SOLUTIONS MANUAL for Game-Theoretical Models in Biology 2nd Edition by Mark Broom and Jan Rychtář .

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3 (0, 0, 1) (0, 1/3, 2/3) (0, 1/2, 1/2)

(1/3, 1/3, 1/3)

(1/10, 3/5, 3/10)

(2/5, 2/5, 1/5)

2.5. (1, 0, 0)

(0, 1, 0)

P 2.6. If a σ-strategist plays in the population αj δpj where pj is a mixed P strategy p = p S , then the σ-strategist plays against Si with probability j i j,i i P α p .

It is thus the same as playing in the population δp̄ . j j j,i 2.7. E[p, q] = pAqT = (−p2 + p3 )q1 + (p1 − p3 )q2 + (−p1 + p2 )q3 = p1 (q2 − q3 ) + p2 (q3 − q1 ) + p3 (q1 − q2 ). Let m = max{q2 − q3 , q3 − q1 , q1 − q2 }. The best reply to q is (1, 0, 0) if q2 − q3 = m; it is (0, 1, 0) if q3 − q1 = m; and it is (0, 0, 1) if q1 − q2 = m.

2.8. (i) generic, (ii) technically non-generic but does not affect the analysis, (iii) non-generic, (iv) generic, (v) technically non-generic but does not affect the analysis, (vi) technically non-generic but does not affect the analysis, (vii) non-generic, (viii) non-generic.

2.9. By Exercise 2.6, the payoffs are the same as if the game is played against an individual playing p = 0.4(0.5, 0.5) + 0.3(1, 0) + 0.3(0.2, 0.8) = (0.56, 0.44).

We thus get E[(0.5, 0.5); Π] = (0.5, 0.5)A(0.56, 0.44)T = 0.4 and similarly for other strategies, giving payoffs 0.56 and 0.464.

2.10. For the matrix games, the mean strategy is as in Exercise 2.9. p = 0.4(0.5, 0.5) + 0.3(1, 0) + 0.3(0.2, 0.8) = (0.56, 0.44). If the opponent is selected based on its probability of playing S1 the mean strategy is pe = 0.5 1 0.2 0.5+1+0.2 (0.5, 0.5) + 0.5+1+0.2 (1, 0) + 0.5+1+0.2 (0.2, 0.8).

Chapter 3 β=1

β=100

0.3

0.3

0.3

0.3

0

0.1

1 2 Generation

3

0

p

0.5

p

0.5

0.1

3.1.

β=10

0.5 p

p

β=0 0.5

0.1

5 Generation

10

0

0.1

50 Generation

100

0

500 Generation

1000

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4 β=1

β=10

β=100 0.5

0.3

0.3

0.3

0.3

0.1

0.1

0

5 Generation

10

p

0.5 p

0.5 p

p

β=0 0.5

0.1

0

10 Generation

20

0

0.1

50 Generation

100

0

3.2.

Note that β here is the same as β + 1 in Exercise 3.1.

500 Generation

1000

3.3.

d p = p ap + b(1 − p) − p ap + b(1 − p) − (1 − p) cp + d(1 − p) dt

= p(1 − p) (a − c)p + (b − d)(1 − p) = p(1 − p)h(p).

d A point p is an asymptotically stable rest point of dynamics dt p = f (p) 0 if and only if f (p) = 0 and f (p) < 0. The conclusion follows by setting f (p) = p(1 − p)h(p). n X

xi ≤ 0 and this inequality x∗i i=1 is satisfied because logarithm is a concave function and Jensen’s inequality !

n n X X xi ≤ ln x1 = ln(1) = 0. implies x∗i ln ∗ x i i=1 i=1 To prove that L is increasing along trajectories, we calculate

n d ln L x(t) X dL x(t) x∗i dxi (t) 1 = = L dt dt x (t) dt i=1 i

n

X T T ∗ = xi A x(t) − x(t)A x(t)

3.4. L(x∗ ) ≥ L(x) can be written as

x∗i ln

i

i=1

T T = x∗ (t)A x(t) − x(t)A x(t) > 0.

1

1

0.8

0.8

0.8

0.6

0.4

0.2

0

3.6.

Best response to s

1

Best response to s

Best response to s

P P 3.5. E[p, SP i] = ji ≤ j pj a j pj aii = aii = E[Si , Si ]. For the second case P E[p, Si ] = j pj aji < j pj aii = aii = E[Si , Si ].

0.6

0.4

0.2

0

0.2

0.4 0.6 0.8 s, representing strategy (s,1−s)

1

0

0.6

0.4

0.2

0

0.2

0.4 0.6 0.8 s, representing strategy (s,1−s)

1

0

0

0.2

0.4 0.6 0.8 s, representing strategy (s,1−s)

1

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5 3.7. The Nash equilibria are a) (0, 1), b) (1/3, 2/3), c) (1, 0), (0, 1), (1/2, 1/2).

All of these are ESSs except (1/2, 1/2) in c).

3.8. No, the best reply to the strategy is to play Paper in the first round and then repeat the opponent’s move from the previous round.

3.9. For trembling hand Nash equilibrium we need E[pε , pe ] ≥ E[qε , pε ] for all q 6= p. If the inequality is satisfied for all small ε > 0, then it is satisfied also for ε = 0.

3.10. For p ∈ (0, 1), we can use the incentive function from Exercise 3.3.

We get h(p) = (a − c)p + (b − d)(1 − p), i.e. h(p) = 0 if and only if p = (d − b)/(a − b − c + d). Also, h0 (p) = (a − b − c − d). Thus, p ∈ (0, 1) is stable if and only if a − b − c + d < 0 and as seen in Section 6.2 these are exactly conditions for (p, 1 − p) (with p ∈ (0, 1)) to be an ESS. In the generic case, a point p = 1 is a stable point of the dynamics if and only if h(1) > 0, i.e. if a > c, which corresponds to (1, 0) being an ESS. Similarly for p = 0.

3.11. See Figure 3.1(a) and compare to Section 6.4.4 to see a specific payoff matrix of an RSP game with no ESS but an asymptotically stable rest point.

Chapter 4 4.1. Following Gintis (2000), the fish play the following game

C D

C 5 8

D

−3 .

0

with D to be the only Nash equilibrium. The cooperation can be achieved if the game is repeated, see Gintis (2000).

4.2. Up to a permutation in the order, there are the following seven ways to divide T = 6 as a sum of N = 3 nonnegative integers: 006, 015, 024, 033, 114, 123, 222. Any such a division yields a mating strategy; and for the strategy to be successful, the strategy has to be based on random permutations of the division (for example, 006 yields a strategy that with probability 1/3 courts for t=6 to a specific female and 0 to the other two).

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