Answer 1

	L	M	H
L	(2,2)	(1, 3)	(0,4)
M	(3,1)	(1,1)	(1,3)
H	(4,0)	(3,1)	(-1, -1)

Source: (as created by the author)

 The table 1 represents the pay-off matrix of the two social network platforms, FB and SC. The first position in each pair in each cell indicates the pay-off to firm FB, whereas, the second position denotes the pay-off to firm SC. Here, both FB and SC are playing the simultaneous game. It implies that both players are moving at the same time. Suppose FB is following the L strategy, SC will choose the H strategy (since 4 >2). Similarly, SC will prefer to play following the H strategy again if FB plays the M strategy (since 3> 1). However, the decision will not remain same while FB follows the H strategy. Altogether, column L will be deleted following the iterated elimination of dominated strategies.  On the other hand, FB would like to follow the H strategy when SC follows the L and M strategies. Conversely, H will not remain the dominating strategy when SC prefers to choose H strategy. Yet, H will be considered as the dominating strategy as players would prefer to play following the H strategy for two times. In this case, the row L will be nullified owing to the iterated dominated strategy.

                                                    Table 2: Iterated pay-off matrix

	M	H
M	(1,1)	(1,3)
H	(3,1)	(-1, -1)

Source: (as created by the author)

Table 2 represents the pay-off values after the application of iterated elimination rule. In this step, the paper will try to derive optimum dominating strategy for the players. Here, SC would prefer to choose H while FB follows M strategy. On the contrary, SC would prefer to M when FB likes H strategy. This implies that there is no dominant strategy for both SC and FB. Here, the middle probabilistic strategy can be applied as to determine the dominant strategy. Suppose probability of 0.5 has been added to both the M and H strategy. Therefore, the new pay-off matrix will be (2,1) and (0,1) which cannot be considered as the dominant strategy.

Answer to question 2

Nash equilibrium denotes the optimum outcome when no one wants deviate from that situation. In this case, (1, 3) and (3,1) are the multiple Nash equilibrium for FB and SC. It has been observed that FB would like to choose H when SC plays either M or H. On the contrary, SC would like to choose M when FB plays either M or H.

Answer to question 3.a

Figure 1: extensive form of game theory

Source: (as created by the author)

After the introduction of subsidy or incentives, FB should be provided with subsidy. On the contrary, SC should be levied with tax to meet the optimum outcome. Therefore, (1,1) can be derived as the optimum outcome under the extensive form of game theory.

Answer to question 3b

If FB moves first, it will move either UP or down. On the other hand, SC as a player 2 will move either left or right.

Figure 2: Backward induction game

Source: (as created by the author)

In backward induction form, the game starts from backward point and will move until reaches the ultimate solution. As B takes the first move, it will have two options (1,1) and (1,3). Afterwards, SC will choose from the rest of the options. Here, (3,1) will be optimum outcome.

Mixed strategies

Answer to question 1

Pure strategy equilibrium refers to one specific strategy that a player will prefer to choose. Player 1 will choose the strategy U rather than D (as 4 >-2) while player 2 follows the L strategy. On the other hand, Player 1 will prefer the strategy D rather than U (since 1> -3) given the player 2’s strategy of R. Therefore, there is no dominant strategy for player 1. On the contrary, player 2 will prefer to choose R (since 3> -4) when player 1 plays strategy U. However, the preference will be altered when player 1 chooses strategy D. Then, player 2 would love to follow strategy L (as 2>1). Altogether, it can be said that there is no equilibrium in pure strategy.

Answer to question 2

	L	R
L	(4,-4)	(-3,3)
R	(-2,2)	(1, 1)

Player 1’s Mixed Strategy

P (-4) + (1 –P) 2 = P (3) + (1 –P) 1

Or, 2 – 6P = 2P +1

Hence, P = 1/8

Player 2’s mixed strategy

q (4) + (1 –q) (-3) = q(-2) + (1 –q) 1

or, 7q -3 = -3q +1

Hence, q = 2/5

Therefore, the equilibrium mixed strategy (1/8, 2/5).

Bibliography

Heifetz, Aviad, and Andrés Perea. “On the outcome equivalence of backward induction and extensive form rationalizability.” International Journal of Game Theory 44, no. 1 (2015): 37-59.

Prokopovych, Pavlo, and Nicholas C. Yannelis. “On the existence of mixed strategy Nash equilibria.” Journal of mathematical economics 52 (2014): 87-97.