Finding the subgame-perfect equilibrium in sequential games with infinite action spaces

Question

Firms 1 and 2 are competing in the same market. Each firm $i$ must choose a quantity $q_{i}$ to supply, and the market price $p$ will depend on their choices according to the inverse demand formula $p(q_{1},q_{2}) = \text{max}[A-(q_{1} + q_{2}), 0]$.

The total cost of production for each firm $i$ is $q_{i}^2$, and so the total profit for firm ${i}$ will be $u_{i}(q_{1},q_{2}) = p(q_{1},q_{2}) \cdot q_{i} - (q_{i})^{2}$

Suppose that firm 1 chooses $q_{1}$ first, and then firm 2 chooses its $q_{2}$ after observing $q_{1}$.

Find : The subgame-perfect equilibrium of this game with perfect information, and compute each firm's expected profit in this equilibrium.

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(What I have tried:)

For any given $q_{2}$, firm 1's best response $q_{1}$ to maximize $u_{1}$ is:

\begin{equation} q_{1}(q_{2}) = \begin{cases} \frac{A-q_{2}}{4} & \text{if } q_{2} < A \\ 0 & \text{if } q_{2} > A \end{cases} \end{equation}

Similarly, for any given $q_{1}$, firm 2's best response $q_{2}$ to maximize $u_{2}$ is:

\begin{equation} q_{2}(q_{1}) = \begin{cases} \frac{A-q_{1}}{4} & \text{if } q_{1} < A \\ 0 & \text{if } q_{1} > A \end{cases} \end{equation}

Therefore, the Nash equilibrium when the two firms choose $q_{1}$ and $q_{2}$ simultaneously and independently are:

$$(q_{1}, q_{2}) = (\frac{A}{5}, \frac{A}{5})$$

With that Nash equilibrium, the expected utilities of firm 1 and 2 are:

$$EU_{1} = EU_{2} = \frac{2A^2}{25}$$

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(Question):

For the subgame-perfect equilibrium, though, how would one use backward induction to find it in this game?

The answer key notes that the subgame-perfect equilibrium is

$$(q_{1}^{s}, q_{2}^{s}) = (\frac{3}{14}A, \frac{11}{56}A)$$

,but how does one arrive at that answer?

Answer 1

In the sequential game, since firm 2 chooses its $q_2$ after observing firm $1$'s $q_1$, firm 2's strategy will be its best response function which is given by:

$q_2(q_1)=\max\left(\dfrac{A-q_1}{4},0\right)$

Given firm $2$'s strategy above, firm $1$ will choose its action (or strategy) $q_1$ by solving the following problem:

\begin{eqnarray*} \max_{q_1\geq 0} & \max(A-q_1-q_2,0)q_1-q_1^2 \\ \text{s.t. } & q_2=\max\left(\dfrac{A-q_1}{4},0\right)\end{eqnarray*}

which is equivalent to solving: \begin{eqnarray*} \max_{0\leq q_1\leq A} & \dfrac{3(A-q_1)}{4}q_1-q_1^2 \end{eqnarray*} Solving it, we get $q_1=\dfrac{3A}{14}$. And therefore, $q_2=\max\left(\dfrac{A-q_1}{4},0\right)=\dfrac{11A}{56}$

Answer 2

You should still use backward induction. Since the game is not finite, you should not look at the penultimate nodes corresponding to different $q_1$ quantities one by one, but rather figure out what the second mover does in response to a general $q_1$ quantity. (Treat $q_1$ as a parameter in the second player's decision problem.) Once you have done this, just like in finite backward induction, player 1 can 'foresee' the action of player 2, and optimize their own choice of action accordingly.