# About the normal form of a repeated game

Consider an infinitely-repeated game with perfect monitoring and pure strategies. The one-shot deviation principle tells us that a strategy profile $$\sigma$$ is a subgame perfect Nash equilibrium if and only if there are no profitable one-shot deviations.

More precisely, let $$h^t$$ denote the history available to players up to the beginning of period $$t$$.

For each history $$h^t$$, let $$a^*=\sigma(h^t)$$. It must be that
$$(1-\delta) u_i(a^*)+\delta U_i(\sigma|_{(h^t,a^*)})\geq (1-\delta) u_i(a_i, a^*_{-i})+\delta U_i(\sigma|_{(h^t,a_i, a^*_{-i})})\quad \forall a_i, \forall i$$ where

• $$\delta$$ is the discount factor

• $$u_i$$ is the stage game payoff

• $$\sigma|_{(h^t,a)}$$ is player $$i$$'s continuation strategy induced by history $$h^t$$ and action $$a$$.

• $$U_i$$ is the continuation payoff

Question: My lecture notes say that in this setting $$U_i(\sigma|_{h^t,a})$$ just depends on $$a$$ and so can be denoted by $$v_i(a)$$. Hence, the normal form of this game with 2 players and 2 actions $$\{0,1\}$$ is

|      (1-delta)u_1(1,1)+delta*v_1(1,1)  | (1-delta)u_1(1,0)+delta*v_1(1,0)
|      (1-delta)u_2(1,1)+delta*v_2(1,1)  | (1-delta)u_2(1,0)+delta*v_2(1,0)
|----------------------------------------|----------------------------------------
|      (1-delta)u_1(0,1)+delta*v_1(0,1)  | (1-delta)u_1(0,0)+delta*v_1(0,0)
|      (1-delta)u_2(0,1)+delta*v_2(0,1)  | (1-delta)u_2(0,0)+delta*v_2(0,0)


I don't understand this step at all. In particular:

(1) The fact that I can write the normal form of a repeated game as above is a consequence of the one-shot deviation principle?

(2) Suppose in the above normal-form game, $$(0,0)$$ is a Nash Equilibrium for given values of $$v_1, v_2$$. What is the relation between such Nash equilibrium and a subgame perfect Nash equilibrium of the repeated game?

(3) Suppose in the above normal-form game, $$(0,0)$$ and $$(0,1)$$ are Nash Equilibria for given values of $$v_1, v_2$$. Does this imply that there are multiple subgame perfect Nash equilibria in the repeated game as well?

The intuition is that, following any history $$h^t$$, the continuation strategy should be a subgame perfect equilibrium $$\sigma|h^t$$, which allows you to predict the value of continuation play ($$V(\sigma|h^t)$$) with certainty. Therefore, when choosing the current-stage strategy, a player know her action $$a$$ counts toward $$h^t$$, which in turn, leads to $$\sigma|h^{t-1}\cup\{a\}$$, generating a certain continuation payoff $$v_i(\sigma|h^{t-1}\cup\{a\})$$. And because other things are already determined, you can simply write $$v_i(a)$$. This is precisely why the continuation payoff is in fact history-strategy dependent.

The normal form says the players are choosing today's strategy $$a$$ by trading off today's payoff $$u_i(a)$$ and (discounted) future valuation $$v_i(a)$$. You can use the normal form because everything can be summarised into a current-stage variable $$a$$.

Edits: Question (1): you are right, there're infinitely many strategies for each player. However, to deal with this repeated game, we conjecture that there is some SPE $$\sigma$$ being played following action $$a$$. One-shot deviation principle makes it clear you never deviate from $$\sigma$$. Think it as some black-box algorithm that automatically choose the best action for you forever, after your play for the current period. In this sense, any future strategies are summarized by the current action $$a$$, and $$U_i(\sigma|(h_{t-1}, a)$$ is function of only $$a$$ but not anything else, so you can relabel it as $$v_i(a)$$.

Question (2): Then the SPE of the repeated game contains $$(0,0)$$ as the first period strategy, e.g., Round 1: play $$0$$; Round 2: if $$(0,0)$$ was played in round 1, play $$a_{(0,0)}$$, otherwise...

Since it's an infinitely repeated game, you should have one SPE where $$(0,0)$$ is played every period. E.g., infinitely repeated prisoner's dilemma, you have (confess,confess) to be played no matter what'' as an SPE, but you can also have grim trigger -- (deny, deny) is a Nash equilibrium in your normal form, and for next period, if (deny, deny) is played, then continue (deny, deny); otherwise, (confess, confess) forever.

Question (3): Yes. From the logic above, you should see there can be infinitely many SPE.

• Thanks DiZ. First, you made me realise that I forgot to add $a$ in the continuation strategy, I've edited my question. Still, I do not understand how we go from $U_i(\sigma| (h_t, a))$ to $\nu_i(a)$. Where do $\sigma$ and $h_t$ go? Are they implicit?
– Star
Commented Jul 10 at 20:34
• I have split my questions into 3 subquestions. If you could answer those, in particular (3), it would be very helpful.
– Star
Commented Jul 10 at 21:23
• Hi Star, thanks for clarifying. Just edited my original answer.
– DiZ
Commented Jul 11 at 2:14
• Thanks, but your answers do not seem to match my questions
– Star
Commented Jul 11 at 17:14
• Sorry for the confusion, I have edited my answers to those questions.
– DiZ
Commented Jul 11 at 18:11