The optimal harvesting problem under price uncertainty

Abstract

In this paper we study the exploitation of a one species forest plantation when timber price is governed by a stochastic process. The work focuses on providing closed expressions for the optimal harvesting policy in terms of the parameters of the price process and the discount factor, with finite and infinite time horizon. We assume that harvest is restricted to mature trees older than a certain age and that growth and natural mortality after maturity are neglected. We use stochastic dynamic programming techniques to characterize the optimal policy and we model price using a geometric Brownian motion and an Ornstein–Uhlenbeck process. In the first case we completely characterize the optimal policy for all possible choices of the parameters. In the second case we provide sufficient conditions, based on explicit expressions for reservation prices, assuring that harvesting everything available is optimal. In addition, for the Ornstein–Uhlenbeck case we propose a policy based on a reservation price that performs well in numerical simulations. In both cases we solve the problem for every initial condition and the best policy is obtained endogenously, that is, without imposing any ad hoc restrictions such as maximum sustained yield or convergence to a predefined final state.

Appendix: Proofs of the results

Proof of Lemma 1

Since prices are positive, we can use the Monotone Convergence Theorem to exchange the order of the expected value and the limit:

$$ \begin{aligned} V({\mathbb{X}}_t,p_t)&= \sup\limits_{\Uppi} \mathbb E_{|p_t} \left[ \lim\limits_{T \to \infty}\sum\limits_{l=t}^{T} \delta^{l-t}p_{l} \pi_l({\mathbb{X}}_l,p_l)\right] \\ &= \sup\limits_{\Uppi} \lim\limits_{T \to \infty} \mathbb E_{|p_t}\left[ \sum\limits_{l=t}^{T} \delta^{l-t}p_{l} \pi_l({\mathbb{X}}_l,p_l)\right]. \end{aligned} $$

By linearity of integrals and using that the total area available is S, we have

$$ \begin{aligned} \sup\limits_{\Uppi} \lim\limits_{T \to \infty} \left[ \sum\limits_{l=t}^{T} \delta^{l-t}p_{t}e^{(l-t)\mu} \pi_l({\mathbb{X}}_l,p_l)\right] &\leq \lim\limits_{T \to \infty} S p_t\left[ \sum\limits_{l=t}^{T} \delta^{l-t}p_{t}e^{(l-t)\mu} \right] \\ &\leq S p_t\frac{1}{1-\delta e^\mu} <\infty. \end{aligned} $$

□

Proof of Theorem 1

To prove that the greedy policy is optimal, we check that the benefit associated with it (Q ^GP) satisfies the Bellman equation (15) from any initial condition. To find the expression of Q ^GP we start by finding the harvests associated to the greedy policy, c _t for all $t\in\mathcal{T}$. If ${\mathbb{X}_t=(\bar{x},x_n,\dots,x_2,x_1)}$ the first n harvests will be:

$$ \left\{\begin{array}{l} c_t=\bar{x}+x_n,\\ c_{t+1}=x_{n-1},\\ \vdots\\ c_{t+n-1}=x_{1}. \end{array}\right. $$

Observe that at time t + n the mature trees are exactly $\bar{x}+x_n$, hence c _t+n  = c _t . From then on, the harvests start repeating with period n:

$$ c_{t+in+j}=c_{t+j}, \quad \hbox{for all}\, i \quad \hbox{and for all}\, j=0,\dots,n-1 \quad \hbox{s.t.} \, (t+in+j)\in {\mathcal{T}}. $$

We consider first the infinite horizon case. With the c _t found above we get,

$$ \begin{aligned} Q_t^{GP}({\mathbb{X}},p_t)&={\mathbb{E}}_{|p_t}\left[ \sum\limits_{i=0}^{\infty}\left(\delta^{in} p_{t+in}\bar{x} + \sum\limits_{j=0}^{n-1}\delta^{in+j}p_{t+in+j}x_{n-j} \right) \right]\\ &= p_t \sum\limits_{i=0}^{\infty}\left((\delta e^\mu)^{in} \bar{x} + \sum\limits_{j=0}^{n-1}(\delta e^\mu)^{in+j} x_{n-j} \right). \end{aligned} $$

(18)

If ${c_t=c\in [0,CA\mathbb{X}_t]}$, then the state at t + 1 is ${\mathbb{X}_{t+1}=(\bar{x}+x_n-c,x_{n-1},\dots,x_1,c)}$ and the corresponding expected benefit when GP is applied from t + 1 onwards is

$$ Q_t^{GP}({\mathbb{X}}_{t+1},p_{t+1}) = p_{t+1} \sum\limits_{i=0}^{\infty}\left((\delta e^\mu)^{in} (\bar{x}+x_n-c) + \sum\limits_{j=0}^{n-2}(\delta e^\mu)^{in+j} x_{n-j-1}+(\delta e^{\mu})^{in+n-1}c \right). $$

Inserting V = Q ^GP into the right hand side of the Bellman equation (15), the argument of the max operator is

$$ \begin{aligned} \Upphi(c)&= p_t c+ \delta {\mathbb{E}}_{|p_t} \left[p_{t+1} \sum\limits_{i=0}^{\infty}\left((\delta e^\mu)^{in} (\bar{x}+x_n-c) + \sum\limits_{j=0}^{n-2}(\delta e^\mu)^{in+j} x_{n-j-1}+(\delta e^{\mu})^{in+n-1}c \right) \right] \\ &= p_tc + p_t \sum\limits_{i=0}^{\infty}\left((\delta e^\mu)^{in+1} (\bar{x}+x_n-c) + \sum\limits_{j=0}^{n-2}(\delta e^\mu)^{in+j+1} x_{n-j-1}+(\delta e^{\mu})^{in+n}c \right). \end{aligned} $$

(19)

The coefficient affecting c in $\Upphi(c)$ is

$$ \begin{aligned} \hbox{coeff}(c)&=p_t \left(1-\sum\limits_{i=0}^\infty (\delta e^\mu)^{in+1}+ \sum\limits_{i=0}^\infty (\delta e^\mu)^{in+n} \right) = p_t \left( \sum\limits_{i=0}^\infty (\delta e^\mu)^{in} -\sum\limits_{i=0}^\infty (\delta e^\mu)^{in+1}\right)\\ &=(1-\delta e^\mu) \sum\limits_{i=0}^\infty (\delta e^\mu)^{in}>0. \end{aligned} $$

As coeff(c) > 0, the maximum in (15) is attained when ${c=CA\mathbb{X}_t=\bar{x}+x_n}$. Inserting this value of c in (19) yields

$$ \begin{aligned} \Upphi(\bar{x}+x_n) &= p_t\left( (\bar{x}+x_n) + \sum\limits_{i=0}^{\infty}\left( \sum\limits_{j=0}^{n-2}(\delta e^\mu)^{in+j+1} x_{n-j-1}+(\delta e^{\mu})^{in+n}(\bar{x}+x_n) \right)\right) \\ &= p_t \sum\limits_{i=0}^{\infty}\left((\delta e^{\mu})^{in}(\bar{x}+x_n)+ \sum\limits_{j=1}^{n-1}(\delta e^\mu)^{in+j} x_{n-j} \right) = Q^{GP}({\mathbb{X}}_t,p_t), \end{aligned} $$

showing that (15) holds and that the GP is optimal.

The proof for the finite horizon case is very similar but more involved. Indeed, let t be expressed as T − (kn + l), where $k=\lfloor \frac{T\,-\, t}{n}\rfloor$ and l takes one value in $\{0,\dots,n-1\}$. This way of expressing t puts in evidence that after completing k cycles there will be still $l\in [0,\dots,n-1]$ time steps to go until reaching the end of the horizon. The expected benefit associated with GP is

$$ Q^{GP}_t({\mathbb{X}}_t,p_t)= p_t\left[\sum\limits_{i=0}^{k-1}\delta^{in}e^{in\mu}\left(\bar{x} +\sum\limits_{j=0}^{n-1}\delta^j e^{j\mu}x_{n-j}\right) + \delta^{kn}e^{kn\mu} \left(\bar{x} +\sum\limits_{j=0}^l \delta^j e^{j\mu}x_{n-j} \right) \right]. $$

Again, we need to check that Q ^GP satisfies (15) and that the maximum is attained for ${c=CA\mathbb{X}_t}$. To this end, it is convenient to divide the study into two cases, depending on the value of l: (a) l > 0 and (b) l = 0 (see the proof of Theorem 2). In both cases, the verification of (15) follows exactly the same lines that the proof in the infinite horizon case. We leave the details to the reader. □

Proof of Lemma 2

Given any initial state ${\mathbb{X}=(\bar{x},x_n,\dots,x_1)}$, it is obvious that al least one of its coordinates must be strictly positive. Without loss of generality, we assume that x _n  > 0. We compare the value function with the benefit delivered by the GP (see (18)) to obtain,

$$ V_t({\mathbb{X}},p_t)\geq Q_t^{GP}({\mathbb{X}},p_t) \geq p_t \sum\limits_{i=0}^\infty (\delta e^{\mu})^{in}x_n=\infty. $$

□

Proof of Theorem 2

To prove that the AP is optimal, we check that the expected benefit associated with it (Q ^AP) satisfies the Bellman equation (15). After some computations we can prove that if ${\mathbb{X}_t=(\bar{x},x_n,\dots,x_1)^T}$ and t = T − (kn + j), with $k=\lfloor \frac{T\,-\,t}{n}\rfloor$ and l takes one value in $\{0,\dots,n-1\}$, the total actualized benefit is

$$ \begin{aligned} Q^{AP}_t({\mathbb{X}}_t,p_t)&={\mathbb{E}}_{|p_t}\left[\delta^l p_{t+l}\left(\bar{x}+\sum\limits_{j=0}^l x_{n-j}\right) + \sum\limits_{i=1}^k \delta^{in+l} p_{t+in+l}S \right]\\ &= p_t\left[(\delta e^\mu)^l \left(\bar{x}+\sum\limits_{j=0}^l x_{n-j}\right) + \sum\limits_{i=1}^k (\delta e^\mu)^{in+l}S \right]. \end{aligned} $$

(20)

For the rest of the proof we divide the study into two cases depending on the value of l: (a) l > 0 and (b) l = 0.

(a) Case t = T − (kn + l), with l > 0. Here, t + 1 = T − (kn + l − 1) with $l-1=0,\dots,n-2$. Denoting c = c _t , we get that ${Q^{AP}_{t+1}(A\mathbb{X}_t+Bc, p_{t+1})}$ can be expressed as

$$ Q^{AP}_{t+1}(A{\mathbb{X}}_t+Bc,p_{t+1}) = p_{t+1}\left[ (\delta e^\mu)^{l-1} \left(\bar{x}+x_n-c+\sum\limits_{j=0}^{l-1} x_{n-j-1}\right) + \sum\limits_{i=1}^k (\delta e^\mu)^{in+j-1}S \right]. $$

Inserting V = Q ^AP into the right-hand side of the Bellman equation (15), the argument of the max operator is

$$ \begin{aligned} \Upphi(c) &= p_tc+\delta{\mathbb{E}}_{|p_t}\left[ -p_{t+1}\left[(\delta e^\mu)^{l-1} \left(\bar{x}+x_n-c+\sum\limits_{j=0}^{l-1} x_{n-j-1}\right) + \sum\limits_{i=1}^k (\delta e^\mu)^{in+l-1}S \right] \right] \\ &=p_tc-p_t\left[(\delta e^\mu )^{l} \left(\bar{x}+x_n-c+\sum\limits_{j=0}^{l-1} x_{n-j-1}\right) + \sum\limits_{i=1}^k (\delta e^\mu )^{in+l}S \right] \\ &= p_tc[1-(\delta e^\mu)^{j}]+Q^{AP}_t({\mathbb{X}}_t,p_t). \end{aligned} $$

As the coefficient of c is negative, the maximum is attained when c = 0 and $\Upphi(0)$ is exactly ${Q^{AP}_t(\mathbb{X}_t,p_t)}$, showing that the Bellman equation (15) holds.

(b) Case t = T − kn. In this case, we have t + 1 = T − [(k − 1)n + n − 1)] and denoting c = c _t Q ^AP _t+1 is expressed as

$$ \begin{aligned} Q^{AP}_{t+1}(A{\mathbb{X}}_t+Bc,p_{t+1})& = p_{t+1}\left[(\delta e^\mu )^{n-1} \left(\bar{x}+x_n-c+\sum\limits_{l=0}^{n-2} x_{n-l-1}+c\right) + \sum\limits_{i=1}^{k-1} (\delta e^\mu )^{in+n-1}S \right] \\ &= p_{t+1}\left[(\delta e^\mu )^{n-1} S + \sum\limits_{i=2}^{k} (\delta e^\mu )^{in-1}S \right] = p_{t+1} \sum\limits_{i=1}^{k} (\delta e^\mu )^{in-1}S. \end{aligned} $$

Inserting V = Q ^AP into the right-hand side of the Bellman equation (15), the argument of the max operator is

$$ \Upphi(c)= p_tc+\delta {\mathbb{E}}_{|p_t} \left[ p_{t+1} \sum\limits_{i=1}^{k} (\delta e^\mu )^{in-1}S\right]. $$

The coefficient of c is positive, and thus, the maximum is attained when $c=\bar{x}+x_n$. So we have,

$$ \Upphi(\bar{x}+x_n)= p_t(\bar{x}+x_n) +p_t \sum\limits_{i=1}^{k} (\delta e^\mu )^{in}S. $$

The right hand side is exactly (20) when j = 0, hence we have ${\Upphi(\bar{x}+x_n)=Q^{AP}_t(\mathbb{X}_t,p_t)}$ and the Bellman equation is satisfied.

In both cases, we have shown that $Q^{AP}_t(\cdot,\cdot)$ satisfies the Bellman equation, hence it is the value function and the proposed policy is optimal.□

Before presenting the proof of Theorem 3 we present a technical proposition that will be used in the proof of the theorem.

Proposition 1

Let $r_{mj}=1-\frac{1-\delta^j}{e^{-m\eta}(1-\delta^j e^{-j\eta})}$ for ${m, j\in \mathbb{N}}$ with $\delta\in (0,1)$ and η > 0. Then ⁴

$$ \frac{\delta(1-e^{-\eta})}{(1-\delta e^{-\eta})}= r_{01} \geq r_{mj},\quad \hbox{for all}\, m\geq0,\, j\geq1 $$

with equality only if m = 0 and j = 1.

Proof

First observe that

$$ r_{01}=1-\frac{1-\delta}{1-\delta e^{-\eta}}=\frac{\delta(1-e^{-\eta})}{1-\delta e^{-\eta}}. $$

Hence the following are equivalent,

$$ \begin{array}{rrcl} &r_{01} &\geq & r_{mj} \\ \Longleftrightarrow: & \frac{1-\delta}{1-\delta e^{-\eta}} &\leq & \frac{1-\delta^{j}}{e^{-m\eta}(1-\delta^{j} e^{-j\eta})}\\ \Longleftrightarrow: & e^{-m\eta} \frac{1-\delta^j e^{-j\eta}}{1-\delta e^{-\eta}} &\leq & \frac{1-\delta^{j}}{1-\delta}\\ \Longleftrightarrow: & e^{-m\eta} \sum\nolimits_{i=0}^{j-1}(\delta e^{-\eta})^i &\leq & \sum\nolimits_{i=0}^{j-1}\delta^i. \end{array} $$

Given that e ^−η < 1 the last inequality is always valid.

The equality can only hold if m = 0, which implies that e ^−mη = 1, and j = 1, which gives that the two sums comprise only one term and are equal to one. □

Proof of Theorem 3

We present a proof under the assumption that at every step we either harvest nothing at all or everything available. The result is still valid without this assumption, but we decided not to present the general proof for two reasons: (i) the general proof follows the same lines that the one presented here, but the calculus are more cumbersome; (ii) due to the linearity of the forestry model, this assumption is equivalent to requiring that the coefficient of c in (15) is never zero, but having a zero coefficient is an event with zero probability.

The main idea of the proof is to consider the role played by c in all the possible expressions of $V_{t}(\cdot,\cdot)$. Of course, characterizing completely every possible expression of $V_{t}(\cdot,\cdot)$ is a titanic task, but we will only be interested in the coefficient affecting c.

After harvesting c _t  = c the state ${\mathbb{X}_{t+1}}$ is

$$ {\mathbb{X}}_t=\left(\begin{array}{c}\bar{x}_t\\ x_{n,t}\\ x_{n-1,t}\\ \vdots\\ x_{1,t} \end{array}\right) \longrightarrow {\mathbb{X}}_{t+1}=A{\mathbb{X}}_t+Bc=\left(\begin{array}{c}\bar{x}_t+ x_{n,t}- c\\ x_{n-1,t}\\ x_{n-2,t}\\ \vdots\\ c \end{array}\right), $$

and the resulting Bellman equation (15) can be stated in terms of ${\mathbb{X}_t}$ and c as follows,

$$ V_t({\mathbb{X}}_t,p_t)=\max\limits_c \{p_tc + \delta {\mathbb{E}}_{|p_t} [V_{t+1}(A{\mathbb{X}}_{t}+Bc,p_{t+1})]\}. $$

To simplify the notation we denote $\Upphi(c)$ the argument of the maximum, this is,

$$ \Upphi(c)=p_tc + \delta {\mathbb{E}}_{|p_t} [V_{t+1}(A{\mathbb{X}}_{t}+Bc,p_{t+1})]. $$

From t + 1 on, two different situations must be considered: (i) nothing is harvested in the next n steps or (ii) the first harvest occurs at t + j ₀ with $j_0\in\{1,\dots,n-1\}$.

In case (i), the state at t + n will be

$$ {\mathbb{X}}_{t+n}=(S-c, c, 0, \dots\dots, 0)^T. $$

It is easy to see that the influence of c extinguishes as the constraint on c _t+n is c _t+n  ≤ S. We do not know the complete expression of $V_{t+1}(\cdot,\cdot)$ but we do know that the coefficient of c in $\Upphi(c)$ will be simply p _t .

If on the contrary, we are in case (ii), there will be harvest at t + j ₀ with $j_0\in\{1,\dots,n-1\}$ and a term ${\delta^{j_0}\mathbb{E}_{|p_t}[- p_{t+j_0}]}$ will be added to the coefficient of c in $\Upphi(c)$,

$$ \Upphi(c)=p_tc +\delta^{j_0} {\mathbb{E}}_{|p_t}[ p_{t+j_0}(\bar{x}_t+x_{n,t}+\dots+ x_{n-j_0,t}-c) +\delta V_{t+j_0+1}(\cdot,\cdot)]. $$

We are left with the task of characterizing the coefficient of c in $V_{t+j_0+1}(\cdot,\cdot)$. We know that c only affects two coordinates of the state ${\mathbb{X}_{t+j_0+1}}$, and these two coordinates are: $x_{t+j_0+1,j_0+1}=c$ and $x_{t+j_{0}+1,1}=\bar{x}_t+x_{n,t}+\dots+ x_{n-j_{0},t}-c$. From now on, we will omit the state variables and will represent the state as

$$ {\mathbb{X}}_{t+j_0+1}= ( 0, \ast, \dots\dots, \ast, c, \ast,\dots\dots, \ast ,\ast -c )^T. $$

(21)

Until $x_{t+j_{0}+1,j_{0}+1}=c$ becomes available for harvesting, the coefficient of c in $\Upphi(c)$ will not be affected by any of the actions taken. These trees will reach maturity exactly at t + n, when the state will be

$$ {\mathbb{X}}_{t+n}=(\ast, c ,\ast ,\dots\dots, \ast ,\ast -c,\dots\dots, \ast)^T, $$

with $x_{t+n,n-j_0}=\ast-c$. Once we reach this time step, two different situations must be considered: (i) nothing is harvested until we reach the time step t + n + j ₀ or (ii) the first harvest occurs at t + m ₁ with $m_1\in\{n,\dots,n+j_0-1\}$.

In case (i), we would have $\bar{x}_{t+n+j_0}=\ast+c$ and $x_{t+n+j_0,n}=\ast-c$ and the influence of c vanishes. We know that the coefficient of c in $\Upphi(c)$ will be simply ${p_t + \delta^{j_0}\mathbb{E}_{|p_t}[- p_{t+j_0}]}$.

In case (ii), there is harvest at t + m ₁ and a new term is added to the coefficient of c in $\Upphi(c)$: ${\delta^{m_1}\mathbb{E}_{|p_t}[p_{t+m_1}]}$. After the harvest, the state will be

$$ {\mathbb{X}}_{t+m_1+1}=(0, \ast,\dots\dots, \ast -c ,\ast ,\dots\dots, \ast ,\ast +c)^T. $$

Again, there will be some time steps where coefficient of c in $\Upphi(c)$ will not be affected by any of the actions taken. In fact, until the fraction of trees $\ast-c$ reaches maturity no new terms of the coefficient are generated.This will happen at t + n + j ₀, when the state would be in this case

$$ {\mathbb{X}}_{t+j_0+n}=(\ast,\ast -c ,\ast ,\dots\dots, \ast ,\ast +c,\dots\dots, \ast)^T. $$

As before, two different situations must be considered: (i) nothing is harvested until we reach the time step t + m + n or (ii) the first harvest occurs at t + m ₁ + j ₁ with j ₁ < n.

In case (i), we will have $\bar{x}_{t+m+n}=\ast-c$ and $x_{t+m+n,n}=\ast+c$ and the influence of c extinguishes. The coefficient of c in $\Upphi(c)$ will be simply: ${p_t + \delta^{j_0}\mathbb{E}_{|p_t}[- p_{t+j_0}]+ \delta^{m_1}\mathbb{E}_{|p_t}[p_{t+m_1}]}$.

In case (ii), there is harvest at t + m ₁ + j ₁ and the term ${\delta^{m_1+j_1}\mathbb{E}_{|p_t}[-p_{t+m_1+j_1}]}$ is added to the coefficient of c.

Observe that the state at the following step (t + m ₁ + j ₁ + 1) is

$$ {\mathbb{X}}_{t+m_1+j_1+1}=(0, \ast,\dots\dots, \ast +c ,\ast ,\dots\dots, \ast ,\ast -c)^T. $$

The state at t + m ₁ + j ₁ + 1 is of the same form that the state at t + j ₀ + 1 (see (21)) and the same reasoning can be applied over and over. In each completed cycle first the term “$\ast+c$” is harvested, the index m _i is generated, with m _i  ≥ m _i-1 + n, and the term ${\delta^{m_i}\mathbb{E}_{|p_t}[p_{t+m_i}]}$ is added to the coefficient of c in $\Upphi(c).$ Secondly,the term “$\ast-c$” is harvested, the index j _i is generated, with 1 ≤ j _i  < n and the term ${\delta^{m_i+j_i}\mathbb{E}_{|p_t}[-p_{t+m_i+j_i}]}$ is added to the coefficient.

The algorithm ends either at the end of the interval or before that if we reach a state of the form

$$ {\mathbb{X}}(t+m_i+n)=\left({\begin{array}{c} \ast-c\\ \ast +c \\ \vdots\\ \vdots\\ \vdots \end{array}}\right) \,\hbox{or}\,\, {\mathbb{X}}(t+m_i+j_i+n)= \left({\begin{array}{c} \ast+c\\ \ast -c \\ \vdots\\ \vdots\\ \vdots \end{array}}\right) $$

and the influence of c extinguishes as ${CA\mathbb{X}(\cdot)}$ is independent of c.

The coefficient of c is the finite sum of terms of the form

$$ \delta^{m_i}{\mathbb{E}}_{|p_t}[p_{t+m_i}]-\delta^{m_i+j_i}{\mathbb{E}}_{|p_t}[p_{t+m_i+j_i}]. $$

(22)

plus possibly one positive term ${\delta^{m_i}\mathbb{E}_{|p_t}[p_{t+m_i}]}$. The proof is completed by showing that the coefficient of c is positive. To this end, we study the sign of expression (22) and its relation with condition (17),

$$ \begin{aligned} 0&\leq \delta^{m_i}{\mathbb{E}}_{|p_t}[p_{t+m_i}]+\delta^{m_i+j_i}{\mathbb{E}}_{|p_t}[-p_{t+m_i+j_i}]\\ \Longleftrightarrow: 0&\leq {\mathbb{E}}_{|p_t}[p_{t+m_i}]+\delta^{j_i}{\mathbb{E}}_{|p_t}[-p_{t+m_i+j_i}]\\ &= e^{-m_i\eta}p_t + (1-e^{-m_i\eta} )\bar{p}- \delta^{j_i} [e^{-(m_i+j_i)\eta}p_t + (1-e^{-(m_i+j_i)\eta} )\bar{p}] \\ &= e^{-m_i\eta}p_t(1-\delta^{j_i}e^{-j_i\eta}) + \bar{p} [ 1-\delta^{j_i}-e^{-m_i\eta}(1-\delta^{j_i}e^{-j_i\eta})] \\ \Longleftrightarrow: \frac{p_t}{\bar{p}}&\geq 1-\frac{1-\delta^{j_i}}{e^{-m_i\eta}(1-\delta^{j_i}e^{-j_i\eta})}. \end{aligned} $$

But the right hand side above is exactly $r_{m_{i}j_{i}}$ of Proposition 1. This proposition together with Condition (17) imply

$$ \frac{p_t}{\bar{p}} \geq r_{01} \geq r_{m_ij}, $$

which proves that (22) is non-negative. Furthermore, we know that the second inequality holds strictly unless m _i  = 0 and j = 1. Hence, the positivity of the coefficient of c in (15) follows for every case, except only for the case where it consists exclusively of one term of the form (22) with m _i  = 0 and j = 1, i.e.,

$$ p_t-{\mathbb{E}}_{|p_t}[\delta p_{t+1}]. $$

(23)

From the previous calculus it is direct to see that (23) is positive if (17) holds strictly. If, on the contrary, (17) holds with equality, there is no influence of c in the value of ${V_t(\mathbb{X}_t,p_t)}$ and we can freely chose the value of c provided that it is feasible. We impose ${c_t=CA\mathbb{X}_t}$, which concludes the proof. □