Effect of exogenous parameter on three player SPNE profits

Question

Consider three agents $A_i$.

• $A_1$ moves first and selects $0\leq q_1\in \mathbb{R}$. $A_2$ moves second and selects $0\leq q_2\in \mathbb{R}$. $A_3$ moves third and selects $0\leq p_1\in \mathbb{R}$ and $0\leq p_2\in \mathbb{R}$. $0\leq x\in \mathbb{R}$ is a parameter.
• $A_1$ tries to maximize $u_1(q_1,p_1, q_2,p_2, x)$, with $\frac{\partial u_1(q_1,p_1, q_2,p_2, x)}{\partial q_1}\geq 0$ and $\frac{\partial u_1(q_1,p_1, q_2,p_2, x)}{\partial p_1}\geq 0$, and independent of $q_2,p_2,x$. $u_1(q_1,p_1, q_2,p_2, x)$ is continuous in all its inputs.
• $A_2$ tries to maximize $u_2(q_1,p_1, q_2,p_2, x)$, with $\frac{\partial u_2(q_1,p_1, q_2,p_2, x)}{\partial q_2}\geq 0$ and $\frac{\partial u_2(q_1,p_1, q_2,p_2, x)}{\partial p_2}\geq 0$, and independent of $q_1,p_1,x$. $u_2(q_1,p_1, q_2,p_2, x)$ is continuous in all its inputs.
• $A_3$ tries to maximize $u_3(q_1,p_1, q_2,p_2, x)$, with $u_3(q_1,p_1, q_2,p_2, x)$ increasing for all $p_1<\bar{p}$ and decreasing for all $p_1\geq \bar{p}$; $u_3(q_1,p_1, q_2,p_2, x)$ decreases in $x$ and $q_1$; and $u_3(q_1,p_1, q_2,p_2\rightarrow \infty, x)=-\infty$; $u_3(q_1,p_1, q_2\rightarrow \infty,p_2, x)=-\infty$. $u_3(q_1,p_1, q_2,p_2, x)$ is continuous in all its inputs, but we do not know more about its properties for $p_2,q_2$.
• We assume that in case that an agent is indifferent between two or more strategies that there exists a decision rule that selects exactly one of these strategies. These rules are known by all agents.

The existence of a unique SPNE is quite clear. However, there could be multiple other equilibria that would lead to the same profit profile (specifically equilibria in which each agent gains the same profits.)

Question: Does $A_3$'s profit decrease in $x$? If yes, why? If not, is there an easy assumption that we need for $u_3(q_1,p_1, q_2,p_2, x)$ behavior with regard to $q_2,p_2$? (I think I miss the forest for the trees) Does the sum of all agents' profit decrease in $x$?

My immediate intuition is yes, because otherwise $A_3$ would be able to pass on all its "losses" to the other agents.

Answer 1

Let's solve by backward induction.

1. 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 $p_2$.

2. As $u_2$ is independent of $p_1, q_1, x$, In stage 2, Player $A_2$ solves: $$ \max_{q_2} u_2(q_2, p_2(q_1, q_2, x)). $$ $u_2$ is increasing in $q_2$ and $p_2$. If $p_2(q_1, q_2, x)$ is increasing in $q_2$ then $A_2$ will set $q_2$ as big as possible, so there is no optimal solution. Else, there might be some optimal level of $q_2$, say $q_2(q_1, x)$.

3. In stage 1, player $A_1$ solves: $$ \max_{q_1} u_1(q_1, \overline{p}). $$ As $u_1$ is increasing in $q_1$ there will probably be no solution to this. So likely this problem is unbounded. If we assume that $q_1 \le \overline{q}_1$ for some fixed value $\overline{q}_1$, then $A_1$ will set $q_1 = \overline{q}_1$.

Going back to the utility of player 3, we therefore see, after substituting: $$ u_3 = u_3(\overline{p}, p_2(\overline{q}_1, q_2(\overline{q}_1, x),x), \overline{q}_1, q_2(\overline{q_1}, x), x). $$ It seems like the change in $u_3$ due to a change in $x$ is ambiguous. As it also influences the optimal choice of $q_2$ (of player $A_2$), and $p_2$ of player $A_3$.