cooperator given that d =1,L ,4 players have observed Z in the past. The probability that
this player meets someone who chooses Z is 1-ρ .
However, since this player chooses Y instead of Z, the transition matrix is now different:
|
г |
1 |
2 |
3 |
4 | |
|
~ |
1 |
0 |
1 |
0 |
0 |
|
A= |
2 |
0 |
1/3 |
2/3 |
0 |
|
3 |
0 |
0 |
1/3 |
2/3 | |
|
4 |
0 |
0 |
0 |
1 |
Again, Pr [2|1] = 1, since this is the case when no one observed a deviation in the past but
someone chooses to play Z today. Also, Pr [4|4] = 1 since the player who deviates today
by choosing cooperation will revert back to playing Z, in the next period (onetime
deviation). Now, consider the second row, i.e., the case when two players observed a
deviation, but only one of them plays Z today (reverting to playing Z, tomorrow). Here,
only one of the three possible pairings includes both players who observed a deviation. In
this case the sanctioning behavior does not spread further. In the other two pairings, it
spreads only to one more player, since only one player plays Z today. Hence, we have Pr
[2|2] = 1/3 and Pr [3|2] = 2/3. The third line is similarly explained.
A2 Off-equilibrium payoffs
Using the matrices above, we can now construct off-equilibrium payoffs in two
contingencies. The first is in equilibrium, when the player deviates for the first time,
choosing Z. The second is off equilibrium, when the player has observed uncooperative
behavior in the previous date (i.e., has seen Z for the first time) and now deviates by
cooperating, choosing Y.
Payoff from a deviation, when everyone follows the sanctioning rule. Suppose that
every player follows the social norm. Define the column vector:
V1
V=
V2
V3
V4
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