Two quasilinear utility functions

Question

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 $y_A=y_B$ ?

Answer 1

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 $y_A \neq y_B$.

Answer 2

Here is an example of the economy in which consumers have identical - increasing, concave 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 $y_A \neq y_B$.

Answer 3

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.