Please answer these questions on strategy and principal-agent problems and show your work so I can compare my solutions against yours - thanks!

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mathsworkmusic | (Level 2) Educator

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There are three long questions here. I will answer #2.

2) Two companies are in competition. The total profits available per year are $10 million dollars. Advertising costs a firm $3 million. If both advertise, the profits are split equally. If only one advertises, that firm gains all profits for that year. If neither advertise the profits are split equally.

a) For a single year in isolation the pay-off matrix for the simultaneous move game is as follows

       A        A'

A   (2,2)    (7,0)

A'  (0,7)    (5,5)

where A = advertise, A' = not advertise

The Nash equilibrium of a game is the shared strategy where both players optimize their strategy taking into account the options of the other player.

The Nash equilibrium is (AA) since the firm 1's pay-off is maximized at the same time as firm 2's pay-off being maximized (the first element is at its maximum over the column, and the second is at its maximum over the row).

b) Now suppose the firms compete for 5 consecutive years. We wish to find the sub-game perfect Nash equilibrium of the game. To find this we go through the game tree backwards. A sub-game perfect Nash equilibrium is a game strategy that implies all sub-games are in Nash equilibrium. A sub-game is any particular path through the nodes of the game decision tree - here, the forks from each node in a particular year go to new nodes for the next year representing the outcomes AA,AA',A'A or A'A'.

Beginning with the final (5th) year, whatever happened in the 4th year the choice AA will be the best for both players as they stand to gain the most on whatever profits they have made in previous years. This same concept applies in the 4th, then 3rd, then 2nd and finally 1st year, meaning that the only sub-game perfect Nash equilibrium for this 5-year game is 11111 where 1 represents the strategy AA (both advertise).

c) We now suppose we have the set-up where the profit each year depreciates by 20%, that is the value next year is 0.8 times the value of last year, and that the game goes on interminably, year after year. We first should notice that after a few years that the cost of advertising ($3 million) will outweigh the available profits, in the first instance when both firms advertise (strategy 1).

The profits available each year progress as follows: 10, 8, 6.4, 5.12.

In the 4th year then, the evenly split profit 5.12/2 = 2.56 is outweighed by the advertising cost (=3). The pay-off matrix for the 4th year in isolation would be

             A                A'

A   (-0.44,-0.44)   (2.12,0)

A'  (0, 2.12)         (2.56,2.56)

whose Nash equilibrium is A'A' (neither firm advertises, since neither can afford to, and copying the other firm yields better returns).

Now, any sub-game perfect Nash equilibrium of this interminable game would include the 4th year optimal decision A'A' (or strategy 4). Also, for every year succeeding the 4th year, A'A' is the preferred strategy, and more and more so with each year as the cost of advertising, even when all the profit is cornered, outweighs the equally shared profit. As the number of years approaches infinity, the profit tends towards zero, as does the (equally) shared profit. For the years preceding the 4th (1st, 2nd and 3rd years), the optimal decision is AA (or strategy 1).

The sub-game perfect Nash equilibrium for this interminable game then is

1114444444...4 (both advertising for the first 3 years, then neither advertising ever again subsequently).

The 'grim trigger' strategy of first not advertising, and then advertising once the competitor starts advertising is not a sub-game equilibrium of this game since the option in the first year would always be to advertise, since this is the optimal option. The particular sub-game that is simply the 1st year's 'game' wouldn't be in Nash equilibrium. For a game strategy to be a sub-game Nash equilibrium, all sub-games must be in Nash equilibrium.


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