I have had an exam (exam is now past and submitted, but I want to now understand the solution without waiting) with the following questions:


Consider two firms playing the following two-stage game:

Firm face the following inverse demand:

$$ P(Q) = \frac{S+s_1 + s_2}{(Q+k)^{\alpha}}, Q=q_1 + q_2. $$

in the first stage, firms can simultaneously lobby to ease trade restrictions by increasing $s_i$, for each unit of $s_i$ each firm pays $\frac{s_i^3}{9}$.

In the second stage they observe each other lobbying choice and set quantity simultaneously. Production costs are zero.

Let $\alpha = 3$ and $k=1$. Show that there is a NE with $S+s_1 +s_2 > S + s_1^{spne} + s_2^{spne}$ and explain why this is NOT a subgame perfect.


SPNE is easy to find by backward induction, we know that in the second stage firms profits are

$$\pi_i = \frac{S'}{(Q+k)^{\alpha}}q_i - \frac{s_i^3}{9},$$

$S'$ is a fixed constant at this stage, so by differentiating we find the best responses and the optimal quantities

$$ q_1 = (q_2 + k)/(\alpha-1)\to q^*_i = \frac{k}{\alpha-1}. $$.

Anticipating this, at stage one, firm will want to maximize

$$\pi_i = \frac{S'}{(Q^*+k)^{\alpha}}q^*_i - \frac{s_i^3}{9},$$

deriving in $s_i$ we find the solution to be

$$ s^*_i = \sqrt{\frac{3q^*_i}{(Q^* +k)^{\alpha}}} $$

Another SPNE?

I have tried finding an NE, but could not. Only thing I can find is what I think is another SPNE in which strategies are:

Both play $(\hat{s}, q_i^*)$ where $\hat{s} > s_i^*$ . If the other player deviates in the first stage, then the other will punish the other in the second stage by producing some

$$ q^p $$

such that $$\pi_1(\hat{s},s_2^* ,q^p,q_2^*) = \pi^{spne}$$ while $$\pi_2(\hat{s},s_2^* ,q^p,q_2^*) < \pi_2(\hat{s},\hat{s} ,q_1^* ,q_2^*) $$ This should be -- assuming there is such a $q^p$ -- a credible threat since the player can get the same as the previous SPNE payoff and effective since it lowers player 2 profits by lowering the demand it receives.

Is this is an equilibrium at all and is it an SPNE or just a NE?

  • 1
    $\begingroup$ What is capital $S$ by the way ? $\endgroup$
    – Lin Jing
    Commented May 27, 2020 at 23:29
  • 1
    $\begingroup$ It should be $q_i^*=\frac{k}{\alpha-2}$. $\endgroup$
    – VARulle
    Commented May 28, 2020 at 0:00

2 Answers 2


Since the choice of $q_i$ can be conditioned on $(s_i,s_j)$, strategies in this game are of the form $(\hat s_i, \hat q_i(s_i,s_j))$. For the given values $\alpha=3$ and $k=1$, the SPNE can be calculated as the profile where $s^*_i=1/3$ and $q^*_i\equiv 1$. Indeed, production levels $q_i^*=1$ are the unique NE in all subgames, independently of the chosen $s_i$-levels in the first stage. Therefore your suggested "other SPNE" is not an SPNE.

But consider the following strategy of player 1: $\hat s_1=1/3$ and $\hat q_1(s_1,s_2)=\left\{ \begin{array}\ 1 & \ldots & s_2 = 1/3+\epsilon \\ 2 & \ldots & s_2 \ne 1/3+\epsilon \end{array} \right\}$, where $\epsilon>0$. Let player 2's strategy be $\hat s_2=1/3+\epsilon$ and $\hat q_2(s_1,s_2)\equiv 1$. Then $(\hat s_1,\hat q_1)$ is a best response to $(\hat s_2,\hat q_2)$, and provided $\epsilon$ is small enough that deviating in the first stage doesn't pay for player 2, $(\hat s_2,\hat q_2)$ is also a best response to $(\hat s_1,\hat q_1)$. Therefore the corresponding strategy profile is a NE with higher total lobbying than in the SPNE (but with the same production levels), benefiting player 1.

However, it is not subgame perfect, since if player 2 deviated in the first stage, player 1 would not carry out his threat to double production in the second stage. Technically speaking, the NE induces production levels $\hat q_1=2$ and $\hat q_2=1$ in all subgames following a deviation of player 2 in the first stage, but these production levels are not in equilibrium in these subgames.

  • $\begingroup$ Your answer is much better than mine, and yes I am wrong here. I upvote for you, while I'll still keep my answer here to make a comparison. $\endgroup$
    – Lin Jing
    Commented May 29, 2020 at 13:47

I'll assume that $S$ is a constant. By the way, the solution for the SPNE is $q_i = \frac{k}{\alpha - 2}$ which is derived from $q_i = (q_j + k)/(\alpha - 1) = ((q_i + k)/(\alpha - 1) + k)/(\alpha - 1)$.

To answer your question, one should know what SPNE does. It is actually a refinement of NE by eliminating non-credible threats. Knowing this, the idea would be that, firm $i$ claims to commit to a production level, and thus maximizes his payoff. However, in the second stage of the game, he can actually be better off by deviating from his commitment. Let's go into the details.

Since $q_j = (q_i + k)/(\alpha - 1)$, substitude this into the player $i$'s payoff function, we get \begin{equation} \pi_i = \dfrac{S+s_i + s_j}{(q_i + \frac{q_i + k}{\alpha - 1} + k)^\alpha}q_i - \dfrac{s_i^3}{9} \quad \Longrightarrow \quad q_i^C = \dfrac{k}{\alpha - 1}. \end{equation}

The best response of player $j$ is to choose $q_j^C = \frac{q_i + k}{\alpha - 1} = \frac{\alpha k}{(\alpha - 1)^2} $.

Substitute the optimal production level with player $i$'s commitment power, you can get $s_i$ and $s_j$ which will have a higher sum than that of SPNE.

However, in the second stage, player $i$ knows that player $j$ will choose $q_j^C=\frac{\alpha k}{(\alpha - 1)^2}$ if he can commit to $q_i^C = \frac{k}{\alpha - 1}$, then what's the best response of player $i$? I'll leave this for you and you'll find out that player $i$'s best response now is different from $q_i^C$. Therefore, this is not an SPNE.

However, as long as player $i$ deviates from $q_i^C$, player $j$ will also adjust his production level, and they will reach the SPNE in the end. Therefore, $q_i^C$ and $q_j^C$ can only be an NE if player $i$ does have the commitment power. A side note is that $q_i^C$ is higher than the production level under SPNE, so player $i$ is better off by being able to commit.

This is closely related to the Stackelberg competition model where one firm commits to a production level by moving first. In such a case, he eliminates the possibility that he will move away from his committed production choice in the second stage. He knows he will deviate in the second stage, so he chooses a way to commit to a certain behavior by eliminating the possibility of deviation in the future. This also sheds some light on self-control problem. The power of commitment benefits us in many cases.

  • $\begingroup$ What do you mean with "Therefore, $q^C_i$ and $q^C_j$ can only be an NE if player $i$ does have the commitment power."? The strategy profile is either a NE or not. $\endgroup$
    – VARulle
    Commented May 29, 2020 at 8:31
  • $\begingroup$ This is only an NE If player $i$ commits to the production level $q_i^C$ and won't change it in the second stage. If he cannot commit, in the second stage, $q_i^C$ is not a best response to $q_j^C$, so he will deviate, and then player $j$ will also deviate, until they reach the original SPNE. $\endgroup$
    – Lin Jing
    Commented May 29, 2020 at 10:49
  • $\begingroup$ It is an NE when player $i$ can commit/stick to $q_i^C$, and $q_j^C$ is player $j$'s best response. It is not an SPNE because $q_i^C$ is not the best response of $q_j^C$. $\endgroup$
    – Lin Jing
    Commented May 29, 2020 at 10:59
  • $\begingroup$ It's a NE if player $i$ can move first. But this is not the case in the given game, so it's not a NE of the given game. Your answer is not an answer to the question asked here. $\endgroup$
    – VARulle
    Commented May 29, 2020 at 11:33
  • $\begingroup$ Yes, moving first definitely generates the commitment power, but commitment power does not necessarily have to be the case of moving first. $\endgroup$
    – Lin Jing
    Commented May 29, 2020 at 13:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.