# Two quasilinear utility functions

Suppose there are two agents $\{A,B\}$ with the following utility functions: $U_A(x_A,y_A)=x_A+f(y_A)$ and $U_B(x_B,y_B)=x_B+g(y_B)$

$f$ and $g$ are strictly increasing and strictly concave. How do I prove that in the set of pareto efficient allocations $y_A=y_B$ ?

• Hi, and welcome to the Economics SE. If you could show some of your work to show what part of the proof you are stuck on, it would allow the community to help you better with your question. Keep in mind that we generally don't answer homework questions unless we see there has been some work involved. – Kitsune Cavalry Nov 3 '15 at 20:51
• Yes, the set of pareto efficient points (Result of maximizing one utility function keeping the other one constant) is: $f'(y_A)=g'(y_B)$ but I do not know how to conclude that both quantities are the same. – Juan Imbett Nov 3 '15 at 20:59
• $f'(y_A)=g'(y_B)$ is a condition that you use to show $y_A = y_B$. I don't think you're trying to conclude $f', g'$ are equal. – Kitsune Cavalry Nov 3 '15 at 21:41
• Yes, that is what I want to show. From the first order conditions I have $f'(y_A)=g'(y_B)$, and I need to conclude that $y_A=y_B$, I think I have to use the fact that they are strictly concave and strictly increasing – Juan Imbett Nov 3 '15 at 21:48

The statement is not true.
Let $x_A + x_B = 1$, $y_A + y_B = 1$.
Let $U_A(x_A,y_A) = x_A + \ln(y_A)$, $U_B(x_B,y_B) = x_B + \ln(y_B + 1)$.
$f \neq g$ are both strictly increasing & concave. For all $z \in [0,1]$ the distributions $$(x_A,y_A) = (z,1), \ (x_B,y_B) = (1-z,0).$$ are Pareto-efficient. This follows from $MRS_A(x_A,y_A) \geq MRS_B(x_B,y_B)$.
In these cases $y_A \neq y_B$.

• Well that explains why I had so much trouble trying to set up a proof. Neat-o. – Kitsune Cavalry Nov 4 '15 at 1:03

Here is an example of the economy in which consumers have identical - increasing, convex and quasi linear utility functions, but consists of PE allocations that satisfy $y_A\neq y_B$:

$u_A(x_A, y_A) = x_A + \log y_A$

$u_B(x_B, y_B) = x_B + \log y_B$

Feasibility requirement is $x_A + x_B = 1$ and $y_A + y_B = 1$.

Allocation in which A consumes everything and B consumes nothing: $(x_A, y_A) = (1,1)$ and $(x_B, y_B) = (0,0)$ is Pareto efficient, but $y_A \neq y_B$.

• Though an unusual form, $u_B(x_B, y_B) = e^{x_B} \cdot y_B$ may be a better here, because for your function $u_B(0,0)$ is undefined. – Giskard Jan 26 '17 at 9:18
• Agreed. In fact in this case any $u_i = x_i + f(y_i)$ will work as long as $f$ is increasing. – Amit Jan 26 '17 at 10:03

EDIT: See denesp's answer.

Here is some work I have so far. Someone can hopefully suggest help or provide a separate full answer.

So we have

$$u_A(x_A, y_A) = x_A + f(y_A)$$ $$u_B(x_B, y_B) = x_B + f(y_B)$$

and we'll say that $x_A + x_B = X$ and $y_A + y_B = Y$

A's MRS:

$$-\frac{dx_A}{dy_A} = \frac{\frac{\partial U_A}{\partial y_A}}{\frac{\partial U_A}{\partial x_A}} = \frac{f'(y_A)}{1} = f'(y_A)$$

B's MRS: similarly, it is by $g'(y_B)$

So we'll get a Pareto efficient allocation when $f'(y_A) = g'(y_B)$

We know a function being strictly concave implies:

$$f((1-\alpha)y_A + \alpha y_B) > (1-\alpha)f(y_A) + \alpha y_B$$

We know a function being strictly increasing implies:

$$y_A > y_B \implies f(y_A) > f(y_B)$$

So take the original MRS condition and substitute in $Y - y_A = y_B$

$$f'(Y - y_B) = g'(y_B)$$

I think you can say that $f$ increasing and concave implies f' strictly decreasing, so I think you can say that

$$f'(Y - y_B) < f'(Y) - f'(y_B)$$

Now suppose that $y_A \neq y_B$ and create a contradiction. Note that $Y > y_B, y_A$ so when taking functions of that, you can take advantage of that somehow.