The stochastic Mitra–Wan forestry model: risk neutral and risk averse cases

Abstract

We extend the classic Mitra and Wan forestry model by assuming that prices follow a geometric Brownian motion. We move one step further in the model with stochastic prices and include risk aversion in the objective function. We prove that, as in the deterministic case, the optimal program is periodic both in the risk neutral and risk averse frameworks, when the benefit function is linear. We find the optimal rotation ages in both stochastic cases and show that they may differ significantly from the deterministic rotation age. In addition, we show how the drift of the price process affects the optimal rotation age and how the degree of risk aversion shortens it. We illustrate our findings for an example of a biomass function and for different values of the model’s parameters.

Appendix

Proof of Lemma 1.   Given any program $\{{\varvec{x}}_t\}$ (not necessarily optimal) and $\{c_t\}$ the corresponding sequence of harvesting, we consider the finite horizon value of problem (3):

$$\begin{aligned} Q_T(\{{\varvec{x}}_t\})\!=\!- p_{0}c_{0} \!+\! \delta \rho _{|p_{0}}[ -p_{1}c_{1} + \ldots +\delta \rho _{|p_{T-2}}[-p_{T-1}c_{T-1} + \delta \rho _{|p_{T-1}}[-p_Tc_T ]]] \end{aligned}$$

Due to the fact that $p_tc_t \ge 0$ for all $t$ (when prices follow a GBM) and the monotonicity of any coherent risk measure we know that $Q_T(\{{\varvec{x}}_t\})\ge Q_{T+1}(\{{\varvec{x}}_t\})$, hence the sequence $Q_T(\{{\varvec{x}}_t\})$ either converges to the limit or diverges to $-\infty$ when $T\rightarrow \infty$. We prove now that $Q_T(\cdot )$ is bounded below for all $T$ and therefore it has a limit. If $\rho =\mathbb E$, we have $\rho _{p_{t-1}}[-p_tc_t]=-e^\mu p_{t-1}c_t$ and hence,

$$\begin{aligned} Q_T(\{{\varvec{x}}_t\})&= - p_{0}c_{0} + \delta \rho _{|p_{0}}[ -p_{1}c_{1} + \ldots +\delta \rho _{|p_{T-2}}[-p_{T-1}(c_{T-1} - \delta e^{\mu } c_T) ]]\\&\vdots \\&= -p_0 \sum _{t=0}^T (\delta e^\mu )^t c_t \ge -p_0 \sum _{t=0}^T (\delta e^\mu )^t = -p_0 \frac{1-(\delta e^\mu )^{T+1}}{1-\delta e^\mu } \end{aligned}$$

where the inequality follows because $c_t\le 1$ for all $t$. If $\delta e^\mu <1$ we get $Q_T> -p_0 \frac{1}{1-\delta e^\mu } >-\infty$ for all $T$. This implies that the sequence $Q_T$ converges when $T$ goes to infinity. This limit is the value associated with program $\{{\varvec{x}}_t\}$ in the infinite time horizon formulation. As the bound does not depend on the particular program we get that the minimal value is also well defined and finite.

When $\rho =\text{ CVaR }_{\alpha }$ the proof is the same, substituting $\delta e^\mu$ by $\delta e^\mu \kappa$.

Proof of Theorem 1. We denote by $\{{\varvec{x}}^*_t\}$ and $\{c^*_t\}$ (resp. $\{{\varvec{x}}_t\}$ and $\{c_t\}$ the program and the harvesting sequence generated by the proposed optimal policy (resp. any alternative optimal policy) from the initial state ${\varvec{x}}^o$.

We treat first the risk neutral case. We let $r$ represent $\delta e^\mu$ and define the auxiliary constants

$$\begin{aligned} \tau _a=\frac{r^a}{1-r^a} ~~\text {with } a=1\dots , n \end{aligned}$$

(13)

and where

$$\begin{aligned} {\varvec{q}}=\Bigg (0,\frac{f_{\theta _{\mathbb {E}}}\tau _{\theta _{\mathbb {E}}}}{\tau _1},\frac{f_{\theta _{\mathbb {E}}}\tau _{\theta _{\mathbb {E}}}}{\tau _2},\dots \dots , \underbrace{\frac{f_{\theta _{\mathbb {E}}}\tau _{\theta _{\mathbb {E}}}}{\tau _{{\theta _{\mathbb {E}}}-1}}}_{{\theta _{\mathbb {E}}}\text{- }th~coord.},f_{\theta _{\mathbb {E}}},\dots \dots ,f_{n} \Bigg ). \end{aligned}$$

We recall that $n$ is the age after which die, hence, it is suboptimal to let trees grow beyond $n$ and without loss of generality, we assume that $c^n_t=x^n_t$ for all $\{x_t\}$. To simplify the calculation below we add the auxiliary $(n+1)-$ coordinate to the state vector ${\varvec{x}}_t$, assuming that $x_{n+1,t}=0$ for all $t$.

Let us first show that

$$\begin{aligned} c_t +\sum _{a=1}^n q_ax_{a,t+1} - \frac{1}{r}\sum _{a=1}^n q_ax_{a,t}\le \sum _{a=1}^{n}q_{a+1}\left( x_{a,t}-\frac{x_{a+1,t}}{r}\right) \qquad \forall t, \end{aligned}$$

(14)

with equality iff $\{{\varvec{x}}_t\}=\{{\varvec{x}}^*_t\}$ and $\{c_t\}=\{c^*_t\}$. We have

$$\begin{aligned}&c_t + \sum _{a=1}^n q_ax_{a,t+1} - \frac{1}{r}\sum _{a=1}^n q_ax_{a,t}~=~c_t + \sum _{a=1}^n q_a\left( x_{a,t+1} - \frac{1}{r}x_{a,t}\right) \\&= \sum _{a=1}^{n} f_a(x_{a,t}-x_{a+1,t+1}) + \sum _{a=1}^{{\theta _{\mathbb {E}}}-1} \frac{f_{\theta _{\mathbb {E}}} \tau _{\theta _{\mathbb {E}}}}{\tau _a} \left( x_{a+1,t+1}-\frac{x_{a+1,t}}{r}\right) \\&+\sum _{a={\theta _{\mathbb {E}}}}^{n-1}f_a\left( x_{a+1,t+1}-\frac{x_{a+1,t}}{r}\right) \\&= \sum _{a=1}^{{\theta _{\mathbb {E}}}-1} \left[ f_a(x_{a,t}-x_{a+1,t+1}) +\frac{f_{\theta _{\mathbb {E}}}\tau _{\theta _{\mathbb {E}}}}{\tau _a}\left( x_{a+1,t+1} \!-\!x_{a,t} \!+\! x_{a,t}\!-\!\frac{x_{a+1,t}}{r} \right) \right] \\&+\sum _{a={\theta _{\mathbb {E}}}}^{n}f_a\left( x_{a,t}\!-\!\frac{x_{a+1,t}}{r}\right) \\&= \sum _{a=1}^{{\theta _{\mathbb {E}}}-1}\left[ \frac{1}{\tau _a}(f_a\tau _a-f_{\theta _{\mathbb {E}}}\tau _{\theta _{\mathbb {E}}})(x_{a,t}-x_{a+1,t+1}) +\frac{f_{\theta _{\mathbb {E}}}\tau _{\theta _{\mathbb {E}}}}{\tau _a}\left( x_{a,t}\!-\! \frac{x_{a+1,t}}{r}\right) \right] \\&+\sum _{a={\theta _{\mathbb {E}}}}^{n}f_a\left( x_{a,t}\!-\!\frac{x_{a+1,t}}{r}\right) \\&= \sum _{a=1}^{{\theta _{\mathbb {E}}}-1} \frac{1}{\tau _a}(f_a\tau _a-f_{\theta _{\mathbb {E}}}\tau _{\theta _{\mathbb {E}}})(x_{a,t}-x_{a+1,t+1}) +\sum _{a=1}^{n}q_{a+1}\left( x_{a,t}-\frac{x_{a+1,t}}{r}\right) \end{aligned}$$

The first sum is always less than or equal to zero, with equality if and only if $x_{a,t}=x_{a+1,t+1}$ for $a<{\theta _{\mathbb {E}}}$ which is equivalent to $c_{a,t}=0$ for all $a<{\theta _{\mathbb {E}}}$. Hence, we have obtained (14).

We focus now in getting a bound independent of ${\varvec{x}}_t$. We claim that

$$\begin{aligned} \sum _{a=1}^{n}q_{a+1}\left( x_{a}-\frac{x_{a+1}}{r}\right) \le \frac{f_{\theta _{\mathbb {E}}}\tau _{\theta _{\mathbb {E}}}}{\tau _1}, \end{aligned}$$

(15)

for every state ${\varvec{x}}$, with equality if and only if $x_a=0$ for all $a>{\theta _{\mathbb {E}}}$. Indeed,

$$\begin{aligned}&\sum _{a=1}^{n}q_{a+1}\left( x_a-\frac{x_{a+1}}{r}\right) ~=~ \sum _{a=1}^{n}q_{a+1}x_a-\sum _{a=2}^{n+1}q_{a}\frac{x_{a}}{r}\\&= q_2x_1+\sum _{a=2}^{n}\left( q_{a+1}-\frac{q_{a}}{r}\right) x_a\\&= \frac{f_{\theta _{\mathbb {E}}}\tau _{\theta _{\mathbb {E}}}}{\tau _1}x_1 + \sum _{a=2}^{{\theta _{\mathbb {E}}}}f_{\theta _{\mathbb {E}}}\tau _{\theta _{\mathbb {E}}} \left( \frac{1}{\tau _{a}}-\frac{1}{r\tau _{a-1}}\right) x_a + \sum _{a={\theta _{\mathbb {E}}}+1}^{n}\left( f_{a}-\frac{f_{a-1}}{r}\right) x_a\\&= \sum _{a=1}^{{\theta _{\mathbb {E}}}}\frac{f_{\theta _{\mathbb {E}}}\tau _{\theta _{\mathbb {E}}}}{\tau _1}x_a + \sum _{a={\theta _{\mathbb {E}}}+1}^{n}\left( f_{a}-\frac{f_{a-1}}{r}\right) x_a, \end{aligned}$$

where in the last step we use the following equality:

$$\begin{aligned} \frac{1}{\tau _a}-\frac{1}{r\tau _{a-1}}=\frac{1}{\tau _1}, \end{aligned}$$

(16)

which follows from (13). To finish the proof of (15) we only need to show that

$$\begin{aligned} f_a-\frac{f_{a-1}}{r}\le \frac{f_{{\theta _{\mathbb {E}}}}\tau _{{\theta _{\mathbb {E}}}}}{\tau _1} \text{ for } \text{ all } a>\theta _{\mathbb {E}}. \end{aligned}$$

(17)

This is evidently true if the left hand side is negative, hence we will only deal with the cases where $f_a-\frac{f_{a-1}}{r}>0$. This means, in particular, that we only need to consider values of $a$ such that $f_{a}-f_{a-1}>0$. Let $a_{M}$ denote $\arg \max {f_a}$. Thanks to Assumption 1 we know that

$$\begin{aligned}&f_{n}-{f_{n-1}}<\dots <f_{a_{M}+1}-f_{a_{M}}<0<f_{a_{M}}- f_{a_{M}-1}<\dots < f_2-f_1\\&\Rightarrow 0<f_a-{f_{a-1}}< f_{a-1}-{f_{a-2}}< \frac{f_{a-1}-f_{a-2}}{r}~~~~~\forall a\le a_{M}\\&\Rightarrow f_a-\frac{f_{a-1}}{r}<f_{a-1}-\frac{f_{a-2}}{r} ~~~~~\forall a\le a_{M}. \end{aligned}$$

Using (16) and the fact that $f_{\theta _{\mathbb {E}}}\tau _{\theta _{\mathbb {E}}}>f_{\theta _{\mathbb {E}}+1}\tau _{\theta _{\mathbb {E}}+1}$ we obtain

$$\begin{aligned} f_{\theta _{\mathbb {E}}+1}-\frac{f_{\theta _{\mathbb {E}}}}{r}< \frac{f_{{\theta _{\mathbb {E}}}}\tau _{{\theta _{\mathbb {E}}}}}{\tau _1}, \end{aligned}$$

which finishes the proof of (17) as well as (15).

Putting (14) and (15) together we readily get

$$\begin{aligned} c_t +\sum _{a=1}^n q_ax_{a,t+1} - \frac{1}{r}\sum _{a=1}^n q_ax_{a,t}\le \frac{f_{{\theta _{\mathbb {E}}}}\tau _{{\theta _{\mathbb {E}}}}}{\tau _1} \qquad \forall t \end{aligned}$$

(18)

In summary, we have that (14) and (18) are fulfilled along any feasible program. Besides, along the proposed optimal program (14) is satisfied with equality for all $t$ and (18) is satisfied with equality for all $t\ge 2$.

To finish the proof we compare the benefits obtained by any program and the proposed optimal program up until some $T$. As prices follow a GBM we know that $\mathbb E_{|p_t}[p_{t+1}]=e^\mu p_t$ and we can write the objective function of (3) as with $r=\delta e^\mu$.

Using (14) to bound the difference corresponding to $t=1$ and (15) for the other terms we get

$$\begin{aligned} \sum _{t=1}^Tr^tp_tc_t-\sum _{t=1}^Tr^tp_tc^*_t \le r^{T-1}\sum _{a=1}^nq_a(x^*_{a,T}-x_{a,T})p_0 \end{aligned}$$

Given that the product $q'({\varvec{x}}^*_T-{\varvec{x}}_T)p_0$ is bounded for all $T$ and that obviously $r^{T-1}\rightarrow 0$ when $T\rightarrow \infty$ we get that

and hence the proposed harvesting policy is optimal.

The proof in the risk averse case follows analogously. Indeed, using that the conditional CVaR is positive homogeneous and that $\text {CVaR}_{p_t}[p_{t+1}]=e^\mu \kappa p_t$ when prices follow a GBM, taking $r=\delta e^\mu \kappa$, we can write the objective function as . Hence, it suffices to take $r=\delta e^{\mu }\kappa$ and substitute $\theta _{\mathbb {E}}$ by $\theta _{\rho }$ throughout the proof.

Proof of Theorem 2.   The function $F_r(a)$ is non-negative and twice differentiable in $[{a_0}, n]$. Its derivative is $F'_r(a)=\frac{r^a}{1-r^a}\left( f'(a)+\frac{f(a)\ln (r)}{1-r^a} \right)$ and, hence,

$$\begin{aligned} F'_r(a)=0\iff \frac{f'(a)}{f(a)}=\frac{\ln (1/r)}{1-r^{a}}. \end{aligned}$$

(19)

A priori, there might be several points satisfying (19). We claim that it is unique. Indeed, evaluating $F''_r$ in $a^*$ such that $F'_r(a^*)=0$ we get

$$\begin{aligned} F''_r(a^*)=\frac{r^{a^*}}{1-r^{a^*}}f''(a^*)-\frac{r^{a^*}\ln ^2(r)}{(1-r^{a^*})^2}~f(a^{*}) <0. \end{aligned}$$

It implies that every zero of the first derivative is a local maximum and, as there cannot be two local maxima without a minimum, we conclude that the first derivative has a unique zero. We denote this point as $a^*_r$ to point out its dependence with the parameter $r$.

To see the variation of $a^*_r$ with respect to $r$ let us consider (19). We know that $F'_r(a)$ is zero only at $a^*_r$ and that $F''_r(a^*_r)<0$, which implies that $F'_r(a)>0$ whenever $a<a^*(r)$. Hence, we know that the graph of $\frac{f'(a)}{f(a)}$ is above the one of $\frac{\ln (1/r)}{1-r^{a}}$ whenever $a<a^*(r)$. While the left hand side of (19) is independent of $r$, its right hand side is decreasing with respect to $r$. From Fig. 3 it is easy to see that $a^*_r$ is increasing with $r$.

Fig. 3

Monotonicity of $a^*(r)$  $(\text{ with } r<r')$