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November 26th, 2014, 09:09 AM  #1 
Newbie Joined: Oct 2014 From: Switzerland Posts: 6 Thanks: 0  Choice under uncertainty
Good evening, I have an exercise on choice under uncertainty. In city A, a ticket for the subway costs c > 0. As the inspectors seldom check the tickets, some people are willing to run the risk and they don't buy tickets. The probability that tickets will be checked is 1 > p > 0. If a passenger is caught without a ticket, they will have to pay a fine f > c. The city council wants to reduce the number of people who don't buy tickets, two alternatives are considered :  either doubling the fine f.  or doubling the control rate i.e. doubling p. People are assumed to be riskaverse, which solution is the best ? Now, I know that between a random investment of payoff W and a riskfree investment of payoff E(W), riskaverse people will choose the riskfree investment. Still, as a start, I have trouble with what my "investment" W should be here. To me W is an investment such that I pay 0 with probability 1  p and f with probability p ; I don't think it's correct though because in that case doubling p or doubling f yields the exact same result ... I'm definitely getting confused; any help or little hint is most welcome !! 
November 26th, 2014, 10:11 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 Math Focus: Mainly analysis and algebra 
You say 0, but the gain is c.

November 26th, 2014, 11:19 AM  #3 
Newbie Joined: Oct 2014 From: Switzerland Posts: 6 Thanks: 0 
So, it's (c)*(1p) + (f)*p ? I don't really see how I am going to use W, whatever it is to choose between the two alternatives... Well, I guess that I could show that u : p,c >(c)*(1p) + (f)*p is concave, it's regular so I can compute the Hessian. Its determinant is 1 < 0 so I can conclude u is concave. From there, I can use that u(E(W)) > E(u(W)). Still, what is this W I'm playing with ? How can I relate that to my initial problem ? Thank you already v8archie ! 
November 26th, 2014, 11:26 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 Math Focus: Mainly analysis and algebra 
I would say that $E_\text{base}(W) = c(1p)  fp$ Then $E_{2p}(W) = c(12p)  2fp$ and $E_{2f}(W) = c(1p)2fp$ Your job is to work out which is the smaller gain for the cheats. 
November 26th, 2014, 11:42 AM  #5 
Newbie Joined: Oct 2014 From: Switzerland Posts: 6 Thanks: 0 
Right. So, E$\displaystyle _{2p}$(W)  E$\displaystyle _{2f}$(W) =  $\displaystyle c \times p$ $\displaystyle \leq$ 0. So, assuming my preference operator has properties that make it possible to apply Von NeumannMorgenstern theorem, I get that W$\displaystyle _{2p}$ $\displaystyle \preceq$ W$\displaystyle _{2f}$. Hence a cheat would rather go for W$\displaystyle _{2f}$, thus the best solution is to take W$\displaystyle _{2p}$. Am I right ? There's something I'm uncomfortable with, though, where did I use the assumption on the riskaversion of the agents ? 
November 26th, 2014, 11:55 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 Math Focus: Mainly analysis and algebra 
By minimising E(W) you minimise the number of people who would take up the gamble. They we would rather just buy a ticket every day. So yes, I think you are right. Not that I'm an economist, nor have I studied this. 
November 26th, 2014, 12:11 PM  #7 
Newbie Joined: Oct 2014 From: Switzerland Posts: 6 Thanks: 0 
Yes, you're right, it makes sense ; I think I don't do everything aright and miss the point on the riskaversion but I don't really see how to use it either. Neither am I, just dabbling in it Well, anyway, thank you very much v8archie, your help definitely stood me in good stead ! 

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