Assignment 5
Z Xiaoyang
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 y. 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. \(\forall \alpha\in\{0,.5,1\}\)? Probability=20n-(c1+c2+c3+…cn)/20n n=3 maxE(Π1)=ae1+1-a/3(e1+e2+e3)-e1/3 dE(Π1)/de1=a+1-a/3+[ae1+1-a/3(e1+e2+e3)](-1/60)-1/3=0 e1=e2=e3=e 1+2a/3+[ae+1-a/33e*](-1/60)-1/3=0 1+2a/3-e/60-1/3=0 when a=0 e=0 when a=0.5 e=20/3 when a=1 e*=10
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 \(\alpha\)?
Probability=(60-E/60) Total catch size E=e1+e2+e3+…en W:total welfare Max: E(W)=E(Π1)+E(Π2)+E(Π3)=ae1+1-a/3(e1+e2+e3)-e2/3+ae1+1-a/3(e1+e2+e3)-e2/3+ae3+1-a/3(e1+e2+e3)-e3/3=a(e1+e2+e3)+(1-a)(e1+e2+e3)-1/3(e1+e2+e3)=E(60-E/60)-1/3E dE(W)/dE=1-e/30-1/3=0 E=20 e=20/3
3 (10 marks)
Explain when and why the equilibrium catch size is different from the socially optimal individual catch size.
Socially optimal: e**=20/3 Epm capture size: when a=0.5, e*=e**. Other conditions are different from the social optimum. During communism, everyone had the obsession not to fish. The result e=0. When laissez-faire, everyone only considers their own interests and catches more fish than the best in society, e=10
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_profit=mean(profit),mean_choice=mean(choice),na.rm=TRUE)%>%
filter(!is.na(mean_profit))%>%
ungroup()%>%
mutate(top_half=case_when(mean_profit>median(mean_profit)~'Yes',
mean_profit<median(mean_profit)~'No'))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(mydf)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.
For data whose average profit is higher than the median (yes), social epm, e=20/3 is higher than when alpha=0, which is about the median value of alpha=0.5, which is basically lower than the value when alpha=1. For data whose average profit is lower than the median value (No), in the effort, when alpha=0 and alpha=0.5, most of the selected e is greater than the social optimal level, and alpha=1 is lower than the social optimal level
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 quadratic relationship, and facet_wrap() the plot by variable alpha.
10 (10 marks)
Describe the results from the plot above.
It can be seen from the figure that the alpha=0 group is likely to be an upward parabola. When the effort increases, the average profit first drops and then rises. The lowest point of average profit is about average profit=13 Alpha=0.5 and group 1 are both parabolas that open downward. The highest point of the average profit with alpha=0.5 is approximately effort=8. The highest value of the alpha=1 group and the average profit is about effort=15, and the maximum value is higher than the alpha=0.5 group