On Stochastic Fishery Games with Endogenous Stage-Payoffs and Transition Probabilities

Abstract

We engineered a stochastic fishery game in which overfishing has a twofold effect: it gradually damages the fish stock inducing lower catches in states High and Low, and it gradually causes the system to spend more time in the latter state with lower landings.

To analyze the effects of this ‘double whammy’ technically, we demonstrate how to determine the set of jointly-convergent pure-strategy rewards supported by equilibria involving threats, under the limiting average reward criterion.

A Appendix

Validation of the claim in Example 3 . A quick and dirty preliminary analysis yielded a first candidate, namely the strategy pair where Player B always plays his second action in High and the first action in Low; Player A always plays his second action. So, A is the player punishing his opponent, and all calculations will induce a threat point reward for B and symmetry then implies that the same reward is the threat point reward for A, as well.

We start with the minimization program. We reduce the four dimensional system to a two dimensional one as follows. We set $q_{2}=qx, q_{4}=q(1-x), q_{5}=(1-q)y, q_{7}=(1-q)(1-y)$ where $0\le x,y,q\le 1.$ The interpretation is that to minimize his opponents rewards even further, A may be allowed to shift an arbitrary weight from the original $q_{4}=1$ to $q_{2},$ hence after the shift we have $q_{2}=qx,q_{4}=q(1-x).$ Similarly, A may shift weight from $q_{7}=1$ to $q_{5},$ such that after the shift we have $q_{5}=(1-q)y, q_{7}=(1-q)(1-y).$

Fig. 3.

The upper manifold represents all rewards to B possible if B plays the fixed strategy: Right (2) in High, left (1) in Low. The horizontal hyperplane indicates level $v_{4,7}=4.4588.$ This justifies the conclusion that B’s rewards are minimized at $x=y=0,$ i.e., Player A should play bottom (2) in both states forever. Partial derivatives are positive on $[0,1]\times [0,1]$.

Fig. 4.

The lower manifold represents all rewards to B on the interval $[0,1]\times$ [0, 1], if Player A plays a fixed strategy of bottom (2) in both states. The horizontal hyperplane indicates level $v_{4,7}.$ Clearly the highest rewards are to be found at $x=y=0$ which means that B should use his right action (2) in High and his left action (1) in Low. It is also easy to see that the partial derivatives are negative on the same interval.

So, the variables to minimize over are x and y, q will result from the analysis of the balance Eq. (1) for given x, y using this transformation of the four original variables. The transformed transition probabilities are

$$\begin{aligned} \begin{array}{l} p_{2}=0.3y+0.4-0.3q\left( 1-x+y\right) \\ p_{4}=0.25y+0.35-0.25q\left( 1-x+y\right) \\ p_{5}=0.2y+0.3-0.2q\left( 1+x-y\right) \\ p_{7}=0.15y+0.25-0.15q\left( 1-x+y\right) \end{array} \end{aligned}$$

The transformed balance equation for this setting is

$$\begin{aligned} \begin{array}{l} 0=q^{2}\left( -0.05x^{2}-0.3xy-0.05x+0.35y^{2}+0.05y+0.1\right) + \\ q\left( 0.3xy-0.2x-0.3y^{2}+0.15y+1.05\right) +\left( -0.05y^{2}-0.2y-0.25\right) \end{array} \end{aligned}$$

Hence, the minimization problem reduces and simplifies to the following.

The other root of the balance equation is real too, but yields nonsense.

There are alternative options to proceed now, but we show a plot of all rewards to B as a function of $x,y\in [0,1]\times [0,1]$ under the conditions of the minimization problem. Figure 3 shows immediately that the rewards of B are minimized for $x=y=0.$

Proceeding in the same fashion for the maximization problem, we generate Fig. 4 showing that maximizes B’s rewards for $x=y=0.$