My Math Forum Choice under uncertainty

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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 risk-averse, which solution is the best ? Now, I know that between a random investment of payoff W and a risk-free investment of payoff E(W), risk-averse people will choose the risk-free 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: 7,685 Thanks: 2665 Math Focus: Mainly analysis and algebra You say 0, but the gain is c. Thanks from WittgensteinFan
 November 26th, 2014, 11:19 AM #3 Newbie   Joined: Oct 2014 From: Switzerland Posts: 6 Thanks: 0 So, it's (-c)*(1-p) + (-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)*(1-p) + (-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: 7,685 Thanks: 2665 Math Focus: Mainly analysis and algebra I would say that $E_\text{base}(W) = c(1-p) - fp$ Then $E_{2p}(W) = c(1-2p) - 2fp$ and $E_{2f}(W) = c(1-p)-2fp$ Your job is to work out which is the smaller gain for the cheats. Thanks from WittgensteinFan
 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 Neumann-Morgenstern 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 risk-aversion of the agents ?
 November 26th, 2014, 11:55 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2665 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. Thanks from WittgensteinFan
 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 risk-aversion 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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