Answers

1.

EP(S)=5p;

EP(H)=2p+2(1-p)=2;

= 5p²+2(1-p)=5p²+2-2p.

Hence,

D(p)=p(-5p²+7p-2).

So, the fixed points are p=0, p=1, p=2/5, which is globally stable. As soon as any small variation around 0 or 1 occurs, the system will evolve towards p=2/5. The graph is not to scale and rough.

2.

We could do the same as in (1). However, let us follow a different procedure that will make life easier. First, we rewrite the matrix, listing only the payoffs for the row player:

	A	B
A	3	1
B	4	0

What we have now is a matrix such that the replicator dynamics does not change if we add a constant to any column. (We shall not prove this but be thankful to the mathematician who did!). This allows us to obtain a matrix in which the bottom row is all zeros, which will simplify calculations; all we need to do is to add -4 to the first column, as the bottom of the second column is already zero. So, we obtain:

	A	B
A	-1	1
B	0	0

Now,

EP(A)= -p+1-p=1-2p;

EP(B)=0

= -2p²+p

Hence,

D(p)=p(2p²-3p+1).

The fixed points are p=0, p=1/2, p=1.

Since for p=1/3, the rate of change is positive and for p=2/3 it is negative, p=1/2 is globally stable. The graph is not to scale and rough.

3. After simplifying the matrix as per exercise 9, we have:

	S	C
S	-5	10
C	0	0

Hence, if Pr(S)=p, we obtain:

EP(S)= -5p+10 (1-p) = -15p + 10;

EP(C)=0;

= -15p²+10p.

Hence,

D(p)=p(15p² -25p +10).

The fixed points are p=0, p=2/3, p=1. The graph is:

p=2/3 is globally stable; the system will end up with 2/3 of the players swerving.

4.

The game is dominance solvable, the equilibrium being (H,H). Since no strongly dominated strategy survives replicator dynamics, only H will remain. So, Pr(H)=1 is globally stable and H will become fixated.

We can verify this the long way. First, let’s simplify the matrix, thus obtaining

	A	B
A	-2	-1
B	0	0

Then, if Pr(A)=p,

EP(A)= -2p-1(1-p)=-1-p

EP(B)=0

= p(-1-p) = -p-p²

Hence,

D(p)=p(-1-p+p+ p²)=p(p+1)(p-1).

The solutions are p=0, p=1, and p= -1. However, no probability is negative and consequently p = -1 must be disregarded. The only question left is whether D(p) is positive or negative between 0 and 1. By substituting, say p=1/2, we see that it’s negative. So, the graph looks, more or less, like this:

Consequently, 0 is globally stable; at the end, A will disappear and B will become fixated, as we discovered above.

5.

After simplification, we obtain

	A	B
A	1	-2
B	0	0

Setting Pr(A)=p,

EP(A)= 3p -2

EP(B)=0

= 3p² -2p

Hence,

D(p)= p(- 3p² +5p– 2) = .

The solutions are p=0, p=1, and p=2/3. By substituting, say, 1/2 for p, we note that the D(p) is negative between 0 and 2/3. The graph is (more or less)

So, 2/3, the interior point, is unstable; A or B will eventually go to fixation.

6.

a. For ρ_A > 1/N to obtain, in replicator dynamics the interior point p* must be less than 1/3. Hence,

p*= (d-b)/(a-c+d-b)<1/3, that is,

3(d-b)< a-c+d-b, or

2(d-b)<a-c.

b. For ρ_A > ρ_B to obtain, it must be that p*<1/2. Proceeding as before, we get d-b<a-c.

7.

p*=1/2, and therefore the relation does not obtain. The fixation probabilities of S and H are the same.