Section #: 01

Students must abide by UVic academic regulations and observe standards of scholarly integrity (i.e. no plagiarism or cheating). Therefore, this assignment must be taken individually and not with a friend, classmate, or group. You are also prohibited from sharing any information about the assignment with others. I affirm that I will not give or receive any aid on this assignment and that all work will be my own. put your name here

What the subjects saw: (page 1)

Consider a lake where \(n \approx 3\) people can catch fish.
In this experiment you will play 20 independent rounds of the same one shot game.
Each round you are put into a new group of \(\approx 3\) people.
In each round your task is to choose how effort to put into fishing \(e_i\).
If the stock of fish \(f\) exceeds the total effort \(f>e_1+e_2+e_3\) then every unit of effort results in one fish caught.
However, the stock of fish is unknown: \(f\) equally likely to be any real number between \(0\) and \(20n\): \(f∼U[0,20n]\)
If the stock of fish is lower than the total effort \(f \leq e_1+e_2+e_3\) the resource is destroyed: all players catch zero fish.
The tradeoff: more effort more fish (if there are enough fish) vs. more effort lower probability there are enough fish.
For example, suppose that the sum of all the individual effort choices is \(e_1+e_2+e_3=45\): what is the probability there is enough fish?

If the total effort is 45, the probability there is enough fish is .25.
The expected profit function for player 1 is:

\[\begin{equation} E[\pi_1]=\left[\alpha e_1+\frac{1-\alpha}{3}(e_1+e_2+e_3)\right]\left(\frac{60-e_1-e_2-e_3}{60}\right)-\frac{e_1}{3} \end{equation}\]

Inside the first [square] brackets is the amount of fish you receive if there are enough fish.
Inside the second (round) brackets is the probability there are enough fish.
The last term is the cost of effort.

The treatments:

Communism (\(\alpha=0\)):
Universal Basic Income (\(\alpha=\frac12\))
laissez-faire (\(\alpha=1\))

With a bit of a stretch this game would also apply to the issue of fossil fuel extraction. In this case, the threat is not the extinction of some species we have over-harvested, but our own extinction. A feature of both interpretations is an unknown tipping point beyond which we can not recover. Regarding climate change, a tipping point may be the result of positive feedback loops such as methane being released from permafrost as it thaws, and a lowering of the planet’s Albedo as we lose snow and ice cover. In both cases if extraction stays below the tipping point the species in question will survive, otherwise extinction. The temptation to extract more comes from the fact that our material well being is increasing in extraction, as long as we stay below the tipping point.

Unfortunately there was a bug in the code, so the feedback that subjects received after each round did not make much sense. Because of this we are going to create a partition of the data allowing us to compare the behaviour of subjects on the basis of how much profit they earned, with the conjecture that some of you figured out what was wrong with the feedback provided, and some of you did not.

1 (10 marks)

Take the derivative of player 1’s expected profit function with respect to \(e_1\) and set the derivative equal to zero. Solve for the symmetric equilibrium where \(e_i=e^{\star}~\forall~i\). What is the equilibrium prediction for catch size for a group of size 3 for the three treatments? i.e. \(\forall \alpha\in\{0,.5,1\}\)

FOC: \(\left[\alpha+\frac{1-\alpha}{3}\right]\left(\frac{60-e_1-e_2-e_3}{60}\right)+\left[\alpha e_1+\frac{1-\alpha}{3}(e_1+e_2+e_3)\right]\left(\frac{-1}{60}\right)-\frac13=0\)
players identical, impose symmetry: \(e_1=e_2=e_3=e\)
\(e^\star=\frac{20\alpha}{1+\alpha}\) (symmetric equilibrium where \(e_1=e_2=e_3=e^\star\))

\(\alpha\)	\(e^{\star}\)
0.0	0.00
0.5	6.67
1.0	10.00

2 (10 marks)

Suppose that a social planner chooses the total amount of effort \(E=e_1+e_2+e_3\). Recall that player i’s consumption (if there are enough fish) is \(\left[\alpha e_1+\frac{1-\alpha}{3}(e_1+e_2+e_3)\right]\). What do you get if you add up the consumption for all three players (again, this is assuming there are enough fish)? Thus, what is the expected welfare in terms of \(E\)? Differentiate expected welfare in terms of \(E\) and set the derivative to zero. Solve for \(E^{\star\star}\), the socially optimal level of total effort. If all players exert the same effort level, what is the socially optimal level of effort per person? Why does the socially optimal level of effort not depend on \(\alpha\)?

If you add up the consumption of the three players (conditional on there being enough fish) you get \(E\), which is the sum of the individual effort levels \(e_1+e_2+e_3\). Recall that one unit of effort results in one unit of fish being caught, assuming there are enough fish.

\(E[W]=E\left(\frac{60-E}{60}\right)-\frac{E}{3}\)
FOC: \(\frac{60-E}{60}-\frac{E}{60}-\frac13=0\)
\(E^{\star\star}=20\)
\(e^{\star\star}=\frac{20}{3}\)

Optimal effort maximizes the size of the “pie” that is created. Conditional on this optimal level of effort, \(\alpha\) just determines how the pie is shared.

3 (10 marks)

Explain when and why the equilibrium catch size is different from the socially optimal individual catch size.

In the communism treatment(\(\alpha=0\)) the equilibrium is to free ride on the effort of others. In the laissez-faire treatment(\(\alpha=1\)) the equilibrium is to over-exploit the common pool resource. In the Universal Basic Income treatment(\(\alpha=.5\)) these two perverse incentives balance out, resulting in an equilibrium that is socially optimal.

4 (10 marks)

Create a new dataframe called subjects using mydf THEN group_by() variables oneid and alpha THEN summarize() creating variables mean_profit by taking the mean() of profit with option na.rm=TRUE and mean_choice by taking the mean() of choice THEN filter keeping only the rows where it is not true that mean_profit is.na() THEN ungroup() THEN mutate() creating variable top_half by testing mean_profit>median(mean_profit). Include a copy of your code below: note eval=FALSE, so the code must be in your .R file as well.

subjects <- mydf%>%
  group_by(oneid,alpha)%>%
  summarize(mean_choice=round(mean(choice),2),
            mean_profit=round(mean(profit,na.rm=TRUE),2))%>%
  filter(!is.na(mean_profit))%>%
  ungroup()%>%
  mutate(top_half=mean_profit>median(mean_profit,na.rm=TRUE))%>%
  arrange(desc(mean_profit))

5 (10 marks)

Use the function datatable() from the library DT to show the dataframe subjects below. Set rownames=FALSE and include options = list(columnDefs = list(list(className = 'dt-center', targets = "_all"))) in your call to datatable().

6 (10 marks)

Add the information from subjects to mydf by performing a left_join() of mydf and subjects THEN get rid of missing values with function na.omit(). Include a copy of your code below: note eval=FALSE, so the code must be in your .R file as well.

mydf <- left_join(mydf,subjects)%>%
  na.omit()

7 (10 marks)

Create boxplots of choice for the three levels of alpha. Add the underlying data with some jittering, and a horizontal white line at the equilibrium level of effort. Facet the plot by variable top_half. Give the y axis a more descriptive name.

8 (10 marks)

Describe the results from the plot above.

Note that the equilibrium predicts that effort should be increasing in \(\alpha\) (the proportion of your catch you get to keep). This appears to be only true for the students where their average profit was in the top half of the class. It is my belief that these are the students that “figured out” the problem with the feedback, whereas for the students in the bottom half behaved the way that they did because they were not getting proper feedback on their choices.

9 (10 marks)

Using dataframe subjects, create a scatterplot of mean_choice vs. mean_profit. Use geom_smooth() with method="lm" and formula=y ~ poly(x, 2) to emphasize the relationship, and facet_wrap() the plot by variable alpha.

10 (10 marks)