Download Final exam, Problem 1 The money pot

Final exam, Problem 1:
The disappearing money pot
• A pot of money increases at a rate of $20 per hour.
– Thus, the amount in the pot is A(t) = $20t dollars after t hours, if it is still
there.
• The pot will be grabbed by someone else (or disappear) at a
random time, uniformly distributed between 0 and 10 hours, unless
you grab it first.
• (a) When should you grab the pot to maximize your EMV?
• (b) What is the answer if the money grows at a rate of $100/hr.?
• Optional hint 1: If the hazard function for the pot disappearing a
time t hours is h(t), then the expected marginal cost of continuing
for a small interval of length dt, from t to t + dt, if the pot is still
there at time t, is A(t)h(t)dt.
• Hint 2: You can also solve this problem without using hazard
functions.
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Final exam, Problem 2
One-armed bandit
• A slot machine has an unknown probability p of paying off an
amount of $1 each time its handle is pulled.
• The prior probability distribution for p is uniform, U[0, 1].
(Thus, the expected payoff on the first pull is $0.50, since
initially E(p) = 0.5.)
• It costs a dime ($0.10) each time you pull the handle.
• If you never win, after how many unsuccessful tries (pulls of
the handle) will the EMV of the next pull (its expected benefit
minus its $0.10 cost) become negative?
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Final exam, Problem 3: Oil drilling
• Your company has the right to drill for oil on a piece of land.
• It costs $200,000 ($0.2M) to drill.
• If you drill and there is oil there, you get $1,000,000 ($1M) worth of
oil, and thus make a profit of $0.8M. Otherwise, if you drill and
there is no oil there, you get $0 return, and thus make a loss of $0.2M.
• If you do not drill, your profit and loss are both $0.
• You want to maximize EMV.
• (a) If the probability that there is oil is p = 0.25, what is the EMV for
drilling? To maximize EMV, should you drill?
• (b) What is the smallest probability of oil, p, for which you should
drill (i.e., for which EMV(drill) is non-negative)?
• (c) Suppose that instead of owning the right to drill, your company
paid $10k for the right (or option) to drill. It must drill this year or
lose the right to drill. As time runs out, does the answer to part (b)
increase, decrease, or remain constant?
Final exam, Problem 4: Expected value
of perfect information
• Suppose that, in Problem 3, it were possible to
buy a perfect test that would reveal with 100%
accuracy whether there is oil.
• Assume p = 0.25 = prior probability of oil
• What is the most you should be willing to pay to
get the results of such a perfect test?
– This is sometimes called the expected value of perfect
information (EVPI)
– You should never pay more than EVPI to get the
results of any test
Solution to Problem 1
• Pot of money: A(t) = kt dollars at t hours, k = 20 or 100
• Money disappears at time uniformly distributed between 0 and 10 hours
• Hazard function for disappearance: h(t) = f(t)/[1 – F(t)] = 0.1/(1 – 0.1t), 0 <
t < 10.
• At time t, continuing for another dt brings expected marginal gain
(dA/dt)dt = k*dt and expected marginal loss A(t)*h(t)dt.
• These are equal when k = h(t)A(t), or k = 0.1kt/(1 - 0.1t), or 1-0.1t = 0.1t or
1 = 0.2t or t = 5 hours. (This does not depend on k.)
• Another way: Stopping at time t has EMV = kt-(t/10)kt = kt*(1-(t/10)) = ktkt2/10. d(EMV)/dt = 0 implies k-2kt/10 = 0 or 1-t/5 = 0 or t = 5. The
second derivative is -2k < 0, so we have found the t that maximizes EMV.
• The answer does not depend on k. The best stopping time is 5 hours,
whether k = $20/hr. or k = $100/hr.
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Solution to Problem 2
• After n failures in n trials, the posterior
probability of failure on the next trial is
(x+1)/(n+2) = (n+1)/(n+2).
• The probability of success is therefore 1(n+1)/(n+2) = ((n+2)-(n+1))/(n+2) = 1/(n + 2).
• The EMV of the next pull after n consecutive
failures is 1/(n + 2) - 0.10.
• This is negative when 1 < 0.10(n + 2) or 10 < n
+ 2, or n > 8. After 8 pulls, EMV < 0.
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Solution to Problems 3 and 4
• Problem 3: EMV(drill) = p*1 + (1-p)(-0.2) = p - 0.2.
– (a) EMV(drill) = 0.25-0.20 = $0.05M if p = 0.25.
– (b) EMV(drill) is positive for p > 0.2.
– (c) Ignore sunk costs: The smallest value of p that
makes EMV(drill) non-negative p > 0.2.
• Problem 4 solution:
– With no test, EMV is 0.25*0.8 + 0.75*(-0.2) = 0.200.15 = $0.05M.
– With a perfect test, the EMV is 0.25*0.8 = $0.20M,
since one will drill only if there is oil.
– The difference is the EVPI: EVPI = $0.15M.