# Pareto Efficient level of Public Good

As part of self study towards an entrance exam, I am solving the following question.

There are two consumers in the economy and two goods, one private and one public. The utilities of the consumers is given by:

$$u_1(x_1,y) = x_1+6\sqrt y$$

$$u_2(x_2,y = x_2 + 10\sqrt y$$

where $$x_i$$ is the quantity of the private good and $$y$$ is the quantity of the public good consumed. Initial endowments of the private goods:

$$\omega_1 = 40, \omega_2 = 60$$.

A unit of private good can be converted one-for-one to a public good.

My attempt:

The Pareto Efficient level of public good is determined by the equation:

$$|MRS_1| + |MRS_2| = MC(G) \quad (1)$$

where $$MRS_1 = \frac{MU_G}{MU_{x_1}}$$, $$MRS_2 = \frac{MU_G}{MU_{x_2}}$$ and $$MC(G)$$ is the marginal cost of providing the public good.

Solving $$(1)$$ with the given utility functions I get:

$$\frac{3}{\sqrt y} + \frac{5}{\sqrt y} = 1$$ $$\implies y = 64$$

Therefore, the Pareto efficient level of public good is $$y=64$$.

The question asks to select the Pareto Inefficient bundle $$(x_1,x_2,y)$$ amongst the following:

A. $$(50,0,50)$$

B. $$(90,0,10)$$

C. $$(0,80,20)$$

D. $$(16,20,64)$$

I think all of A, B and C are Pareto Inefficient.

The social welfare maximizing outcome you computed (by equating the sum of marginal benefits to marginal cost) is just one of the Pareto efficient outcomes. Although maximization of joint utilities is a sufficient condition for Pareto efficiency, it is not a necessary condition. Pareto efficiency is defined by the lack of Pareto improvement -- a reallocation of resources to make at least someone better off without making any other worse off.

Option C is not Pareto efficient because a Pareto improvement exists. At $$(0,80,20)$$, $$$$u_1(0,20)=0+6\sqrt{20}\approx 26.83 \qquad u_2(80,20)=80+10\sqrt{20}\approx124.72.$$$$ Observe that at this bundle, consumer 2's MU for the good $$y$$ is $$5/\sqrt{20}\approx1.11$$ and his MU for $$x$$ is $$1$$. Therefore, increasing one unit of good $$y$$ at the expense of one less unit of $$x$$ is utility enhancing for consumer 2. At the same time, one extra $$y$$ benefits consumer 1 as well. So we can easily verify that $$(0,79,21)$$ is a Pareto improvement over $$(0,80,20)$$: $$$$u_1(0,21)=0+6\sqrt{21}\approx27.495 \qquad u_2(79,21)=79+10\sqrt{21}\approx124.83$$$$ where both consumers experience higher utilities.

In a similar way, you should be able to verify that no Pareto improvements exist for options A, B, and D.

Consider the following maximisation problem: $$\begin{eqnarray*} \max_{x_1, x_2, y} & x_1 + 6\sqrt{y} + \alpha(x_2 + 10 \sqrt{y}) \\ \text{s.t.} & x_1 + x_2 +y = 100 \\ \text{and} & x_1\geq 0, x_2 \geq 0,y \geq 0\end{eqnarray*}$$ which can be re-written as: $$\begin{eqnarray*} \max_{x_1, x_2, y} & x_1 + \alpha x_2 + 2\beta \sqrt{y} \\ \text{s.t.} & x_1 + x_2 +y = 100 \\ \text{and} & x_1\geq 0, x_2 \geq 0,y \geq 0\end{eqnarray*}$$ where $$\beta = 3 + 5\alpha$$.

Clearly, for any $$\alpha > 0$$, any solution to the above problem is Pareto efficient. We can now find the set of solutions to the above problem:

$$\begin{eqnarray*} (x_1^*, x_2^*, y^*)(\alpha) \in \begin{cases} \{(100 - \beta^2,0,\beta^2)\} & \text{if } \alpha < 1 \\ \{(x_1, x_2, y)\in\mathbb{R}^3_+|x_1+x_2=36,y=64\} & \text{if } \alpha = 1 \\ \left\{\left(0,100-\frac{\beta^2}{\alpha^2},\frac{\beta^2}{\alpha^2}\right)\right\} & \text{if } \alpha > 1 \end{cases} \end{eqnarray*}$$

Now we can check for Pareto efficiency of the given options:

• Option (a) $$(50,0,50)$$ is Pareto efficient. For that consider $$\beta = \sqrt{50}$$ or when $$\alpha = \dfrac{\sqrt{50} -3}{5}$$.
• Option (b) $$(90,0,10)$$ is Pareto efficient. For that consider $$\beta = \sqrt{10}$$ or when $$\alpha = \dfrac{\sqrt{10} -3}{5}$$.
• Option (d) $$(16,20,64)$$ is Pareto efficient. For that consider $$\alpha = 1$$.
• Option (c) $$(0,80,20)$$ is not Pareto efficient, because $$(0,75,25)$$ is Pareto superior.

Besides the solutions obtained above, there are two more Pareto efficient allocations: $$\lim_{\alpha\rightarrow 0^+}(x_1^*, x_2^*, y^*)(\alpha) = (91, 0, 9)$$ and $$\lim_{\alpha\rightarrow \infty}(x_1^*, x_2^*, y^*)(\alpha) = (0, 75, 25)$$. These are the solutions to the following problems: $$\begin{eqnarray*} \max_{x_1, x_2, y} & x_1 + 6\sqrt{y} \\ \text{s.t.} & x_1 + x_2 +y = 100 \\ \text{and} & x_1\geq 0, x_2 \geq 0,y \geq 0\end{eqnarray*}$$ and $$\begin{eqnarray*} \max_{x_1, x_2, y} & x_2 + 10\sqrt{y} \\ \text{s.t.} & x_1 + x_2 +y = 100 \\ \text{and} & x_1\geq 0, x_2 \geq 0,y \geq 0\end{eqnarray*}$$ respectively.

To see how to find the set of efficient allocations graphically, you can watch this playlist: https://youtube.com/playlist?list=PLUJGfL_499TJIHv2nw4rn40BbQIe8DChz