6
Let's first determine the sets of actions of the players.
An action of player 1 is simply a bid $x_1 \in \mathbb{R}_+$.
An action of player 2 is a function: $f_2: \mathbb{R}_+ \to \mathbb{R}_+$ that determines for every action $x_1$ of Player $1$ an action $x_2 = f_2(x_1) \in \mathbb{R}_+$. Let us denote by $F_2$ the set of all actions of player 2.
Let's now ...
6
You are only required to get Nash equilibria and not sequentially rational/subgame perfect equilibria. Hence Player 2's actions at information sets that do not occur (that do not reflect Player 1's actual strategy) do not need to be best responses. All you have to make sure is that no one is better off by deviating.
In case 4., regardless of what Player 2 ...
4
If the capacities $q_1$ and $q_2$ are the same, say $\overline{q}$ you can rule out some cases.
The best response functions are given by:
$$
\begin{align*}
\hat q_1 &= \min\left\{\frac{M - (1-\theta) \hat q_2^L - \theta \hat q_2^H - m}{2}, \overline{q}\right\},\\
\hat q_2^L &= \min\left\{\frac{M - \hat q_1 - m}{2}, \overline{q}\right\},\\
\hat q_2^H &...
3
I think you are a bit confused. The game has the following structure.
nature draws $c_1, c_2 \sim U[0,1]$, which is only revealed to the sellers and not to the buyer.
The buyer proposes a price $p \in [0,1]$
Sellers decide to sell the good at price $p$. The good is sold if at least one seller agrees with the price.
You need to solve this using backwards ...
3
A payoff pair $(x,y)$ is Pareto efficient if it is not Pareto dominated. This means that there does not exist another payoff pair $(x', y')$ such that $x < x'$ and $y < y'$. (Depending on the exact definition, sometimes of of the two inequalities can be weak).
All payoffs in the payoff matrix except for $(C,\beta), (B, \beta)$ and $(C, \delta)$ are ...
3
$BA'$ doesn't work because $A'$ is not a best response to $X$, which is a best response to $B$.
3
There are three classes of equilibria of this game.
The first class is sequential:
\begin{equation}
(s_1,s_2)=(y,r)
\end{equation}
and the beliefs are
\begin{equation}
\mu_1(a)=\mu_1(b)=\mu_2(a\mid y)=\mu_2(b\mid y)=\frac12.
\end{equation}
The second class is not sequential, but weak perfect Bayesian:
\begin{equation}
(s_1,s_2)=(x,l)
\end{equation}
and the ...
3
Let's solve by backward induction.
In stage 3, player $A_3$ solves:
$$
\max_{p_1, p_2} u_3(p_1, p_2, q_1, q_2, x).
$$
Given the assumptions, $A_3$ will set: $p_1 = \overline{p}$ and in general $p_2$ will be a function of the $q_1, q_2$ and $x$, denoted by $p_2(q_1, q_2, x)$. For a solution to exist, you have to assume that $u_3$ is not unbounded in $p_2$.
...
3
Let me first try to answer Statement 2:
The agent's utility profiles are continuous in x
Let's try to flesh out the problem. Let $S_1$ and $S_2$ be compact, convex, strategy sets of the players (say subsets of $\mathbb{R}^n$) (for simplicity) . Let $u_1(s_1, s_2, x)$ and $u_2(s_1, s_2,x)$ be the continuous utility functions of the two players.
Let's solve ...
2
This game need not have any equilibrium as defined here with the lexicographic structure. Consider a situation where the payoff depends only on the action $x$ of player $1$ with $u_1(x)=\sin x$ and $u_2(x)=x$. There are infinitely many maxima of $u_1$, but no largest maximizer. So one cannot select a best choice for player $2$ among those that are optimal ...
2
Condition 3 can be written in the following way: $a^\ast \in B$ iff for all $i$ and all $j$
$$
a^\ast_{i,j} = 1 \iff u_i(a_i^\ast + \{a_{i,j} = 1\}, a^\ast_{-i}) \ge u_i(a_i^\ast + \{a_{i,j} = 0\}, a^\ast_{-i})
$$
As there are no ties, this can be written as:
$$
a^\ast_{i,j} = 1 \to u_i(a_i^\ast + \{a_{i,j} = 1\}, a^\ast_{-i}) \ge u_i(a_i^\ast + \{a_{i,j} = ...
1
One "natural" approach would be to base the price on the expected discounted revenue stream. Writing $R_t$ as the (random) revenue in period $t$ of one share, this then would be:
$$
p_t = \mathbb{E}\left(\sum_{t = 0}^T \frac{R_t}{(1+r)^t}\right) = \sum_{t = 0}^T \frac{\mathbb{E}(R_t)}{(1+r)^t}.
$$
In addition, you might want to lower this price a ...
1
Given that the system does converge, i.e. $a_i^t=a_i^{t+1}=a_i^*$ for all $i$ after some $T<\infty$, and that $a_i^{t+1}\in BR_i(a_{-i}^{t})$, it follows that $a_i^*\in BR_i(a_{-i}^*)$ for all $i$. Hence $a^*$ is a Nash equilibrium of the underlying game.
1
With the help of a former fellow student I was able to find out some details.
The anecdote was about the following three companies:
Macy's, Inc.
Robert Campeau
Federated (covered in the article about Macy's)
Some news coverage exists:
https://mindyourdecisions.com/blog/2011/09/20/when-game-theory-backfires-a-case-study-of-robert-campeaus-takeover-bid/
...
1
This is probably an old question. However there is a book on PDE's and game theory.
Game Theory and Partial Differential Equations, Pablo Blanc and Julio Daniel Rossi.
If you found other books on this matter, I would like to know as well.
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