Solution to (i)
Underlining finds the pure NE:
Player 1
L
Player 2
C R
U 5,5 0,0 0,6
M 0,0 3,3 0,0
D 6,0 0,0 1,1
You have now found that there are 2 pure NE (M,C) and (D,R).
Notice that the mixed strategy (0 U + 0 .
9 M + 0 .
1 D) strictly dominates pure strategy U for
Player 1, and (0 L + 0 .
9 C + 0 .
1 R) strictly dominates pure strategy L for Player 2. As strictly dominated strategies cannot be part of a mixed or pure NE, we can forget about them while searching for more NE. Note there are many other mixed strategies that dominate actions U and
L, but you only need one to strictly dominate in order to rule out a pure strategy. Now you are essentially left with the 2x2 sub-grid (ignoring row U and column L). Indeed, both the pure NE are in here. There is a nondegenerate mixed NE which you find simply from the 2x2 sub-grid where:
S
1
= 0 U +
1
4
M +
3
4
D and S
2
= 0 L +
1
4
C +
3
4
R
So there are 3 NE.
Solution to (ii)
The definition of SPNE is that we have NE play in all subgames. Here there are 2 classes of subgame to consider. Those that start in t=2 (many), and the one that starts in t=1 (just one, the whole game). When you have a finitely repeated game, as we do here as we have 2 periods
(and 2 < ∞ !!), use backwards induction to solve for SPNEs. Starting from t=2, we see that the game above is the subgame that will be played today, and then that’s it, there is no tomorrow.
Hence, by definition of SPNE, we must have a NE played in t=2. So one of the 3 NE we found in part (i) must occur in t=2.
Knowing this, and the fact we were told to look only for SPNE that include (U,L) in t=1, we can narrow our search. Further, the game is symmetric (in payoffs) and they have a common discount factor, so we can save time by just looking at the row player (the maths for the column
1

player will be identical).
Obviously SPNE strategies must include (U,L) in t=1 as we were told to look for this. So as
Player 1, we should have U as our strategy in t=1. And we must have a NE played in t=2 by the above reasoning, i.e. either M, D, or the mixed S
1 of part (i). But we need to be careful now. If we just specify some simple strategy like: in t=1 play U, then in t=2 play M, that will not work, because Player 2’s best response to this is to play R, not L in t=1. What we need is a “carrot and stick” strategy to ensure Player 2’s cooperation in t=1. One carrot and stick would be ( X
1 referring to Player 1’s strategy, and X
2 to Player 2’s):
X
1
= Play U in t=1, then if (U,L) happened in t=1, play C in t=2, otherwise in t=2 play D.
X
2
= Play L in t=1, then if (U,L) happened in t=1, play M in t=2, otherwise in t=2 play R.
What Player 1 is saying here is: You better play do L in t=1, because otherwise I will punish you with the “bad” NE (D,R), but if you behave, I will reward you with the “good” NE (M,C).
Player 2 has the analogous argument for his strategy. The threat is credible here because the threat is a NE. The strategy I just defined will be a NE of the subgame starting in t=1 if:
Payoff from X i given X
− i is played ≥ Payoff from optimal deviation given X
− i is played
5 + 3 δ ≥ 6 + δ
δ ≥ 1
2
If this holds, then X
1 and X
2 are NE for the subgame starting at t=1. As we already said that the strategies are NE for the t=2 subgames, because X
1 and X
2 specify NE play in t=2 in all circumstances. Hence we have found the condition for which the strategies X
1 and X
2 are SPNE
(NE in all subgames).
Solution to (iii)
There is of course another SPNE using all the reasoning above, but this time the mixed NE
( S
1
, S
2
) found in part (i) is used as the punishment (the stick), and the (M,C) as the credible reward (carrot). In this SPNE the strategies will be:
Y
1
= Play U in t=1, then if (U,L) happened in t=1, play C in t=2, otherwise in t=2 play S
1
.
Y
2
= Play L in t=1, then if (U,L) happened in t=1, play M in t=2, otherwise in t=2 play S
2
.
Note that the expected payoff in the stage game from ( S
1
, S
2
) is done by: 3 .
1
4
.
1
4
+ 1 .
3
4
.
3
4
=
3
4
.
3
4
, 3
4
. The payoff calculation is
Y = ( Y
1
, Y
2
) is SPNE if:
5 + 3 δ ≥ 6 +
3
4
δ
δ ≥ 4
9
2

Try to find a third SPNE where (D,R) is the carrot and ( S
1
, S
2
) is the stick. Think. Why will this fail?
Solution to (iv)
Clearly the second SPNE we found was easier to satisfy as we only needed
δ ≥
δ ≥ 4
9
, rather than
1
2
. This is because the first SPNE, with strategies X had a carrot worth 3 and a stick worth
1, whereas the second SPNE Y has a carrot worth 3 and a stick worth
3
4
. Hence, in the second
SPNE there are some less patient people with δ ∈ 4
9
, 1
2 who will not deviate to get 6 in t=1 because the stick/punishment is larger in t=2 (giving only
3
4
), whereas in the first SPNE they would deviate as the stick/punishment is less severe (giving 1), but the carrot is the same.
3