<blockquote>
  <p>Let $L$ now rise. What happens to $u_n$ and $Y_n$?</p>
</blockquote>

<p>From the equation, if $L$ rises, everything else equals, then $u_n$ will rise (that is, the fraction will be smaller).</p>

<p>$Y_n$ will also rise and that is a fairly reasonable conclusion - Extra work force will lead to more outputs. 
Now, the misleading conclusion is in regard to the markup. Consider now the labour market. When demand (work force) increases and all else remains equal (supply side - firms offering job opportunities), wages will lower and thus that setup will have a new equilibrium - with lower wages and more work force.</p>

<p>This is my interpretation of your question, although the answer may not be 100% accurate, or an answer that actually answers your question.</p>

CC BY-SA 3.0
2016-09-05T00:56:47.040Z
You're either assuming labor inelasticity (s.t. n=1 is always optimal in eq.)  or that the optimal labor choice is dependent upon the size of the labor force. Neither of those two assumptions is necessarily true here. Thus, $Y_n$ doesn't necessarily increase.
123
CC BY-SA 3.0
2016-01-08T19:19:59.203Z
2016-01-08T19:19:59.203Z
gsilvapt
<p>To me the structural rate of unemployment was defined by $$\frac{1}{1+m} = F(u_n,z)$$ where $m$ is the markup and $z$ a catch-all variable. We could equivalenty get this in terms of $Y_n$ by letting $u_n = 1 - \frac{Y_n}{L}$, $L$ being the labor force.</p>

<p>Let $L$ now rise. What happens to $u_n$ and $Y_n$? From the equation, if $L$ rises but everything else must remain constant, then $Y_n$ must rise in order to still satisfy the equation, by the same amount. But then $u_n = 1 - Y_n/L$ is unchanged? Does this make sense? Why is the natural rate of unemployment unchanged when we just increased the amount of people willing to supply labor?</p>

<p>EDIT: The markup is the one that firms use to set prices, with respect to wages, $\text{prices} = \text{wages} \cdot (1+m)$, F is a function which is negative in $u_n$ (more $u_n$ means smaller $F$-value given constant $z$) and positive in $z$, and $Y_n$ is the structural rate of output.</p>

CC BY-SA 3.0
2016-09-05T00:52:46.780Z
At a minimum, you need to clearly define variables here and define how you're indexing those variables before anyone can provide a sensible answer to your question. The mathematical interpretation of your question is rather simple: fixing all other terms and allowing L to increase will cause $U_n$ to increase. In the limit of L, $U_n$ converges to 1.
123
<labor-economics>
What changes the structural rate of output/unemployment?
Andaor
# What changes the structural rate of output/unemployment?

(Asked by: username_0 on 2016-01-06)

To me the structural rate of unemployment was defined by $$\frac{1}{1+m} = F(u\_n,z)$$ where $m$ is the markup and $z$ a catch-all variable. We could equivalenty get this in terms of $Y\_n$ by letting $u\_n = 1 - \frac{Y\_n}{L}$, $L$ being the labor force.

Let $L$ now rise. What happens to $u\_n$ and $Y\_n$? From the equation, if $L$ rises but everything else must remain constant, then $Y\_n$ must rise in order to still satisfy the equation, by the same amount. But then $u\_n = 1 - Y\_n/L$ is unchanged? Does this make sense? Why is the natural rate of unemployment unchanged when we just increased the amount of people willing to supply labor?

EDIT: The markup is the one that firms use to set prices, with respect to wages, $\text{prices} = \text{wages} \cdot (1+m)$, F is a function which is negative in $u\_n$ (more $u\_n$ means smaller $F$-value given constant $z$) and positive in $z$, and $Y\_n$ is the structural rate of output.

**Tags:** labor-economics
> **Comments:**
> **username_1**: At a minimum, you need to clearly define variables here and define how you're indexing those variables before anyone can provide a sensible answer to your question. The mathematical interpretation of your question is rather simple: fixing all other terms and allowing L to increase will cause $U_n$ to increase. In the limit of L, $U_n$ converges to 1.  

---

## 1 answers

### Answer by username_2 on 2016-01-08. Score: 1

> Let $L$ now rise. What happens to $u\_n$ and $Y\_n$?

From the equation, if $L$ rises, everything else equals, then $u\_n$ will rise (that is, the fraction will be smaller).

$Y\_n$ will also rise and that is a fairly reasonable conclusion - Extra work force will lead to more outputs.
Now, the misleading conclusion is in regard to the markup. Consider now the labour market. When demand (work force) increases and all else remains equal (supply side - firms offering job opportunities), wages will lower and thus that setup will have a new equilibrium - with lower wages and more work force.

This is my interpretation of your question, although the answer may not be 100% accurate, or an answer that actually answers your question.

> **Comments:**
> **username_1**: You're either assuming labor inelasticity (s.t. n=1 is always optimal in eq.)  or that the optimal labor choice is dependent upon the size of the labor force. Neither of those two assumptions is necessarily true here. Thus, $Y_n$ doesn't necessarily increase.  

---

