How can I build a fixed point argument for Nash Equilibrium in $\mathbb{L}^2$ Banach space?

Consider the sets $$Y_1,Y_2\subset\mathbb{L}^2(\Omega,\mathbb{F}_{s},\mathbb{P})$$. I am looking for some $$y_1$$ measurable with respect to $$Y_1$$ and $$y_2$$ measurable with respect to $$Y_2$$ s.t. the pair $$(y^*_1,y^*_2)$$ constitutes a Nash Equilibrium. In particular, the pair $$(y^*_1,y^*_2)\in \mathcal{Y}= \Pi_{i=1}^2Y_i$$ is the cartesian product of the strategic choices of evry agent.

I present a setup similar to Vayanos and Wang of $$2011$$ model, where the set of strategic actions of every agent is in the intercept of their demand functions. Consider that two agents, indexed by $$i=\{1,2\}$$, have the following utility functions $$g_1(y_1)=-d_1(y_1)p(y_1,y_2)+\mathbb{E}[d_1(y_1)\bar{x}+\bar{z}_1|Y_1]-\gamma_1\mathbb{V}ar[d_1(y_1)\bar{x}+\bar{z}_1|Y_1]$$ and $$g_2(y_2)=-d_2(y_2)p(y_1,y_2)+\mathbb{E}[d_2(y_2)\bar{x}+\bar{z}_2|Y_2]-\gamma_2\mathbb{V}ar[d_2(y_2)\bar{x}+\bar{z}_2|Y_2]$$

where $$(\bar{x},\bar{z}_i)$$ are i.d and $$Y_i$$ measurable and also it holds that $$\bar{x}\sim N(x,\sigma_{x}^2)$$, $$\bar{z}_i\sim N(0,\sigma_{z_i}^2)$$, $$\gamma_i\in (0,+\infty)$$, $$\mathbb{C}ov(\bar{x},\bar{z_i})\neq 0$$, $$\mathbb{C}ov(\bar{z_1},\bar{z_2})\neq 0$$ and $$g_i:Y_i\subset\mathbb{L}^2\rightarrow \mathbb{R}$$ where $$i\in\{1,2\}$$. For the demand functions hold that $$d_i(y_i)=\frac{y_i-p(y_1,y_2)}{2\gamma_i\mathbb{V}ar[\bar{x}|Y_i]}$$ where $$y_i$$ is measurable with respect to $$Y_i$$ and $$p(y_1,y_2)=w_1y_1+w_2y_2+w_3y$$ s.t $$w_1+w_2+w_3=1$$

Every agent needs to solve the following problem $$y^*_i=\max_{y_i}g_i(y_i;y_{-i})$$ where $$y_{-i}$$ is the strategic choice of agent $$-i$$ (in our case the agents are two so equilevantely $$y^*_1=\max_{y_1}g_i(y_1;y_{2})$$) So, I will obtain two solutions for $$y_1$$ and $$y_2$$ and by solving the system of them, I result to the pair of $$(y_1^*,y_2^*)$$

$$\textbf{The solutions of the agents' problems}$$

By solving the maximization problem of agent-$$1$$, we obtain: $$$$y_1(y_2)=\frac{1}{1+\omega_1}P_1+\frac{\omega_1\omega_2}{(1-\omega_1)(1+\omega_1)}y_2+\frac{\omega_1\omega_3}{(1-\omega_1)(1+\omega_1)}y$$$$ where $$P_1=\mathbb{E}[\bar{x}|Y_1]-2\gamma_1\mathbb{C}ov(\bar{x},\bar{z}_1|Y_1)$$

Similarly, by solving the maximization problem of agent-$$2$$ $$$$y_2(y_1)=\frac{1}{1+\omega_2}P_2+\frac{\omega_1\omega_2}{(1-\omega_2)(1+\omega_2)}y_1+\frac{\omega_2\omega_3}{(1-\omega_2)(1+\omega_2)}y$$$$ where $$P_2=\mathbb{E}[\bar{x}|Y_2]-2\gamma_2\mathbb{C}ov(\bar{x},\bar{z}_2|Y_2)$$

By solving the system of $$(y_1,y_2)$$ we obtain that $$y_1^*=\frac{(1-\omega_1)(1-\omega_2^2)}{1-\omega_1^2-\omega_2^2}P_1+\frac{\omega_1(1-\omega_1)\omega_2}{1-\omega_1^2-\omega_2^2}P_2+\frac{\omega_1\omega_3}{1-\omega_1^2-\omega_2^2}y$$

and

$$y_2^*=\frac{\omega_1(1-\omega_2)\omega_2}{1-\omega_1^2-\omega_2^2}P_1+\frac{(1-\omega_1^2)(1-\omega_2)}{1-\omega_1^2-\omega_2^2}P_2+\frac{\omega_2\omega_3}{1-\omega_1^2-\omega_2^2}y$$

$$\textbf{My question}$$

My question is, how can I build the Banach fixed point argument for the aforementioned problem where the set of strategic choices is in the intercept of the demand functions?

• Probably Banach but how about first letting us see the problem – Meowdog Jul 29 '20 at 8:36
• Well, tis is a thought to write it down, but it will need some time. I will try to simplify it, since I can not write the general form...it is not so fancy... – Nav89 Jul 29 '20 at 8:47
• Untill, I find some way to simplify the problem and write it, if anybody has some information, or has worked in something similar, I would appreciate to know it! thank you in advance! – Nav89 Jul 29 '20 at 8:59
• No I don't think so, I might have a look at it later – Meowdog Jul 30 '20 at 22:30
• So, the solution of your best responses $(y_1^*,y_2^*)$ is if fact a weighthed average of $P_1,P2$ and $y$. This is interested! But I can not tell, how can you apply your fixed point theorem argument, whtever it is... – Hunger Learn Aug 3 '20 at 6:44

Well, jsut to help your post and may someone else come and make clear what you need to check. Every agent needs to solve the following problem $$y^*_i=\max_{y_i}g_i(y_i)$$ So you will obtain a two solutions for $$y_1$$ and $$y_2$$ and by solving the system of them, you take the pair of $$(y_1^*,y_2^*)$$. The question is why sould this be a Nash Equilibrium and is this the unique equilibrium pair of strategies? The answer is deinitely yes for the second part of my-your question, and it is implied by the lienarity of the derivative of $$g_i(y_i)$$ with respect to $$y_i$$. So we talk about a unique solution. Why this is a N.E. or how the Banach fixed point theorem is going to be applied here, it is something else that I can not tell, because I do not know. I hope someone else could have a full answer, but in first place, this is a part of your question to be answered.