关于finite repeated PD

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For a finite repeated game with only one NE in each stage (repeated twice). eg.PD.
Can we find a NE which is not SPE?

I found that, if a PD with only 2 strategies for each player (cooprate and defect), then there is only one NE which is SPE.
But for a 3 strategies PD(only one NE in each stage), there exist a NE which is not SPE.
Is it correct?

Thanks!

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关键词：Repeated finite repeat nite ATE only game repeated correct defect

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REPEAT两次，NE 但是不是SPNE，那就要考虑NON CREDIBLE threat.
NE注重的是最终的结果，SPNE注重的是每一个阶段。

crozame 发表于 2011-11-7 22:31
REPEAT两次，NE 但是不是SPNE，那就要考虑NON CREDIBLE threat.
NE注重的是最终的结果，SPNE注重的是每一个 ...

谢谢
那就是说对于2*2的PD，因为没有credible threat 所以他的唯一NE 就是SPE。
但是对于3*3且只有1个NE in each stage(defect,defect)的game，也应该没有credible threat 吧？但是如果在stage1play nonNE strategy say(cooprate, defect) ,and in stage2 play NE (d,d) if stage1 paly (c,d).
play the worst strategy otherwise. 好像是NE 但不是SPE.

多次重复，PD，里面基本上都是NON-CREDIBLE THEAT。不用考虑CREDIBLE的情况啊。因为A威胁B后，B就会按照A的去做。但是，如果B没有按照A的去做，A也不会实施处罚，因为实施处罚会降低收益。但是最终的结果是NE。
抱歉，上面的分析在两次重复博弈中不管用，因为不论第一个阶段怎么做，在第二个阶段都会D。





一个比较重要的NE但是不是SPNE的概念是，不在EQUILIBRIUM PATH上面的SUBGAME 可以不是NE。


画出树图，比较清楚

额。。重复了，请编辑掉这个。。。

The answer is No
For PD,
there is only on NE in the stage game.
And if the stage game only have one unique NE, then for any finite repeated game,
the repeated game has a unique sub game perfect outcome:the NE of the stage game will played in every period.