Solutions to the example questions on the Replicator Dynamics page. Each question is worth 25 marks.
(a) Fitnesses. [3]
(b) Replicator equation. [7]
(c) Stable populations. [3]
Setting \(\dot{x} = 0\) gives the stable populations \(x = 0\), \(x = 1\) and \(x = \tfrac{2}{3}\).
(d) Which is approached. [6]
For \(0 < x < \tfrac{2}{3}\) the factor \((2 - 3x) > 0\), so \(\dot{x} > 0\); for \(\tfrac{2}{3} < x < 1\) it is negative, so \(\dot{x} < 0\). From any interior start the dynamics therefore approach \(x = \tfrac{2}{3}\), while \(x = 0\) and \(x = 1\) are repelling. This is exactly where the class settled: about two-thirds of the room playing Dig.
(e) ESS. [6]
Yes, \(x = \tfrac{2}{3}\) is an ESS. It is an interior stable population to which the dynamics return: from part (d), \(\dot{x} > 0\) for \(x < \tfrac{2}{3}\) and \(\dot{x} < 0\) for \(x > \tfrac{2}{3}\), so the population returns to it after any small perturbation. Equivalently, by the invasion condition, at \(x = \tfrac{2}{3}\) the two strategies are equally fit (\(f_D = f_S\)). A small invasion that raises \(x\) above \(\tfrac{2}{3}\) makes \(f_D - f_S = 2 - 3x < 0\), so the now more numerous Dig type earns less and is selected against; an invasion that lowers \(x\) makes \(f_D > f_S\) and Dig recovers. Either way the invading mix earns a lower fitness than the resident, so it cannot spread: \(x = \tfrac{2}{3}\) is evolutionarily stable.
(a) Replicator equation. (Bookwork.) [3]
(b)(i) Interpretation. [2]
Each fitness is a base value of \(1\) (the payoff to not crashing) plus a term increasing in the proportion of others using the same side, since matching the majority lowers the chance of a collision.
(b)(ii) Replicator equation. [6]
(b)(iii) Stable populations. [4]
Setting \(\dot{x} = 0\) gives the stable populations \(x = 0\), \(x = 1\) and \(x = \tfrac{1}{2}\).
(b)(iv) Stability and ESS. [8]
Examining the sign of \(\dot{x} = x(1 - x)(2x - 1)\):
The stable populations \(x = 0\) (everyone drives right) and \(x = 1\) (everyone drives left) are approached and are the evolutionarily stable strategies; the mixed population \(x = \tfrac{1}{2}\) is a stable population in the sense \(\dot{x} = 0\) but is repelling, so it is not an ESS.
(b)(v) Two stable populations. [2]
The game has two stable populations the dynamics can settle on, \(x = 0\) and \(x = 1\), separated by the unstable population \(x = \tfrac{1}{2}\). Which side everyone ends up on is determined by history: whichever side starts in the majority is reinforced and takes over. Because \(x = \tfrac{1}{2}\) is unstable, any tiny imbalance grows, so the symmetric mixed population is never observed: a population settles on a single convention, all driving left or all driving right.
(a) Fitnesses and average fitness. [5]
Each fitness is the corresponding row of \(A x\):
The average fitness is
since \(A\) is antisymmetric and every term cancels in pairs.
(b) Replicator equations. [4]
With \(\phi = 0\), the replicator equation \(\dot{x}_i = x_i(f_i - \phi)\) reduces to
(c) Interior stable population. [5]
At an interior stable population every proportion is positive and each \(\dot{x}_i = 0\), so we need \(f_R = f_P = f_S = 0\). From \(f_R = 0\) we get \(x_S = x_P\), and from \(f_P = 0\) we get \(x_R = x_S\). Hence \(x_R = x_P = x_S\), and with \(x_R + x_P + x_S = 1\) this gives the unique interior stable population \(x_R = x_P = x_S = \tfrac{1}{3}\).
(d) A constant of motion. [7]
Along any interior trajectory,
Hence \(\ln H\), and so \(H = x_R x_P x_S\) itself, keeps the same value all the way along a trajectory. We can use this to see how the dynamics behave. Suppose we start in the interior, so all three proportions are positive and \(H > 0\). If one of the proportions ever dropped to zero, the product \(H = x_R x_P x_S\) would be zero too; but \(H\) cannot change, so this never happens and the trajectory stays away from the edges of the simplex. Equally, the dynamics cannot drift in to the centre and stop there, because the centre is the only fixed point and any other starting value of \(H\) is different from its value there.
The trajectory is therefore boxed in: it can never settle down (the only resting point is the centre, which it does not start at) and it can never reach the edges. With nowhere to go, it simply loops back on itself, tracing a closed curve around the centre \(\left(\tfrac{1}{3}, \tfrac{1}{3}, \tfrac{1}{3}\right)\) and repeating the same cycle forever. A small perturbation away from the centre neither grows nor shrinks; it just circulates. The interior population is therefore stable, in that nearby populations stay nearby, but not asymptotically stable, since the dynamics never return to it.
(e) Not an ESS. [4]
An evolutionarily stable strategy must be asymptotically stable under the replicator dynamics. From part (d) the interior population is only stable, not asymptotically stable, so \(\left(\tfrac{1}{3}, \tfrac{1}{3}, \tfrac{1}{3}\right)\) is not an ESS. This reflects the cyclic structure of the game: Rock beats Scissors, Scissors beats Paper, and Paper beats Rock. Whenever one strategy becomes common the strategy that beats it gains, so the population chases its own tail through Rock, Paper and Scissors without ever settling.
(a) The function is non-negative. [5]
Write \(V(x) = -\sum_i x^{*}_i \ln\!\dfrac{x_i}{x^{*}_i}\). Since \(\ln\) is concave and the weights \(x^{*}_i\) sum to one, Jensen's inequality gives
Hence \(V(x) \ge 0\). Equality in Jensen's inequality holds only when the ratios \(x_i/x^{*}_i\) are all equal; since both \(x\) and \(x^{*}\) sum to one this forces \(x = x^{*}\).
(b) Its rate of change. [9]
Differentiating along a trajectory, and using \(\sum_i x^{*}_i = 1\),
The replicator dynamics give \(\dot{x}_i = x_i\bigl(f_i - \phi\bigr)\), with \(f_i = (A x)_i\) the fitness of strategy \(i\) and \(\phi = x A x\) the average fitness, so \(\dot{x}_i / x_i = f_i - \phi\). Therefore
since \(\sum_i x^{*}_i f_i = x^{*} A x\) and \(\phi = x A x\).
(c) Asymptotic stability. [6]
For \(x \neq x^{*}\) near \(x^{*}\) the evolutionarily stable strategy condition gives \(x^{*} A x > x A x\), so \(x^{*} A x - x A x > 0\) and hence \(\tfrac{d}{dt} V < 0\). Together with part (a), \(V\) is positive away from \(x^{*}\), zero at \(x^{*}\), and strictly decreasing along every nearby trajectory. It is therefore a Lyapunov function, and \(x^{*}\) is asymptotically stable: nearby populations not only stay close but converge to \(x^{*}\).
(d) Contrast with Rock-Paper-Scissors. [5]
In the Rock-Paper-Scissors game of Question 3 the interior population is only neutrally stable, not an evolutionarily stable strategy, so the inequality \(x^{*} A x > x A x\) fails: the antisymmetry of the payoff matrix makes \(x^{*} A x = x A x\) for every \(x\). The same calculation then gives \(\tfrac{d}{dt} V = 0\), so the cross-entropy neither decreases nor increases. The natural conserved quantity there is \(H = x_R x_P x_S\), with \(\tfrac{d}{dt}\ln H = 0\). Because a constant of motion is preserved rather than a Lyapunov function decreasing, the trajectories cannot approach the centre; they trace closed orbits around it. The decreasing \(V\) of an evolutionarily stable strategy gives convergence, whereas the conserved \(H\) of Rock-Paper-Scissors gives perpetual cycling.
Marking exercise 1 (Question 2(b)(ii) and (iii)).
The replicator equation compares a strategy's fitness with the average fitness of the population, \(\phi = x f_L + (1 - x) f_R\), not with the rival's fitness:
as in the solution above. The missing factor \((1 - x)\) silently deletes the stable population \(x = 1\), and the transcript's list of stable populations, "nobody on the left" and "half and half", should be absurd on its face: in a symmetric coordination game, everyone driving on the left must be a stable population exactly as everyone driving on the right is. The wrong answer breaks the symmetry of the game, which is the kind of sense-check that catches this without any algebra. A fair mark is [2] of [6] for (ii), for the correct fitness difference, and [2] of [4] for (iii), for the two fixed points that survive.
Marking exercise 2 (Question 1(e)).
The final answer is right and the justification is not. The pivotal claim runs the implication backwards: an evolutionarily stable strategy is asymptotically stable under the replicator dynamics, but a stable population need not be an ESS: the interior population of the Rock-Paper-Scissors game in Question 3 is stable (trajectories stay nearby) yet is not an ESS. So stability alone cannot certify an ESS, and the question explicitly asks for the invasion argument: at \(x^{*} = \tfrac{2}{3}\) a small invading group of either pure strategy earns strictly less than the resident mixture, which is what makes \(x^{*}\) evolutionarily stable. An examiner gives method marks for \(x^{*}\) and withholds the rest: a right answer justified by a false theorem is not a right answer. A fair mark is [3] of [6].