# Walrasian equilibrium with quasi linear function

There is a two-person exchange economy

Each agent has the following utility $$u_i(x_i,y_i)=v(x_i)+y_i$$ for agent $$i=\{A,B\}$$

Assume that $$v$$ is strictly concave and increasing function that has a continuous first derivative. $$v(0)=0$$ and $$v(x)<1$$.

Agent A has the endowment $$(1,10)$$. And agent B has the endowment $$(0,10)$$.

For each Pareto efficient allocation, suggest how we might change the endowments so that the Pareto efficient allocation in the question is a walrasian equilibrium.

I found the Pareto optimal allocation set as

$$v’(x_A)=v’(1-x_B)$$ $$y_A+y_B=20$$

I also found the Walrasian equilibrium set as $$\{(x^*_A, y^*_A)=(1/2, 10+\frac{P_x}{2P_y}), (x^*_B, y^*_B)=(1/2, 10-\frac{P_x}{2P_y})\}$$ with the Walrasian equilibrium price ratio

$$\frac{P_x}{P_y}= min\{v’(x_A),v’(x_B)\}$$

If $$y^*_A>0$$ Then $$\frac{P_x}{P_y}= v’(x_A)$$

If $$y^*_A=0$$ Then $$\frac{P_x}{P_y}> v’(x_A)$$ so, $$x^*_A> x^*_A$$

I could only found Walrasian and Pareto optimal allocations. But I am not sure. And I don’t understand the questions. How can I show this question. All helps will be appreciated. Thanks a lot.

*duplicated question

First thing to notice, when is it that $$v'(x)=v'(1-x)$$?, this is only possible if $$x=1-x$$ or $$x=1/2$$. This follows from the strict concavity of $$v$$, $$v'$$ is decreasing so $$v'(x)>v'(y)$$ whenever $$x. Therefore, the Pareto set is

$$P=\{(1/2,y,1/2,20-y):0\leq y\leq 20\}\cup \{(1,20,0,0)\}\cup \{(0,0,1,20)\}$$

Note that the allocations $$(1,20,0,0)$$ and $$(0,0,1,20)$$ are also efficient.

Now, for you question.

For each Pareto efficient allocation, suggest how we might change the endowments so that the Pareto efficient allocation in the question is a walrasian equilibrium.

Let $$E$$ be the set of possible endowments

$$E=\{(x_A,y_A,1-x_A,20-y_A): 0\leq x_A\leq 1, 0\leq y_A\leq 20\}$$

For each $$e\in E$$ let $$W(e)$$ be the Walrasian equilibrium when the agents start with endowment $$e$$. Your question can be formally stated as follows

For all $$p\in P$$ find a $$e_p\in E$$ such that $$W(e_p)=p$$

Here is a quick thought, what if the agents start at $$p$$? That is, if $$e_p=p$$ then what is $$W(e_p)$$? It should be $$p$$.