<p>You could add up the equations
\begin{align}
\delta(1-\underline{S})S_j=e_i(S_i+S_j)^2 \\
\delta(1-\underline{S})S_i=e_j(S_i+S_j)^2
\end{align}
to get
\begin{align}
\delta(1-\underline{S})(S_i+S_j)=(e_i+e_j)(S_i+S_j)^2
\end{align}
which simplifies to
\begin{align}
\delta(1-\underline{S})=(e_i+e_j)(S_i+S_j).
\end{align}
This gives you $(S_i+S_j)$. From the initial equations
\begin{align}
\delta(1-\underline{S})S_j=e_i(S_i+S_j)^2 \\
\delta(1-\underline{S})S_i=e_j(S_i+S_j)^2
\end{align}
you can also get
$$
\frac{S_j}{e_i} = \frac{(S_i+S_j)^2}{\delta(1-\underline{S})} = \frac{S_i}{e_j}, 
$$
which gives you
$$
\frac{S_j}{S_i} = \frac{e_i}{e_j}.
$$
Now you have both the sum and ratio of the variables $S_j,S_i$. Calculating their individual values should be a straightforward matter.</p>

CC BY-SA 3.0
2017-07-22T09:29:36.713Z
@Alfa have you consider accepting the answer?
luchonacho
CC BY-SA 3.0
2017-08-24T19:01:28.707Z
@luchonacho yes, I accept the answer. It gives me the solution.
Alfa
CC BY-SA 3.0
2017-08-24T21:29:45.063Z
@Alfa To accept an answer on Stackexchange you have to press the green check sign to the left of the answer.
Giskard
CC BY-SA 3.0
2017-08-25T08:26:40.327Z
@Alfa ahaha funny. Yes, you need to formally accept it :)
luchonacho
CC BY-SA 3.0
2017-07-08T08:06:57.753Z
Thanks a lot!! I appreciate your taking time answering my question.
Alfa
CC BY-SA 3.0
2017-07-07T13:27:51.983Z
2017-07-07T14:00:21.383Z
2017-07-07T14:00:21.383Z
Giskard
<p>I have difficulties deriving the non-cooperative Nash Equilibrium of this problem.</p>

<p>The objective function is to maximize the expected total rent over the two periods, that is:</p>

<p>\begin{align}
\max_{S_i} (1-e_i S_i + P_i(S_i, S_j) \delta(1-\underline{S}))
\end{align}</p>

<p>with $P_i(S_i, S_j)=\frac {S_i}{S_i+S_j}$, and $i,j \in \{A,B\}$, and $e_A=1-d$ and $e_B=1+d$</p>

<p>$S_i$ denotes the quality of the service, $e_i$ is a parameter that captured the spending needs in $i$, $P_i(S_i, S_j)$ is the re-election probability, $\delta$ is the discount factor and $\underline{S}$ is the minimum service quality.</p>

<p>The solution of the Nash Equilibrium of the game is
\begin{align}
(S_A^*, S_B^*)=\Big(\frac{\delta (1+d)(1-\underline{S})}{4}, \frac{\delta (1-d)(1-\underline{S})}{4}\Big)
\end{align}</p>

<p>I have tried but I don't get that same solution. </p>

<p>This is what I have done: 
I solve the maximization problem 
\begin{align}
\frac {\partial{U_i}}{\partial{S_i}} = -e_i+\dfrac{\delta(1-\underline{S})S_j}{(S_i+S_j)^2}\stackrel{!}{=}0
\end{align}</p>

<p>and get the best respond function for $i$ and $j$ respectively as follows
\begin{align}
\delta(1-\underline{S})S_j=e_i(S_i+S_j)^2 \\
\delta(1-\underline{S})S_i=e_j(S_i+S_j)^2
\end{align}</p>

<p>I impose symmetry $S_i=S_j=S^*$ to the reaction function and I get this: 
\begin{align}
\delta(1-\underline{S})S^*&amp;=e_i(2S^*)^2 \\
S^*&amp;=\dfrac{\delta(1-\underline{S})}{4e_i}
\end{align}</p>

<p>I am not sure though.</p>

<game-theory><nash-equilibrium>
Non-cooperative Nash Equilibrium in political game
Alfa
# Non-cooperative Nash Equilibrium in political game

(Asked by: username_0 on 2017-07-06)

I have difficulties deriving the non-cooperative Nash Equilibrium of this problem.

The objective function is to maximize the expected total rent over the two periods, that is:

\begin{align}
\max\_{S\_i} (1-e\_i S\_i + P\_i(S\_i, S\_j) \delta(1-\underline{S}))
\end{align}

with $P\_i(S\_i, S\_j)=\frac {S\_i}{S\_i+S\_j}$, and $i,j \in \{A,B\}$, and $e\_A=1-d$ and $e\_B=1+d$

$S\_i$ denotes the quality of the service, $e\_i$ is a parameter that captured the spending needs in $i$, $P\_i(S\_i, S\_j)$ is the re-election probability, $\delta$ is the discount factor and $\underline{S}$ is the minimum service quality.

The solution of the Nash Equilibrium of the game is
\begin{align}
(S\_A^\*, S\_B^\*)=\Big(\frac{\delta (1+d)(1-\underline{S})}{4}, \frac{\delta (1-d)(1-\underline{S})}{4}\Big)
\end{align}

I have tried but I don't get that same solution.

This is what I have done:
I solve the maximization problem
\begin{align}
\frac {\partial{U\_i}}{\partial{S\_i}} = -e\_i+\dfrac{\delta(1-\underline{S})S\_j}{(S\_i+S\_j)^2}\stackrel{!}{=}0
\end{align}

and get the best respond function for $i$ and $j$ respectively as follows
\begin{align}
\delta(1-\underline{S})S\_j=e\_i(S\_i+S\_j)^2 \\
\delta(1-\underline{S})S\_i=e\_j(S\_i+S\_j)^2
\end{align}

I impose symmetry $S\_i=S\_j=S^\*$ to the reaction function and I get this:
\begin{align}
\delta(1-\underline{S})S^\*&=e\_i(2S^\*)^2 \\
S^\*&=\dfrac{\delta(1-\underline{S})}{4e\_i}
\end{align}

I am not sure though.

**Tags:** game-theory, nash-equilibrium

---

## 1 answers

### Answer by username_1 on 2017-07-07. Score: 4 | Accepted answer

You could add up the equations
\begin{align}
\delta(1-\underline{S})S\_j=e\_i(S\_i+S\_j)^2 \\
\delta(1-\underline{S})S\_i=e\_j(S\_i+S\_j)^2
\end{align}
to get
\begin{align}
\delta(1-\underline{S})(S\_i+S\_j)=(e\_i+e\_j)(S\_i+S\_j)^2
\end{align}
which simplifies to
\begin{align}
\delta(1-\underline{S})=(e\_i+e\_j)(S\_i+S\_j).
\end{align}
This gives you $(S\_i+S\_j)$. From the initial equations
\begin{align}
\delta(1-\underline{S})S\_j=e\_i(S\_i+S\_j)^2 \\
\delta(1-\underline{S})S\_i=e\_j(S\_i+S\_j)^2
\end{align}
you can also get
$$
\frac{S\_j}{e\_i} = \frac{(S\_i+S\_j)^2}{\delta(1-\underline{S})} = \frac{S\_i}{e\_j},
$$
which gives you
$$
\frac{S\_j}{S\_i} = \frac{e\_i}{e\_j}.
$$
Now you have both the sum and ratio of the variables $S\_j,S\_i$. Calculating their individual values should be a straightforward matter.

> **Comments:**
> **username_0**: Thanks a lot!! I appreciate your taking time answering my question.  
> **username_2**: @Alfa have you consider accepting the answer?  
> **username_0**: @luchonacho yes, I accept the answer. It gives me the solution.  
> **username_1**: @Alfa To accept an answer on Stackexchange you have to press the green check sign to the left of the answer.  
> **username_2**: @Alfa ahaha funny. Yes, you need to formally accept it :)  

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