Assume you are interested in buying a used vehicle C_1. You are also considering

Question

Assume you are interested in buying a used vehicle C_1. You are also considering of taking it to a qualified mechanic and then decide whether to buy it or not. The cost of taking it to the mechanic is $100. C_1 can be in good shape (quality q^+) or bad one (quality q^-). The mechanic might help to indicate what shape the vehicle is in. C_1 costs $3,000 to buy and its market value is $4,000 if in good shape; if not, $1, 400 in repairs will be needed to make it in good shape. Your estimate is that C_1 has a 70% chance of being in good shape. Assume that the utility function depends linearly on the vehicle's monetary value. a. Calculate the expected net gain from buying C_1, given no test. b. We also have the following information about whether the vehicle will pass the mechanic's test: P(pass(C_1)|q^+ (C_1)) = 0.8 P(pass(C_1)|q^- (C_1)) = 0.35 Use Bayes' theorem to calculate the probability that the car will pass/fail the test and hence the probability that it is in good/bad shape given what the mechanic will tell you. c. What is the best decision given either a pass or a fail? What is the expected utility in each case? d. What is the value of optimal information for the mechanic's test? Will you take C_1 to the mechanic or not?

Explanation / Answer

Probability of good shape = 0.7

Gain if car is in good shape = 1000, loss if not = 3000+1400-4000 = 400

So, expected net gain = 0.7*1000 - 0.3*400 = 700-120 = 580

P(q+|pass) = P(pass|q+)*p(q+) / P(pass)

P(pass) = P(pass|q+)*P(q+) + P(pass|q-)*P(q-) = 0.8*0.7 + 0.35*0.3 = 0.56+0.105 =0.665
P(fail) = 1 - 0.665 = 0.335

So, P(q+|pass) = 0.8*0.7 / 0.665 = 0.8421

P(q-|pass) = 0.35*0.3 / 0.665 = 0.1579

P(q+|fail) = 0.2*0.7 / 0.335 = 0.4179

P(q-|fail) = 0.65*0.3 /0.335 = 0.5820

If it is a pass, it is better to buy as probability of good quality is higher given a pass. If it is a fail, still buying can be considered as probability of q+ is high.

It is not of much use to take C1 to the mechanic as a fail test may of may not indicate the quality.

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