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

<p>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 <span class="math-container">$i$</span> 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. </p>

<p>Since <span class="math-container">$q_j = (q_i + k)/(\alpha - 1)$</span>, substitude this into the player <span class="math-container">$i$</span>'s payoff function, we get 
<span class="math-container">\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}</span></p>

<p>The best response of player <span class="math-container">$j$</span> is to choose <span class="math-container">$q_j^C = \frac{q_i + k}{\alpha - 1} = \frac{\alpha k}{(\alpha - 1)^2} $</span>. </p>

<p>Substitute the optimal production level with player <span class="math-container">$i$</span>'s commitment power, you can get <span class="math-container">$s_i$</span> and <span class="math-container">$s_j$</span> which will have a higher sum than that of SPNE. </p>

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

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

<p>This is closely related to the <em>Stackelberg competition</em> 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. </p>

CC BY-SA 4.0
2020-05-29T08:31:49.247Z
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.
VARulle
CC BY-SA 4.0
2020-05-29T10:49:27.017Z
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.
Lin Jing
CC BY-SA 4.0
2020-05-29T10:59:01.750Z
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 $q_j^C$.
Lin Jing
CC BY-SA 4.0
2020-05-29T11:33:10.040Z
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.
VARulle
CC BY-SA 4.0
2020-05-29T13:23:39.463Z
Yes, moving first definitely generates the commitment power, but commitment power does not necessarily have to be the case of moving first.
Lin Jing
CC BY-SA 4.0
2020-05-30T08:20:16.983Z
I think it does. Not chronologically, but in the game tree. How else would you model perfect commitment in a game tree?
VARulle
CC BY-SA 4.0
2020-05-28T01:10:15.963Z
2020-05-28T01:17:37.700Z
2020-05-28T01:17:37.700Z
Lin Jing
<p>Since the choice of <span class="math-container">$q_i$</span> can be conditioned on <span class="math-container">$(s_i,s_j)$</span>, strategies in this game are of the form <span class="math-container">$(\hat s_i, \hat q_i(s_i,s_j))$</span>. For the given values <span class="math-container">$\alpha=3$</span> and <span class="math-container">$k=1$</span>, the SPNE can be calculated as the profile where <span class="math-container">$s^*_i=1/3$</span> and <span class="math-container">$q^*_i\equiv 1$</span>. Indeed, production levels <span class="math-container">$q_i^*=1$</span> are the unique NE in all subgames, independently of the chosen <span class="math-container">$s_i$</span>-levels in the first stage. Therefore your suggested "other SPNE" is not an SPNE.</p>

<p>But consider the following strategy of player 1: <span class="math-container">$\hat s_1=1/3$</span> and <span class="math-container">$\hat q_1(s_1,s_2)=\left\{
\begin{array}\ 1 &amp; \ldots &amp; s_2 = 1/3+\epsilon \\ 2 &amp; \ldots &amp; s_2 \ne 1/3+\epsilon \end{array}
\right\}$</span>, where <span class="math-container">$\epsilon&gt;0$</span>. Let player 2's strategy be <span class="math-container">$\hat s_2=1/3+\epsilon$</span> and <span class="math-container">$\hat q_2(s_1,s_2)\equiv 1$</span>. Then <span class="math-container">$(\hat s_1,\hat q_1)$</span> is a best response to <span class="math-container">$(\hat s_2,\hat q_2)$</span>, and provided <span class="math-container">$\epsilon$</span> is small enough that deviating in the first stage doesn't pay for player 2, <span class="math-container">$(\hat s_2,\hat q_2)$</span> is also a best response to <span class="math-container">$(\hat s_1,\hat q_1)$</span>. 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.</p>

<p>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 <span class="math-container">$\hat q_1=2$</span> and <span class="math-container">$\hat q_2=1$</span> in all subgames following a deviation of player 2 in the first stage, but these production levels are not in equilibrium in these subgames.</p>

CC BY-SA 4.0
2020-05-29T13:47:46.400Z
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.
Lin Jing
CC BY-SA 4.0
2020-05-29T12:12:47.910Z
2020-05-29T12:12:47.910Z
VARulle
<p>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:</p>

<p><strong>GAME</strong></p>

<p>Consider two firms playing the following two-stage game: </p>

<p>Firm face the following inverse demand:</p>

<p><span class="math-container">$$
P(Q) = \frac{S+s_1 + s_2}{(Q+k)^{\alpha}}, Q=q_1 + q_2.
$$</span></p>

<p>in the first stage, firms can simultaneously lobby to ease trade restrictions by increasing <span class="math-container">$s_i$</span>, for each unit of <span class="math-container">$s_i$</span> each firm pays <span class="math-container">$\frac{s_i^3}{9}$</span>.</p>

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

<p>Let <span class="math-container">$\alpha = 3$</span> and <span class="math-container">$k=1$</span>. Show that there is a NE with <span class="math-container">$S+s_1 +s_2 &gt; S + s_1^{spne} + s_2^{spne}$</span> and explain why this is NOT a subgame perfect.</p>

<p><strong>SPNE</strong></p>

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

<p><span class="math-container">$$\pi_i = \frac{S'}{(Q+k)^{\alpha}}q_i - \frac{s_i^3}{9},$$</span></p>

<p><span class="math-container">$S'$</span> is a fixed constant at this stage, so by differentiating we find the best responses and the optimal quantities</p>

<p><span class="math-container">$$
q_1 = (q_2 + k)/(\alpha-1)\to  q^*_i = \frac{k}{\alpha-1}.
$$</span>.</p>

<p>Anticipating this, at stage one, firm will want to maximize</p>

<p><span class="math-container">$$\pi_i = \frac{S'}{(Q^*+k)^{\alpha}}q^*_i - \frac{s_i^3}{9},$$</span></p>

<p>deriving in <span class="math-container">$s_i$</span> we find the solution to be</p>

<p><span class="math-container">$$
s^*_i = \sqrt{\frac{3q^*_i}{(Q^* +k)^{\alpha}}}
$$</span></p>

<p><strong>Another SPNE?</strong></p>

<p>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:</p>

<p>Both play <span class="math-container">$(\hat{s}, q_i^*)$</span> where <span class="math-container">$\hat{s} &gt; s_i^*$</span> . If the other player deviates in the first stage, then the other will punish the other in the second stage by producing some</p>

<p><span class="math-container">$$ q^p $$</span></p>

<p>such that <span class="math-container">$$\pi_1(\hat{s},s_2^* ,q^p,q_2^*) = \pi^{spne}$$</span> while
<span class="math-container">$$\pi_2(\hat{s},s_2^* ,q^p,q_2^*) &lt; \pi_2(\hat{s},\hat{s} ,q_1^* ,q_2^*) $$</span> This should be -- assuming there is such a <span class="math-container">$q^p$</span> -- 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.</p>

<p>Is this is an equilibrium at all and is it an SPNE or just a NE?</p>

CC BY-SA 4.0
2020-05-27T23:29:20.617Z
What is capital $S$ by the way ?
Lin Jing
CC BY-SA 4.0
2020-05-28T00:00:34.757Z
It should be $q_i^*=\frac{k}{\alpha-2}$.
VARulle
<game-theory><nash-equilibrium>
NE equilibrium with lobbying of cournot producers
asderello
# NE equilibrium with lobbying of cournot producers

(Asked by: username_0 on 2020-05-27)

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:

**GAME**

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**

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?

**Tags:** game-theory, nash-equilibrium
> **Comments:**
> **username_1**: What is capital $S$ by the way ?  
> **username_2**: It should be $q_i^*=\frac{k}{\alpha-2}$.  

---

## 2 answers

### Answer by username_2 on 2020-05-29. Score: 4 | Accepted answer

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.

> **Comments:**
> **username_1**: 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.  

---

### Answer by username_1 on 2020-05-28. Score: 0

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.

> **Comments:**
> **username_2**: 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.  
> **username_1**: 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.  
> **username_1**: 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 $q_j^C$.  
> **username_2**: 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.  
> **username_1**: Yes, moving first definitely generates the commitment power, but commitment power does not necessarily have to be the case of moving first.  
> **username_2**: I think it does. Not chronologically, but in the game tree. How else would you model perfect commitment in a game tree?  

---

