<p>Starting from </p>

<p>$$Y = A\left( \alpha L^{\rho} + (1-\alpha)K^{\rho} \right)^{\frac{1}{\rho}}$$</p>

<p>$$...\implies MP_L = \alpha A\cdot\left( \alpha + (1-\alpha)\left(\frac {K}{L}\right)^{\rho} \right)^{\frac{1}{\rho}-1}$$</p>

<p>assume that $\rho &lt;0$ (which gives an elasticity of substitution smaller than unity). Then we can write</p>

<p>$$MP_L = \frac A {\left( \alpha  + (1-\alpha)\left(\frac {L}{K}\right)^{|\rho|} \right)^{\frac{1}{|\rho|}+1}}$$</p>

<p>So </p>

<p>$$L\to 0 \implies MP_L \to \frac A { \alpha  ^{\frac{1}{|\rho|}+1}} &gt;0 $$</p>

<p>although not infinite.</p>

CC BY-SA 3.0
2017-06-01T09:10:30.617Z
Thanks. The math folks got the same result (see [here](https://math.stackexchange.com/questions/2304218/infinite-lhopital-loop)). However, they also find that $MP_L=0$ is a solution too, which is what the referred slide 15 in my answer states. Any idea how this square up with your answer? In any case, it is interesting to learn that the Inada conditions do not hold for the general non-Cobb-Douglas CES!
luchonacho
CC BY-SA 3.0
2017-06-01T14:11:46.327Z
@luchonacho "Solution" to what? "Solution" in what sense? We go from the function to the derivative and calculate the limit of the latter. The limit is unique so there is no other "solution". What am I missing?
Alecos Papadopoulos
CC BY-SA 3.0
2017-06-02T08:55:59.317Z
In the sense that $MP_L = \alpha \left(\frac{Y}{L}\right)^{1-\rho}$. So as $L \rightarrow 0$, so does $Y$. So you end up with an undefined expression, where you can apply the L'Hopital. According to the post I referred to, one solution of such "infinite L'Hopital loop" is that $f'(0)=0$. Since the function $f(x)=Y$ is of the type described in that question, I fail to see why the second solution does not apply, even if it does not show up in your particular proof.
luchonacho
CC BY-SA 3.0
2017-06-02T09:06:04.407Z
L' hopital _may_ provide a solution. In our case, since the direct calculation of limit gives us a definite answer, we conclude that the possible other solution just doesn't apply.
Alecos Papadopoulos
CC BY-SA 3.0
2017-06-02T09:12:51.987Z
Interesting. So you are basically saying that using an indirect method adds a possible solution. Or perhaps more puzzling, even if I have a general problem with two solutions, an specific problem, belonging to the family of the general problem, might only have one solution. Any intuition behind this? Seems quite an extraordinary result.
luchonacho
CC BY-SA 3.0
2017-06-02T09:18:25.510Z
It is not. A lot of economic models end up being a solution to a quadratic polynomial with one root positive and one root negative. And we just discard the negative root because it doesn't make sense for an economic quantity to be negative. Lesson: mathematical tools are more general than the needs for which they are put to use.
Alecos Papadopoulos
CC BY-SA 3.0
2017-05-31T21:13:10.050Z
2017-05-31T21:13:10.050Z
Alecos Papadopoulos
<p>Ok, rather embarrassing, but the problem was in the formula for $Y_0$. In effect, for $\rho&gt;0$, it is true that</p>

<p>$$ Y_{_{L=0}} = AK(1-\alpha)^{\frac{1}{\rho}} $$</p>

<p>However, for the case of $\rho&lt;0$, the limit of $Y$ is zero. The proof is trivial. Rewrite the CES as:</p>

<p>$$ Y = AL\left(\alpha + (1-\alpha)\left(\frac{L}{K}\right)^{-\rho} \right)^{\frac{1}{\rho}} $$</p>

<p>Then, it is trivial to see that:</p>

<p>$$ \lim_{L\rightarrow 0} Y = A0 \: \alpha^{\frac{1}{\rho}} = 0 $$</p>

<p>With this, we can see that $\Delta Y = Y &gt; 0$. This solves the inconsistency.</p>

<hr>

<p>PS: as it turns out, <strong>the marginal product of labour at $L=0$ is not infinity</strong>. Apparently, the Inada conditions <a href="http://econpapers.repec.org/article/eeeecolet/v_3a81_3ay_3a2003_3ai_3a3_3ap_3a361-363.htm" rel="nofollow noreferrer">do not hold</a> for the CES with $\rho&lt;0. Actually, <strong>the marginal product of labour seem to be zero!</strong> (check slide 15 <a href="http://www2.warwick.ac.uk/fac/soc/economics/staff/jchavezcotrado/ec92a_phdmacro/recitation1.pdf" rel="nofollow noreferrer">here</a>). This is a major revelation to me.</p>

CC BY-SA 3.0
2017-05-31T14:03:39.300Z
2017-05-31T14:03:39.300Z
luchonacho
<p>I am getting some "weird results". I find that in a CES, with short term fixed capital and elasticity of substitution smaller than one, <strong>it is optimal for firms to hire zero labour, which seems at odd with the idea that the marginal product of labour is always positive</strong> (except in the extreme case of Leontief, where $\rho = -\infty$).</p>

<p>Assume a simple CES production function:</p>

<p>$$Y = A\left( \alpha L^{\rho} + (1-\alpha)K^{\rho} \right)^{\frac{1}{\rho}}$$</p>

<p>Assume that capital is fixed in the short term, so the firm only optimises over  $L$. Importantly, <strong>the firm must pay the cost of capital anyway</strong>. Therefore, as long as $p&gt;0$, the firm is always better off by producing something. Thus, the key question is how much labour to hire. </p>

<p>Under competitive markets, optimal labour comes from:</p>

<p>\begin{equation}
\frac{\partial Y}{\partial L} \equiv \alpha A^{\rho} \left(\frac{Y^*}{L^*}\right)^{1-\rho}  = \left(\frac{w}{p}\right) 
\end{equation}</p>

<p>Importantly, notice that MPL is (i) <strong>always positive</strong>, (ii) for $\rho&lt;1$ (including negative values) and $L^*=0$, <strong>it is infinity!</strong>. In other words, whenever there is imperfect substitution of factors, and the wage rate is non-infinity, <strong>you want to produce with some labour</strong>.</p>

<p>Replacing output into the above yields a single equation with only one unknown ($L_i^*$), which solution is:</p>

<p>$$ L^* = K \Omega^{\frac{1}{\rho}}  $$</p>

<p>where </p>

<p>$$ \Omega = \left(\dfrac{(1-\alpha)}{\left(\frac{w}{A p \alpha}\right)^{\frac{\rho}{1-\rho}}-\alpha}\right) $$</p>

<p>Now, discard the parameterisations where $\Omega&lt;0$. These are corner solutions where $L^*=0$.</p>

<p>So far so good. Except for some strange result when $\rho&lt;0$. To see this, let us replace optimal labour into the production function. After some rearranging, you get:</p>

<p>$$ Y^* = AK\left(\alpha\Omega + (1-\alpha)\right)^{\frac{1}{\rho}} $$</p>

<p>Now, compare this with the case of zero labour. Let us call this $Y_0$. Here, output is:</p>

<p>$$ Y_0 = AK(1-\alpha)^{\frac{1}{\rho}} $$</p>

<p>Now, <strong>we want to know under which parameterisation the firm is better off producing with zero labour</strong>. Profits with optimal labour and with zero labour are respectively:</p>

<p>$$ \pi = pY^* - rK - wL^* $$</p>

<p>$$ \pi_0 = pY_0 - rK $$</p>

<p>Thus, the firm will produce with no labour <em>iff</em> $\pi_0&gt; \pi$, which is:</p>

<p>$$ wL^* &gt; p(Y-Y_0) $$</p>

<p>So, if the labour costs do not compensate for extra profits, the firm will choose $L^*=0$. Here is where the problem arises: <strong>for $\rho&lt;0$, $Y-Y_0$ is always negative!</strong> Which means firm will produce with zero labour.</p>

<p>To see this, let us find the conditions under which $\Delta Y=Y-Y_0&lt;0$. This is, </p>

<p>$$ AK\left[\left( \alpha \Omega + (1-\alpha)\right)^{\frac{1}{\rho}} - (1-\alpha)^{\frac{1}{\rho}}  \right] &lt;0 $$</p>

<p>$$ \left( \alpha \Omega + (1-\alpha)\right)^{\frac{1}{\rho}} &lt; (1-\alpha)^{\frac{1}{\rho}} $$</p>

<p>Noting that $\rho&lt;0$ will invert the components, you get:</p>

<p>$$ \frac{1}{\left( \alpha \Omega + (1-\alpha)\right)} &lt; \frac{1}{1-\alpha} $$</p>

<p>Which is clearly true for:</p>

<p>$$ 0&lt; \alpha\Omega $$</p>

<p>So, for "normal" parameterisations where $\Omega&gt;0$, we can see that $Y-Y_0&lt;0$ is always true. </p>

<p>I have confirmed this in R. The following code produces delta in output = -15:</p>

<pre><code>rm(list=ls())

rho       &lt;- -0.5
alpha     &lt;- .6
A &lt;- 1
p &lt;- 1
w &lt;- 0.7

K &lt;- 2.756729
omega = (1-alpha)/((w/(alpha*A*p))^(rho/(1-rho))-alpha)
L = K*omega^(1/rho)

ll = alpha*L^rho
kk = (1-alpha)*K^rho
tr = (1/rho)
Y = A*(ll+kk)^tr
Y0 = A*(kk)^tr

delta_output = Y-Y0

delta_profits = p*(Y-Y0)-w*L
</code></pre>

<p>if I change $\rho$ to 0.5, you get the desired result (that using labour does increases output).</p>

<p>You can also see the plot of delta output <a href="http://www.wolframalpha.com/input/?i=plot+(0.6*x%5E(-0.5)%2B1)%5E(-2)+-+1%5E(-2)+between+0+and+100" rel="nofollow noreferrer">here</a>. It is clear that the delta output is always negative, and asymptotically zero for $x=\infty$, which is when $L^*=0$.</p>

<microeconomics><elasticity-of-substitution>
Is optimal labour zero when (i) capital fixed and (ii) elasticity of substitution less than 1?
luchonacho
# Is optimal labour zero when (i) capital fixed and (ii) elasticity of substitution less than 1?

(Asked by: username_0 on 2017-05-30)

I am getting some "weird results". I find that in a CES, with short term fixed capital and elasticity of substitution smaller than one, it is optimal for firms to hire zero labour, which seems at odd with the idea that the marginal product of labour is always positive (except in the extreme case of Leontief, where $\rho = -\infty$).
Assume a simple CES production function:
$$Y = A\left( \alpha L^{\rho} + (1-\alpha)K^{\rho} \right)^{\frac{1}{\rho}}$$
Assume that capital is fixed in the short term, so the firm only optimises over  $L$. Importantly, the firm must pay the cost of capital anyway. Therefore, as long as $p>0$, the firm is always better off by producing something. Thus, the key question is how much labour to hire. 
Under competitive markets, optimal labour comes from:
\begin{equation}
\frac{\partial Y}{\partial L} \equiv \alpha A^{\rho} \left(\frac{Y^*}{L^*}\right)^{1-\rho}  = \left(\frac{w}{p}\right) 
\end{equation}
Importantly, notice that MPL is (i) always positive, (ii) for $\rho<1$ (including negative values) and $L^*=0$, it is infinity!. In other words, whenever there is imperfect substitution of factors, and the wage rate is non-infinity, you want to produce with some labour.
Replacing output into the above yields a single equation with only one unknown ($L_i^*$), which solution is:
$$ L^* = K \Omega^{\frac{1}{\rho}}  $$
where 
$$ \Omega = \left(\dfrac{(1-\alpha)}{\left(\frac{w}{A p \alpha}\right)^{\frac{\rho}{1-\rho}}-\alpha}\right) $$
Now, discard the parameterisations where $\Omega<0$. These are corner solutions where $L^*=0$.
So far so good. Except for some strange result when $\rho<0$. To see this, let us replace optimal labour into the production function. After some rearranging, you get:
$$ Y^* = AK\left(\alpha\Omega + (1-\alpha)\right)^{\frac{1}{\rho}} $$
Now, compare this with the case of zero labour. Let us call this $Y_0$. Here, output is:
$$ Y_0 = AK(1-\alpha)^{\frac{1}{\rho}} $$
Now, we want to know under which parameterisation the firm is better off producing with zero labour. Profits with optimal labour and with zero labour are respectively:
$$ \pi = pY^* - rK - wL^* $$
$$ \pi_0 = pY_0 - rK $$
Thus, the firm will produce with no labour iff $\pi_0> \pi$, which is:
$$ wL^* > p(Y-Y_0) $$
So, if the labour costs do not compensate for extra profits, the firm will choose $L^*=0$. Here is where the problem arises: for $\rho<0$, $Y-Y_0$ is always negative! Which means firm will produce with zero labour.
To see this, let us find the conditions under which $\Delta Y=Y-Y_0<0$. This is, 
$$ AK\left[\left( \alpha \Omega + (1-\alpha)\right)^{\frac{1}{\rho}} - (1-\alpha)^{\frac{1}{\rho}}  \right] <0 $$
$$ \left( \alpha \Omega + (1-\alpha)\right)^{\frac{1}{\rho}} < (1-\alpha)^{\frac{1}{\rho}} $$
Noting that $\rho<0$ will invert the components, you get:
$$ \frac{1}{\left( \alpha \Omega + (1-\alpha)\right)} < \frac{1}{1-\alpha} $$
Which is clearly true for:
$$ 0< \alpha\Omega $$
So, for "normal" parameterisations where $\Omega>0$, we can see that $Y-Y_0<0$ is always true. 
I have confirmed this in R. The following code produces delta in output = -15:
rm(list=ls())

rho       <- -0.5
alpha     <- .6
A <- 1
p <- 1
w <- 0.7

K <- 2.756729
omega = (1-alpha)/((w/(alpha*A*p))^(rho/(1-rho))-alpha)
L = K*omega^(1/rho)

ll = alpha*L^rho
kk = (1-alpha)*K^rho
tr = (1/rho)
Y = A*(ll+kk)^tr
Y0 = A*(kk)^tr

delta_output = Y-Y0

delta_profits = p*(Y-Y0)-w*L

if I change $\rho$ to 0.5, you get the desired result (that using labour does increases output).
You can also see the plot of delta output here. It is clear that the delta output is always negative, and asymptotically zero for $x=\infty$, which is when $L^*=0$.

**Tags:** microeconomics, elasticity-of-substitution


---

## 2 answers

### Answer by username_0 on 2017-05-31. Score: 3 | Accepted answer

Ok, rather embarrassing, but the problem was in the formula for $Y\_0$. In effect, for $\rho>0$, it is true that

$$ Y\_{\_{L=0}} = AK(1-\alpha)^{\frac{1}{\rho}} $$

However, for the case of $\rho<0$, the limit of $Y$ is zero. The proof is trivial. Rewrite the CES as:

$$ Y = AL\left(\alpha + (1-\alpha)\left(\frac{L}{K}\right)^{-\rho} \right)^{\frac{1}{\rho}} $$

Then, it is trivial to see that:

$$ \lim\_{L\rightarrow 0} Y = A0 \: \alpha^{\frac{1}{\rho}} = 0 $$

With this, we can see that $\Delta Y = Y > 0$. This solves the inconsistency.

---

PS: as it turns out, **the marginal product of labour at $L=0$ is not infinity**. Apparently, the Inada conditions [do not hold](http://econpapers.repec.org/article/eeeecolet/v_3a81_3ay_3a2003_3ai_3a3_3ap_3a361-363.htm) for the CES with $\rho<0. Actually, **the marginal product of labour seem to be zero!** (check slide 15 [here](http://www2.warwick.ac.uk/fac/soc/economics/staff/jchavezcotrado/ec92a_phdmacro/recitation1.pdf)). This is a major revelation to me.


---

### Answer by username_1 on 2017-05-31. Score: 3

Starting from

$$Y = A\left( \alpha L^{\rho} + (1-\alpha)K^{\rho} \right)^{\frac{1}{\rho}}$$

$$...\implies MP\_L = \alpha A\cdot\left( \alpha + (1-\alpha)\left(\frac {K}{L}\right)^{\rho} \right)^{\frac{1}{\rho}-1}$$

assume that $\rho <0$ (which gives an elasticity of substitution smaller than unity). Then we can write

$$MP\_L = \frac A {\left( \alpha + (1-\alpha)\left(\frac {L}{K}\right)^{|\rho|} \right)^{\frac{1}{|\rho|}+1}}$$

So

$$L\to 0 \implies MP\_L \to \frac A { \alpha ^{\frac{1}{|\rho|}+1}} >0 $$

although not infinite.

> **Comments:**
> **username_0**: Thanks. The math folks got the same result (see [here](https://math.stackexchange.com/questions/2304218/infinite-lhopital-loop)). However, they also find that $MP_L=0$ is a solution too, which is what the referred slide 15 in my answer states. Any idea how this square up with your answer? In any case, it is interesting to learn that the Inada conditions do not hold for the general non-Cobb-Douglas CES!  
> **username_1**: @username_0 "Solution" to what? "Solution" in what sense? We go from the function to the derivative and calculate the limit of the latter. The limit is unique so there is no other "solution". What am I missing?  
> **username_0**: In the sense that $MP_L = \alpha \left(\frac{Y}{L}\right)^{1-\rho}$. So as $L \rightarrow 0$, so does $Y$. So you end up with an undefined expression, where you can apply the L'Hopital. According to the post I referred to, one solution of such "infinite L'Hopital loop" is that $f'(0)=0$. Since the function $f(x)=Y$ is of the type described in that question, I fail to see why the second solution does not apply, even if it does not show up in your particular proof.  
> **username_1**: L' hopital _may_ provide a solution. In our case, since the direct calculation of limit gives us a definite answer, we conclude that the possible other solution just doesn't apply.  
> **username_0**: Interesting. So you are basically saying that using an indirect method adds a possible solution. Or perhaps more puzzling, even if I have a general problem with two solutions, an specific problem, belonging to the family of the general problem, might only have one solution. Any intuition behind this? Seems quite an extraordinary result.  
> **username_1**: It is not. A lot of economic models end up being a solution to a quadratic polynomial with one root positive and one root negative. And we just discard the negative root because it doesn't make sense for an economic quantity to be negative. Lesson: mathematical tools are more general than the needs for which they are put to use.  

---

