Skip to main content
Economics

Return to Answer

deleted 2 characters in body
Source Link
Lin Jing
Lin Jing
  • 319
  • 1
  • 6

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} \Longrightarrow q_i^C = \dfrac{k}{\alpha - 1}. \end{equation}\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. And

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 equilibriumNE 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 deviatingdeviation in the future. This also sheds some light on self-control problem. The power of commitment benefits us in many cases.

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} \Longrightarrow 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. And 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 equilibrium 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 deviating in the future. This also sheds some light on self-control problem. The power of commitment benefits us in many cases.

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.

Source Link
Lin Jing
Lin Jing
  • 319
  • 1
  • 6

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} \Longrightarrow 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. And 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 equilibrium 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 deviating in the future. This also sheds some light on self-control problem. The power of commitment benefits us in many cases.