<p>The statement is not true. <br>
Let $x_A + x_B = 1$, $y_A + y_B = 1$. <br>
Let $U_A(x_A,y_A) = x_A + \ln(y_A)$, $U_B(x_B,y_B) = x_B + \ln(y_B + 1)$. <br>
$f \neq g$ are both strictly increasing &amp; 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 $MRS_A(x_A,y_A) \geq MRS_B(x_B,y_B)$. <br>
 In these cases $y_A \neq y_B$.</p>

CC BY-SA 3.0
2015-11-04T01:03:08.587Z
Well that explains why I had so much trouble trying to set up a proof. Neat-o.
Kitsune Cavalry
CC BY-SA 3.0
2015-11-03T22:44:02.090Z
2015-11-03T23:11:44.640Z
2015-11-03T23:11:44.640Z
Giskard
<p>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 <span class="math-container">$y_A\neq y_B$</span>:</p>
<p><span class="math-container">$u_A(x_A, y_A) = x_A + \log y_A$</span></p>
<p><span class="math-container">$u_B(x_B, y_B) = x_B + \log y_B$</span></p>
<p>Feasibility requirement is <span class="math-container">$x_A + x_B = 1$</span> and <span class="math-container">$y_A + y_B = 1$</span>.</p>
<p>Allocation in which A consumes everything and B consumes nothing: <span class="math-container">$(x_A, y_A) = (1,1)$</span> and <span class="math-container">$(x_B, y_B) = (0,0)$</span> is Pareto efficient, but <span class="math-container">$y_A \neq y_B$</span>.</p>

CC BY-SA 3.0
2017-01-26T09:18:32.040Z
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
CC BY-SA 3.0
2017-01-26T10:03:26.290Z
Agreed. In fact in this case any $u_i = x_i + f(y_i)$ will work as long as $f$ is increasing.
Amit
CC BY-SA 4.0
2017-01-26T08:40:15.593Z
2023-04-12T11:46:37.283Z
2023-04-12T11:46:37.283Z
Amit
<p><strong>EDIT:</strong> See denesp's answer.</p>

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

<p>So we have</p>

<p>$$u_A(x_A, y_A) = x_A + f(y_A)$$
$$u_B(x_B, y_B) = x_B + f(y_B)$$</p>

<p>and we'll say that $x_A + x_B = X$ and $y_A + y_B = Y$</p>

<p><strong>A's MRS:</strong></p>

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

<p><strong>B's MRS:</strong> similarly, it is by $g'(y_B)$</p>

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

<p>We know a function being strictly concave implies:</p>

<p>$$f((1-\alpha)y_A + \alpha y_B) &gt; (1-\alpha)f(y_A) + \alpha y_B$$</p>

<p>We know a function being strictly increasing implies:</p>

<p>$$y_A &gt; y_B \implies f(y_A) &gt; f(y_B)$$</p>

<p>So take the original MRS condition and substitute in $Y - y_A = y_B$</p>

<p>$$f'(Y - y_B) = g'(y_B)$$</p>

<p>I <em>think</em> you can say that $f$ increasing and concave implies f' strictly decreasing, so I <em>think</em> you can say that</p>

<p>$$f'(Y - y_B) &lt; f'(Y) - f'(y_B)$$</p>

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

CC BY-SA 3.0
2015-11-03T22:25:11.213Z
2015-11-04T01:03:36.287Z
2015-11-04T01:03:36.287Z
Kitsune Cavalry
<p>Suppose there are two agents $\{A,B\}$ with the following utility functions: $U_A(x_A,y_A)=x_A+f(y_A)$ and $U_B(x_B,y_B)=x_B+g(y_B)$</p>

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

CC BY-SA 3.0
2015-11-03T20:51:33.190Z
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
CC BY-SA 3.0
2015-11-03T20:59:00.537Z
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
CC BY-SA 3.0
2015-11-03T21:41:41.630Z
$f'(y_A)=g'(y_B)$ 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
CC BY-SA 3.0
2015-11-03T21:48:38.623Z
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
<microeconomics><pareto-efficiency>
Two quasilinear utility functions
Juan Imbett
# Two quasilinear utility functions

(Asked by: username_0 on 2015-11-03)

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 $y\_A=y\_B$ ?

**Tags:** microeconomics, pareto-efficiency
> **Comments:**
> **username_1**: 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.  
> **username_0**: 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.  
> **username_1**: $f'(y_A)=g'(y_B)$ 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.  
> **username_0**: 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  

---

## 3 answers

### Answer by username_2 on 2015-11-03. Score: 5

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 $y\_A \neq y\_B$.

> **Comments:**
> **username_1**: Well that explains why I had so much trouble trying to set up a proof. Neat-o.  

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### Answer by username_3 on 2017-01-26. Score: 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 $y\_A \neq y\_B$.

> **Comments:**
> **username_2**: 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.  
> **username_3**: Agreed. In fact in this case any $u_i = x_i + f(y_i)$ will work as long as $f$ is increasing.  

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### Answer by username_1 on 2015-11-03. Score: 0

**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.


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