The solutions to some or all of the problems from Resnik’s Choices that we’re doing this semester are posted on a page on the course web site for the section of this course that I taught in Fall 2006 (go to http://web.ku.edu/~utile/courses/rct2, and then click on “Solutions to problems from Resnik’s Choices”). Since we’re using that page as a reference in class this semester, here I’ll just note corrections, different solutions, and additional information (and I’ll put ‘same as Fall 2006’ otherwise).
The following proof differs from the Fall 2006 proof in appealing directly to the definition of an equilibrium pair rather than the minimax equilibrium test.
# | claim | justification |
1 | Suppose row i is a dominant row and column j is a dominant column. | assumption for proving conditional |
2 | Row i is a dominant row. | 1 |
3 | v_{ij} is at least as large as any other value in column j. | 2 |
4 | The row player has no incentive to unilaterally deviate from the outcome at row i, column j. | 3 |
5 | Column j is a dominant column. | 1 |
6 |
v_{ij} is at least as small as any other value in row i. |
5 |
7 |
The column player has no incentive to unilaterally deviate from the outcome at row i, column j. |
6 |
8 |
(R_{i}, C_{j}) is an equilibrium pair. |
4 and 7, definition of equilibrium pair |
9 |
If row i is a dominant row and column j is a dominant column, then (R_{i}, C_{j}) is an equilibrium pair. |
1–8 |
Notice that lines 1–9 prove a conditional (the one asserted in line 9), lines 10–18 prove its converse (the conditional asserted in line 18), and line 19 asserts the resulting biconditional. Notice also that lines 10–18 do not draw on anything established in lines 1–9. This is because the proof of the converse of the conditional is independent of the proof of the conditional. In fact, you could prove the converse first (i.e., do lines 10–18 first), and then the conditional (lines 1–9), and then still be justified in asserting line 19.
top of p. 135—tables 5-16 and 5-17:
(Instead of verifying that the mentioned strategy pair is the equilibrium strategy pair, we’ll just verify that the mentioned strategy pair is an equilibrium strategy pair.)
The following remarks concerning tables 5-16 and 5-17 differ from those in the Fall 2006 notes by not relying on the shortcut used there of just being concerned with equality between the expected utilities of the row player’s pure strategies, and equality between the expected utilities of the column player’s pure strategies. We’ll get to that shortcut a little later.
table 5-16:
To verify that [(½ R_{1}, ½ R_{2}), (½ C_{1}, ½ C_{2})] is an equilibrium pair, we need to verify that neither the row player nor the column player can do better by unilaterally changing his strategy.
First, consider the game from the row player’s point of view. If the column player is playing strategy (½ C_{1}, ½ C_{2}), then the expected utility of the game for the row player, with his strategy (p R_{1}, (1 – p) R_{2}), is (p)[(½)(1) + (½)(–1)] + (1 – p)[(½)(–1) + (½)(1)], or p(½ – ½) + (1 – p)(–½ + ½), or p(0) + (1 – p)(0), or 0. Because this expected utility does not depend on p, in that the p drops out of the expression as we simplify it, the row player’s expected utility is unaffected by his own strategy, and thus he has no incentive to unilaterally deviate from whatever strategy he happens to be playing. This goes for (½ R_{1}, ½ R_{2}) as much as for any other strategy.
Second, consider the game from the column player’s point of view. If the row player is playing strategy (½ R_{1}, ½ R_{2}), then the expected utility of the game for the column player, with her strategy (q C_{1}, (1 – q) C_{2}), is (q)[(½)(1) + (½)(–1)] + (1 – q)[(½)(–1) + (½)(1)], or q(½ – ½) + (1 – q)(–½ + ½), or q(0) + (1 – q)(0), or 0. Because this expected utility does not depend on q, in that the q drops out of the expression as we simplify it, the column player’s expected utility is unaffected by her own strategy, and thus she has no incentive to unilaterally deviate from whatever strategy she happens to be playing. This goes for (½ C_{1}, ½ C_{2}) as much as for any other strategy.
Since neither the row player nor the column player can do better by unilaterally changing his or her strategy, [(½ R_{1}, ½ R_{2}), (½ C_{1}, ½ C_{2})] is an equilibrium pair.
To compute the value of the game—same as Fall 2006.
table 5-17:
To verify that [(½ R_{1}, ½ R_{2}), (½ C_{1}, ½ C_{2})] is an equilibrium pair, we need to verify that neither the row player nor the column player can do better by unilaterally changing his strategy.
First, consider the game from the row player’s point of view. If the column player is playing strategy (½ C_{1}, ½ C_{2}), then the expected utility of the game for the row player, with his strategy (p R_{1}, (1 – p) R_{2}), is (p)[(½)(22) + (½)(–18)] + (1 – p)[(½)(–18) + (½)(22)], or p(11 – 9) + (1 – p)(–9 + 11), or p(2) + (1 – p)(2), or 2. Because this expected utility does not depend on p, in that the p drops out of the expression as we simplify it, the row player’s expected utility is unaffected by his own strategy, and thus he has no incentive to unilaterally deviate from whatever strategy he happens to be playing. This goes for (½ R_{1}, ½ R_{2}) as much as for any other strategy.
Second, consider the game from the column player’s point of view. If the row player is playing strategy (½ R_{1}, ½ R_{2}), then the expected utility of the game for the column player, with her strategy (q C_{1}, (1 – q) C_{2}), is (q)[(½)(22) + (½)(–18)] + (1 – q)[(½)(–18) + (½)(22)], or q(11 – 9) + (1 – q)(–9 + 11), or q(2) + (1 – q)(2), or 2. Because this expected utility does not depend on q, in that the q drops out of the expression as we simplify it, the column player’s expected utility is unaffected by her own strategy, and thus she has no incentive to unilaterally deviate from whatever strategy she happens to be playing. This goes for (½ C_{1}, ½ C_{2}) as much as for any other strategy.
Since neither the row player nor the column player can do better by unilaterally changing his or her strategy, [(½ R_{1}, ½ R_{2}), (½ C_{1}, ½ C_{2})] is an equilibrium pair.
To compute the value of the game—same as Fall 2006.
problems on p. 137:
I mentioned that x_{1} + x_{2} + . . . x_{k} = 1. Equally worth mentioning is that 0 ≤ x_{1}, x_{2}, . . . x_{k} ≤ 1. I should have also mentioned that these conditions needn’t be stated unless asked for.
To verify that [(½ R_{1}, ½ R_{2}), (½ C_{1}, ½ C_{2})] is an equilibrium pair, we need to verify that neither the row player nor the column player can do better by unilaterally changing his strategy.
First, consider the game from the row player’s point of view. If the column player is playing strategy (½ C_{1}, ½ C_{2}), then the expected utility of the game for the row player, with his strategy (p R_{1}, (1 – p) R_{2}), is (p)[(½)(a) + (½)(–b)] + (1 – p)[(½)(–b) + (½)(a)], or p(½(a – b)) + (1 – p)(½(a – b)), or ½(a – b). Because this expected utility does not depend on p, in that the p drops out of the expression as we simplify it, the row player’s expected utility is unaffected by his own strategy, and thus he has no incentive to unilaterally deviate from whatever strategy he happens to be playing. This goes for (½ R_{1}, ½ R_{2}) as much as for any other strategy.
Second, consider the game from the column player’s point of view. If the row player is playing strategy (½ R_{1}, ½ R_{2}), then the expected utility of the game for the column player, with her strategy (q C_{1}, (1 – q) C_{2}), is (q)[(½)(a) + (½)(–b)] + (1 – q)[(½)(–b) + (½)(a)], or q(½(a – b)) + (1 – q)(½(a – b)), or ½(a – b). Because this expected utility does not depend on q, in that the q drops out of the expression as we simplify it, the column player’s expected utility is unaffected by her own strategy, and thus she has no incentive to unilaterally deviate from whatever strategy she happens to be playing. This goes for (½ C_{1}, ½ C_{2}) as much as for any other strategy.
Since neither the row player nor the column player can do better by unilaterally changing his or her strategy, [(½ R_{1}, ½ R_{2}), (½ C_{1}, ½ C_{2})] is an equilibrium pair.
To compute the value of the game—same as Fall 2006.
C_{1} | C_{2} | |
R_{1} | 2 | 1 |
R_{2} | 1 | 2 |
section 5-3c:
table 5-25:
EU_{Row}
= p[q(3) + (1 – q)(1)] + (1 – p)[q(–7) + (1 – q)(4)]
= p(3q + 1 – q) + (1 – p)(–7q + 4 – 4q)
= p(2q + 1) + (1 – p)(–11q + 4)
= p(2q + 1) + (–11q + 4) – p(–11q + 4)
= p(2q + 1 + 11q – 4) + (–11q + 4)
= p(13q – 3) + (–11q + 4)
= (13q – 3)p + (–11q + 4)To make the coefficient of p equal to 0, we proceed as follows:
13q – 3 = 0
13q = 3
q = 3/13EU_{Column}
= q[p(3) + (1 – p)(–7)] + (1 – q)[p(1) + (1 – p)(4)]
= q(3p – 7 + 7p) + (1 – q)(p + 4 – 4p)
= q(10p – 7) + (1 – q)(–3p + 4)
= q(10p – 7) + (–3p + 4) – q(–3p + 4)
= q(10p – 7 + 3p – 4) + (–3p + 4)
= q(13p – 11) + (–3p + 4)
= (13p – 11)q + (–3p + 4)To make the coefficient of q equal to 0, we proceed as follows:
13p – 11 = 0
13p = 11
p = 11/13EU(game): same as Fall 2006
table 5-26:
EU_{Row}
= p[q(4) + (1 – q)(20)] + (1 – p)[q(5) + (1 – q)(–3)]
= p(4q + 20 – 20q) + (1 – p)(5q – 3 + 3q)
= p(–16q + 20) + (1 – p)(8q – 3)
= p(–16q + 20) + (8q – 3) – p(8q – 3)
= p(–16q + 20 – 8q + 3) + (8q – 3)
= p(–24q + 23) + (8q – 3)
= (–24q + 23)p + (8q – 3)To make the coefficient of p equal to 0, we proceed as follows:
–24q + 23 = 0
–24q = –23
q = 23/24EU_{Column}
= q[p(4) + (1 – p)(5)] + (1 – q)[p(20) + (1 – p)(–3)]
= q(4p + 5 – 5p) + (1 – q)(20p – 3 + 3p)
= q(–p + 5) + (1 – q)(23p – 3)
= q(–p + 5) + (23p – 3) – q(23p – 3)
= q(–p + 5 – 23p + 3) + (23p – 3)
= q(–24p + 8) + (23p – 3)
= (–24p + 8)q + (–23p – 3)To make the coefficient of q equal to 0, we proceed as follows:
–24p + 8 = 0
–24p = –8
p = 8/24
p = 1/3EU(game): same as Fall 2006
section 5-3d:
section 5-4a:
section 5-4c:
section 5-4e:
The expected gain from becoming a cooperator is as follows:
(1/2)(1/4)(3/4 – 1/4) – (3/4)(1/4)(1 – 1/2)
= (1/8)(2/4) – (3/16)(1/2)
= 2/32 – 3/32
= –1/32Since the expected gain from remaining a cheater positive, and the expected “gain” becoming a cooperator is negative, it is more rational to remain a cheater. (This is not surprising, considering that there is a fairly high value for q (the possibility that a cheater can exploit a cooperator) and a low value for r (the probability of being able to cooperate with a fellow cooperator).
Note, by the way, that the foregoing approach just compares the utility gains, relative to the status quo, of entering in the specified scenario as either a cheater or a cooperator. You can also compare the total utilities (not just utility gains) to be expected from either course of action; if you did that, then you would just add the value of u to each of the two values. Since u = 1/4, the total utility of remaining a cheater is 9/32 + 1/4, or 9/32 + 8/32, or 17/32; and the total utility of becoming a cooperator is –1/32 + 1/4, or –1/32 + 8/32, or 7/32.