<p>You don't want a worker <span class="math-container">$x$</span> and a firm <span class="math-container">$y$</span> to be able to give up on who they were previously matched with, match, and be better off. Currently, the worker receives <span class="math-container">$w(x)$</span> and the firm <span class="math-container">$\pi(y)$</span>. If they would be better off together, they need to generate more surplus than <span class="math-container">$w(x)+\pi(y)$</span>. This requires <span class="math-container">$f(x,y)&gt;w(x)+\pi(y)$</span>. That <span class="math-container">$f(x,y)\leq w(x)+\pi(y)$</span> for all <span class="math-container">$x$</span> and <span class="math-container">$y$</span> means this is not possible, and the matching must be stable.</p>

CC BY-SA 4.0
2023-10-11T17:42:10.487Z
I understand your point which implies that the match would be stable. However, why does the paper claim that the outcome  $(\mu,\pi,w)$ is stable? Would not a stable outcome require that the firm actually pay a competitive wage? For example, if worker $i$ is matched with a firm and makes a huge surplus, but worker $i$ is paid $w_i = 0$ the match may be stable but the outcome may not be. Yet the paper I quoted claims that the outcome is stable. I have seen one other paper make the same claim so I am not sure whether it is a mistake or not.
AYC
CC BY-SA 4.0
2023-10-11T18:00:25.197Z
Whether a worker being paid $0$ is compatible with a stable outcome will depend on the details. Maybe you could add all relevant definitions and the actual reference to the paper.
Michael Greinecker
CC BY-SA 4.0
2023-10-10T22:48:02.137Z
2023-10-10T22:48:02.137Z
Michael Greinecker
<p>I am reading a paper that discusses mathcing of workers to firms in the labor market (see the excerpt below). Here an outcome <span class="math-container">$(\mu,\pi,w)$</span> consists of a matching function pairing workers to firms, a profit vector for firms, and a wage vector for workers. The text states that the outcome is stable if there is no blocking pair and that this happens if <span class="math-container">$w(x) + \pi(y) &gt;= f(x,y) \forall (x,y)$</span>. My question is: how does this condition ensure a stable outcome?</p>
<p>I understand that this condition requires that the sum of the wage and profit in the existing pair be larger than any pair. And I understand that this means that a firm can always increase wages  to keep their preferred worker from deviating to another firm. But that implies that the matching function mu is stable. I don't understand what in the above condition states that the wage and profit vectors in the stable outcome are actually selected to be stable?</p>
<p>Thank you!</p>
<p><a href="https://i.sstatic.net/Ma5vF.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Ma5vF.png" alt="Excerpt in Question" /></a></p>

<labor-economics><matching>
Stable Outcomes and Matching Models for the Labor Market
AYC
# Stable Outcomes and Matching Models for the Labor Market

(Asked by: username_0 on 2023-10-10)

I am reading a paper that discusses mathcing of workers to firms in the labor market (see the excerpt below). Here an outcome $(\mu,\pi,w)$ consists of a matching function pairing workers to firms, a profit vector for firms, and a wage vector for workers. The text states that the outcome is stable if there is no blocking pair and that this happens if $w(x) + \pi(y) >= f(x,y) \forall (x,y)$. My question is: how does this condition ensure a stable outcome?

I understand that this condition requires that the sum of the wage and profit in the existing pair be larger than any pair. And I understand that this means that a firm can always increase wages to keep their preferred worker from deviating to another firm. But that implies that the matching function mu is stable. I don't understand what in the above condition states that the wage and profit vectors in the stable outcome are actually selected to be stable?

Thank you!

[![Excerpt in Question](https://i.sstatic.net/Ma5vF.png)](https://i.sstatic.net/Ma5vF.png)

**Tags:** labor-economics, matching


---

## 1 answers

### Answer by username_1 on 2023-10-10. Score: 0

You don't want a worker $x$ and a firm $y$ to be able to give up on who they were previously matched with, match, and be better off. Currently, the worker receives $w(x)$ and the firm $\pi(y)$. If they would be better off together, they need to generate more surplus than $w(x)+\pi(y)$. This requires $f(x,y)>w(x)+\pi(y)$. That $f(x,y)\leq w(x)+\pi(y)$ for all $x$ and $y$ means this is not possible, and the matching must be stable.

> **Comments:**
> **username_0**: I understand your point which implies that the match would be stable. However, why does the paper claim that the outcome  $(\mu,\pi,w)$ is stable? Would not a stable outcome require that the firm actually pay a competitive wage? For example, if worker $i$ is matched with a firm and makes a huge surplus, but worker $i$ is paid $w_i = 0$ the match may be stable but the outcome may not be. Yet the paper I quoted claims that the outcome is stable. I have seen one other paper make the same claim so I am not sure whether it is a mistake or not.  
> **username_1**: Whether a worker being paid $0$ is compatible with a stable outcome will depend on the details. Maybe you could add all relevant definitions and the actual reference to the paper.  

---

