关键词 > MATH/MTHE339
MATH/MTHE 339: Evolutionary Game Theory Homework Assignment 3
发布时间：2022-04-23
Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: daixieit
MATH/MTHE 339: Evolutionary Game Theory
Homework Assignment 3
2022
1. (5 marks) Consider the following symmetric game. Find all evolutionarily stable strategies (pure and/or mixed).
P2
|
|
|
|
|
|
|
|
4, 9 |
0, 3 |
|
|
9, 4 |
0, 0 |
0, 3 |
|
|
3, 0 |
3, 0 |
|
2. (10 marks total) Imagine an infinite population of players that pair up randomly and play matches against one another. A match consists of n rounds (where n is a finite positive integer). In each round, the players play the following game:
P2
|
|
C |
D |
|
C |
3, 3 |
|
|
D |
|
|
The payoff to a player for a match is the sum of their payoffs over the n rounds of that match (Note: you
do not need to discount future payoffs in this question). We define the following three strategies: “Always
Cooperate“ (denoted AllC) where C is played in every round;“Always Defect” (denoted AllD) where D
is played in every round; and“Tit-For-Tat” (denoted TFT) where C is played in the first round, and in all subsequent rounds, whatever your partner played the round before is played.
a. (2 marks) Construct a payoff matrix for a match. Your answer should be a symmetric 3 × 3 matrix.
b. (3 marks) Can you eliminate any strictly dominated strategies? Can you eliminate any weekly dom-
inated strategies? How do your answer to these questions depend on n?
c. (5 marks) Find all evolutionarily stable strategies. If applicable, explain how they might depend on n.
3. (10 marks total) Imagine a population of reproducing organisms that live in a habitat. We will suppose
that our habitat consists of an infinite number of discrete sites, each of which can support a single breeding
adult. We will refer to these sites as breeding spots, and the individuals occupying them as breeders. In
each year, breeders produce a large number of offspring (asexually), and then die. Breeders have a strategy
that determines which proportion of their offspring will remain at home, and which proportion of their
offspring will disperse to look for breeding spots elsewhere. These dispersed and non-dispersed offspring
then compete with one another to occupy breeding spots (vacated by the death of the breeders) in the
following year. The benefit to be gained by dispersal is the potential to colonize and take over a non-natal
breeding spot. The success of a dispersal strategy will be measured in terms of the expected number of
offspring that successfully become breeders in the following year.
Let u ∈ (0, 1] be the proportion of an individual’s offspring that will disperse. Assuming that each breeder produces a large number of children (C ), then we have, for an individual playing strategy u:
(1 − u)C = # of breeder’s offspring that remain at their natal breeding spot, uC = # of breeder’s offspring that disperse to another random breeding spot.
We assume that dispersal is costly. Thus, for every dispersed offspring, the probability that they survive
the trip is assumed to be s ∈ [0 , 1]. Offspring that do not disperse will be assumed to survive with probability
1. (Note that this does not mean that non-dispersers will successfully occupy a breeding spot in the next
year). Once dispersal happens, all the offspring compete with one another to occupy the vacant breeding
spot in the site where they happen to have landed. We will assume that dispersal is random, meaning a
dispersed offspring, if it survives, is equally likely to land at any other breeding spot. We can imagine that
at every breeding spot, there is some number of offspring that have not dispersed, and some number of
offspring that have arrived from elsewhere. One of these offspring, selected randomly, will become the
breeder for that particular breeding spot in the following year.
We imagine that in our population there is a single mutant breeder playing strategy u, whereas every other breeder plays the resident strategy v ∈ (0, 1]. We will refer to the site where the mutant lives as the mutant breeding spot. All other sites are referred to as resident breeding spots since they are occupied by individuals playing the resident strategy v . We want to get a handle on the relative success of strategy u, so we need to keep track of how successful our mutant offspring are at securing breeding spots for the following year.
a. (1 mark) How many offspring in total (on average) will disperse to the mutant breeding spot? How
many offspring in total (on average) will disperse to a resident breeding spot?
b. (1 mark) How many offspring in total (on average) will be competing at the mutant breeding spot? How many offspring in total (on average) will be competing at a resident breeding spot?
c. (1 mark) What is the probability that one of the mutant individual’s offspring will occupy the mutant
breeding spot the following year? What is the probability that one of the mutant offspring will occupy a
resident breeding spot in the following year?
d. (1 mark) Let E(i,j) be the expected number of offspring that successfully secure a breeding spot
for the following year for an individual playing strategy i in a population playing strategy j . Using the
previous results, find E(u,v) and E(v,v) and simplify.
e. (2 marks) Find the optimal dispersal strategy v ∗ in terms of s by solving the equation
∂E(u,v)
∂u u=v=v ∗
f. (4 marks) Show that the strategy found in the previous part, v∗, is in fact an ESS.
