne i}{\sum } c^{j}\left( \Xi \right) }{1+\delta \int \limits _{\Theta }\mu \left( \theta \right) f\left( \theta \right) \xi ^{i}\left( \theta \right) \text {d}\theta }\right] +\delta \int \limits _{\mathcal {H}}\kappa ^{i}\left( \Xi \left( \cdot ~|\,\,\,\eta \right) \right) \left[ \int \limits _{\Theta }\phi \left( \eta |\theta \right) \xi ^{i}\left( \theta \right) \text {d}\theta \right] \text {d}\eta \end{aligned}$$

$$\begin{aligned} +\delta \ln \left( B\right) +\delta \int \limits _{\Theta }\mu \left( \theta \right) f\left( \theta \right) \xi ^{i}\left( \theta \right) \text {d}\theta \cdot \ln \left\{ \frac{\left[ 1-\underset{j\ne i}{\sum }c^{j}\left( \Xi \right) \right] \delta \int \nolimits _{\Theta }\mu \left( \theta \right) f\left( \theta \right) \xi ^{i}\left( \theta \right) \text {d}\theta }{1+\delta \int \nolimits _{\Theta }\mu \left( \theta \right) f\left( \theta \right) \xi ^{i}\left( \theta \right) \text {d}\theta }\right\} . \end{aligned}$$

(59)

Isolating all terms that are multiplicatively separable expressions of $\ln \left( k\right)$ these must satisfy,

$$\begin{aligned} \left[ \int \limits _{\Theta }f\left( \theta \right) \xi ^{i}\left( \theta \right) \text {d}\theta \right] \cdot \ln \left( k\right) =\ln \left( k\right) +\delta \left[ \int \limits _{\Theta }\mu \left( \theta \right) f\left( \theta \right) \xi ^{i}\left( \theta \right) \text {d}\theta \right] \cdot \ln \left( k\right) , \end{aligned}$$

which is an expression that complies with function $f\left( \theta \right) ,$ given by (55), verifying the validity of (10). Substituting the expression given by (9) into (59), verifies Eqs. (11) and (12), proving the proposition. $\square$