<blockquote>
  <p>The question is a reformulation of an incomplete version.</p>
</blockquote>

<p>Consider the following dynamic dividing a dollar game where agent 1 claims $x(t)$ of the dollar and agent 2 $y(t)$ (<a href="http://www.tau.ac.il/~fersht/Papers/1989%20Simultaneous%20Moves%20Concession%20Game.pdf" rel="nofollow">paper</a>). </p>

<p>\begin{align}
&amp;\max_{\{u(t)\}^T_{t=0}} e^{-rT}x(T)\\[2mm]
&amp;\max_{\{v(t)\}^T_{t=0}} e^{-rT}y(T)\\[2mm]
\text{s.t. }&amp; \dot x(t) = -u(t)\in[0,1-x(t)]\\
&amp; \dot y(t) = -v(t)\in[0,1-y(t)]\\
&amp; x(0) + y(0) &gt; 1
\end{align}</p>

<p>where $y(T) = 1 - x(T)$ with $T:=\min\{t \in \mathbb{R}_{++} : x(t) + y(t) = 1\}$ and $r &gt; 0$ is the time preference rate. Optimal feedback control is given by $u(t) = r(1-x(t))$ and $v(t) = r(1-y(t))$. This solution is somehow motivated by guess and verify. </p>

<ul>
<li>Can we obtain the solution by some first order conditions (e.g. maximum principle)?</li>
</ul>

<h2>Transformation of objective function</h2>

<p>I tried to somehow transform the objective function into a common functional.
Can someone confirm the following:
\begin{align}
&amp;1 + \int^T_0{-re^{-rt}dt} = 1 + \left[e^{-rt}\right]^T_0 = 1 + e^{-rT} - e^0 = e^{-rT}\\
&amp; x(0) - \int^T_0{u(t)dt} = x(0) + \int^T_0{\dot x(t) dt} = x(T)\\
\Longrightarrow \qquad&amp; e^{-rT} x(0) + \int^T_0{e^{-rt}u(t)(r - e^{rt})dt} = e^{-rT}x(T)
\end{align} </p>

<ul>
<li>What next?</li>
</ul>

CC BY-SA 3.0
2016-05-23T09:00:19.770Z
Ok, I see your points. Actually it's a dividing a dollar game, and $x(T)$ is the final claim of player 1 and $y(T) = 1-x(T)$ the final claim of player 2. He (player 2) maximizes $\max_v y(T)$ s.t. $\dot y = -v$ where $x(0) + y(0) > 1$ given. The markovian solution is given in the paper, however it is more or less guess and verify. I wanted to construct the solution by some first order conditions. Should I restate the entire question and delete this thread?
clueless
CC BY-SA 3.0
2016-05-11T17:51:14.800Z
It seems something is missing: for any given r,T, x(0), you are maximizing x(T), which is done by setting u=0. Right?
Fix.B.
CC BY-SA 3.0
2016-05-11T20:04:37.570Z
Are you looking for a single $u$ or for a sequence of varying $u$'s along the $t$-index?
Alecos Papadopoulos
<dynamic-games><dynamic-optimization>
Dynamic dividing a dollar game
clueless
# Dynamic dividing a dollar game

(Asked by: username_0 on 2016-05-11)

> The question is a reformulation of an incomplete version.

Consider the following dynamic dividing a dollar game where agent 1 claims $x(t)$ of the dollar and agent 2 $y(t)$ ([paper](http://www.tau.ac.il/~fersht/Papers/1989%20Simultaneous%20Moves%20Concession%20Game.pdf)).

\begin{align}
&\max\_{\{u(t)\}^T\_{t=0}} e^{-rT}x(T)\\[2mm]
&\max\_{\{v(t)\}^T\_{t=0}} e^{-rT}y(T)\\[2mm]
\text{s.t. }& \dot x(t) = -u(t)\in[0,1-x(t)]\\
& \dot y(t) = -v(t)\in[0,1-y(t)]\\
& x(0) + y(0) > 1
\end{align}

where $y(T) = 1 - x(T)$ with $T:=\min\{t \in \mathbb{R}\_{++} : x(t) + y(t) = 1\}$ and $r > 0$ is the time preference rate. Optimal feedback control is given by $u(t) = r(1-x(t))$ and $v(t) = r(1-y(t))$. This solution is somehow motivated by guess and verify.

* Can we obtain the solution by some first order conditions (e.g. maximum principle)?

## Transformation of objective function

I tried to somehow transform the objective function into a common functional.
Can someone confirm the following:
\begin{align}
&1 + \int^T\_0{-re^{-rt}dt} = 1 + \left[e^{-rt}\right]^T\_0 = 1 + e^{-rT} - e^0 = e^{-rT}\\
& x(0) - \int^T\_0{u(t)dt} = x(0) + \int^T\_0{\dot x(t) dt} = x(T)\\
\Longrightarrow \qquad& e^{-rT} x(0) + \int^T\_0{e^{-rt}u(t)(r - e^{rt})dt} = e^{-rT}x(T)
\end{align}

* What next?

**Tags:** dynamic-games, dynamic-optimization

> **Comments:**
> **username_1**: It seems something is missing: for any given r,T, x(0), you are maximizing x(T), which is done by setting u=0. Right?  
> **username_2**: Are you looking for a single $u$ or for a sequence of varying $u$'s along the $t$-index?  
> **username_0**: Ok, I see your points. Actually it's a dividing a dollar game, and $x(T)$ is the final claim of player 1 and $y(T) = 1-x(T)$ the final claim of player 2. He (player 2) maximizes $\max_v y(T)$ s.t. $\dot y = -v$ where $x(0) + y(0) > 1$ given. The markovian solution is given in the paper, however it is more or less guess and verify. I wanted to construct the solution by some first order conditions. Should I restate the entire question and delete this thread?  

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## 0 answers

