First mover advantage stackelberg case.

Question

So in the Stackelberg problem, there is usually a first mover advantage for the leading firm (where the first firm to enter has a competitive advantage). Is there ever a case where the leader firm produces less than a follower firm?

Answer 1

Yes, if you are willing to consider Nash equilibria that are not subgame perfect Nash equilibria.

For example, suppose that the follower firm, let's call it $F_{2}$, can always choose a quantity $y_{2}$ in response to the first firm's quantity $y_{1}$ such that the price falls to zero. In that case, the following strategy profile is a Nash equilibrium:

• For $F_{2}$, if $y_{1} = 0$, then choose $y_{2}$ equal to the monopolist's solution. If $y_{1} > 0$, then choose $y_{2}$ such that the price falls to zero.
• For $F_{1}$, choose $y_{1} = 0$.

$F_{1}$ has no incentive to deviate since all quantities that it might choose yield a profit of zero. $F_{2}$ has no incentive to deviate since along the path of play it gets the profit that it would get in a monopoly.

Of course, this profile being a Nash equilibrium hinges on the fact that $F_{2}$ is threatening to let the price drop to zero if $F_{1}$ chooses a non-zero quantity, and it is therefore not a subgame perfect NE.

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A related question has also been asked in this post: First mover advantage in a Stackelberg game, but that example consider price instead of quantity competition.