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. T. Joy

What the subjects saw: (page 1)

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

The treatments:

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. \(\forall \alpha\in\{0,.5,1\}\)?

in a symmetric equilibrium the e*=e1=e2=e3

find f' of (ae+(1-a)(e)/3)((60-e)/60)-e/3 and set it to zero
       0 = e(-1/60)+(a+(1-a)/3)(1-e/20)-1/3
       0 = 2a/3 - ea/30 - e/30
       e = 20a - ea
       e = 20a/(1+a)

The equilibrium catch rate when 
     alpha=0,   20(0)/1 = 0 so e = 0.
     alpha=.5,  20(.5)/1.5 = 20/3 so e = 20/3.
     alpha=1,   20/2  = 10 so 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 \(\alpha\)?

in a symmetric equilibrium the e*=e1=e2=e3

When you take the derivative of the function and set it to zero 
     f  = e - e^2/20 -e/3
     f' = 1 - e/10 - 1/3 = 0
    2/3 = e/10 
    so e = 20/3 (per person) and E=3e so E=20 (total effort).
     
    Alpha does not effect the treatment because your actions are in the best interest of society, not yourself. This means that regardless of the treatment you will have the same level of effort. Mathematically this is because (ae-3e/3-3ae/3) which equals e because ae - ae + e = e.

3 (10 marks)

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

The catch size is different from the socially optimal catch size when you get to keep all the fish you catch or when you don’t keep any and receive part of the combined social catch. When you keep all the fish you catch, your incentive is to catch as many fish as you can as other people will be doing the same. This typically leads to a destruction of the resource (tragedy of the commons). when you only receive a piece of the combined social catch, you will catch less than the socially optimal amount as it costs effort to catch fish and your incentive is to free ride off of other peoples catch.

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, na.rm=TRUE), 
             mean_choice = mean(choice)) %>%   
             filter(!is.na(mean_profit)) %>%    
   ungroup() %>%
   mutate(top_half =  mean_profit>median(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(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. On the left, the three boxplots are for the people who had a less than median profit. The three boxplots to the right are for the people who had higher than median profit (played the game maximizing their benifit). for profit winners, when alpha=0 choice was low(median of roughly 5), when alpha=.5 choice was in between alpha=0 and 1(median of roughly 7), when alpha=1 choice was high (median of roughly 9)

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 marks)

Describe the results from the plot above. The data from alpha=0 suggests that people with a lower mean_choice had a higher mean_profit. The data from alpha=1 suggests that people with a higher mean_choice had a higher mean_profit, the data from alpha=.5 suggests that people with a mean_choice between 5-7 had a higher mean_profit. The data from plots alpha=0 and alpha=1 is much more linear than the alpha=.5 this could be because it was more difficult to understand what was socially optimal when you directly receive some but not all of your catch. peoples mean choice was mostly between 5 and 10 for all treatments, although there much more variability in alpha=.5.