4. Dynamic Games Below is a game matrix that shows a typical prisoners’ dilemma.

Question

4. Dynamic Games

Below is a game matrix that shows a typical prisoners’ dilemma. In the static version of the game, the unique Nash equilibrium is for each firm to charge $4.

Suppose this prisoners’ dilemma game is played an infinite number of times. Suppose further that Firm 2 uses a tit-for-tat strategy: It initially chooses $6 and, for each future round, chooses the price that Firm 1 chose in the previous period.

If Firm 1 plays $4 in the first round and $6 in all later rounds, its combined payoff over the first four rounds is ____________. If, instead, Firm 1 plays $6 in the first round and in all later rounds, its combined payoff over the first four rounds is __________. If Firm 1 plays a tit-for-tat strategy its combined payoff over the first four rounds is _________.

The strategy combination of Firm 1 playing tit-for-tat and Firm 2 playing tit-for-tat is a Nash equilibrium. Is it a Nash equilibrium for Firm 2 to play tit for tat and for Firm 1 to adopt a strategy of always playing $6. [YES   NO]. Explain: _______________________________________________________________________________

Now suppose that the game ends after four rounds, and this is common knowledge. What can you say about any Nash equilibria in this game?

______________________________________________________________________________________________________________________________________________________________.

Firms 1 and 2 produce identical products and both operate with zero cost. The inverse industry demand function is P = 6 – Q. If the firms play a simultaneous move (Cournot) game, each has a best response function, qi = 3 – 0.5qj (i,j = 1,2) and each firm produces an output of 2.

(i) Suppose now that Firm 1 is a Stackelberg leader. Verify that it will produce q1 = 3. What output will Firm 2 produce?

(ii) Construct a game tree with Firm 1 choosing either q1 = 2 or q1 = 3 as the first mover. Assume that if Firm 1 chooses q1 = 2 then Firm 2 also chooses the Nash-Cournot quantity, q2 = 2. Also suppose that if Firm 1 chooses q1 = 3 then Firm 2 chooses one of three quantities: q2 = 2 (Nash-Cournot), q2 = 3 (Stackelberg leader) or q2 = 1.5 (Stackelberg follower). Use P = 6 – Q to calculate the profits for each firm with the different production combinations. Add the payoffs to the game tree below and identify the subgame perfect (sequential) Nash equilibrium in the game tree below.

Explanation / Answer

Consider the given game here both the players have 2 possible strategies, “Charge $6” and “Charge $4”.

So, if this game will be played 1 time then there is an unique NE, which is (“Charge $4”, “Charge $4”)=(3, 3).

Since here for both players the dominant strategy is, ““Charge $4”.

Now, suppose the “F-2” will use “tit-for-tat”, strategy and if “F1” will start with ““Charge $4”, then will keep playing with “Charge $6”, then in the 1st stage the outcome will be (“Charge $4”, “Charge $6”)=(8,1) , in the 2nd stage it will be (“Charge $6”, “Charge $4”)=(1, 8), and in the last 2 stage they both will play “Charge $6” with pay off (6, 6).

Since in the 1st stage “F1” will play “4”, so in the 2nd stage “F2” will play “4” and since from the 2nd stage “F1” will keep playing “6” irrespective of “F2”, => form the 3rd stage both of them will continue with playing “6”.

So, the “total pay off” for “f1” = 8+1+6+6=21 and for “F2”=1+8+6+6=21 at the end of the 4th stage.

Now, let’s assume that “F1” will start with “6” and will keep playing with “6” irrespective of “F2”, so here since “F1” will continue with playing “6” so “F2” will also continue with playing “6”, so in each and every round the unique outcome be (“Charge $6”, “Charge $6”)=(6, 6).

So, the payoff of “F1” and the end of the 4 stage is given by, “6+6+6+6=24” and same goes for “F2” here.

Now, suppose “F1” will also play “tit-for tat”, => will start with playing “6” and in future will play what the “F2” will play in the previous stage.

So, here also in each and every stage only 1 strategy will be played, => (“Charge $6”, “Charge $6”)=(6, 6).

Since both of them will start with “6”, => both of them will also play the same strategy in the latter stage and this will keep continue. So here also both of them will get “24” as a total payoff at the end of the 4 stage.

“YES”, if “F1” play “tit-for-tat”, then the optimum for “F2” is do the same, => it’s a “NE”. her, but if “F1” will play only “6” irrespective of “F2” then “tit-for-tat” is not the optimum choice for “F2”, so under the 2nd case it’s not a “NE”.

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