<p>This is not the elasticity of substitution formula. The correct one is, between factors, $K,L$</p>

<p>$$\sigma = \frac{\mathrm{d}\log \left(\frac{K}{L}\right)}{\mathrm{d}\log \left(\frac{F_L}{F_K}\right)}$$</p>

<p>This expression can conveniently be worked as follows: </p>

<p>$$\frac{\mathrm{d}\log \left(\frac{K}{L}\right)}{\mathrm{d}\log \left(\frac{F_L}{F_K}\right)} = \frac {(L/K)[d(K/L)]}{(F_K/F_L)[d(F_{L}/F_{K})]}$$</p>

<p>$$=\frac {LF_L}{KF_K} \cdot \frac{d(K/L)}{d(F_{L}/F_{K})}$$</p>

<p>In the Cobb-Douglas case we have</p>

<p>$$ \frac {F_L}{F_L} = \frac {1-a}{a}\frac {K}{L} $$</p>

<p>so</p>

<p>$$d\left(\frac{F_{L}}{F_{K}}\right) = \frac {1-a}{a}d\left(\frac{K}{L}\right)$$</p>

<p>Substituting both expressions we have</p>

<p>$$\sigma = \frac {L}{K}\frac {1-a}{a}\frac {K}{L} \cdot \frac{d(K/L)}{[(1-a)/a]d(K/L)}=1.$$</p>

CC BY-SA 4.0
2018-09-07T16:17:20.457Z
2018-09-07T16:17:20.457Z
Alecos Papadopoulos
<p>Good night, I'm reading the Romer's macroeconomy book in the page 42, "A complication" section title. The begin of third paragraph say:</p>

<blockquote>
  <p>This is not a general property of production functions, however. <strong>With Cobb–Douglas production, the elasticity of substitution between inputs is 1</strong>.  ...</p>
</blockquote>

<p>I understand that the factors elasticity of sustitution is defined like $El_x(y)=\frac{d\,\ln(y)}{d\,\ln(x)}$ in this case will be $El_K(L)=\frac{d\,\ln(L)}{d\,\ln(K)}=\frac{d L}{d K}\frac{K}{L}$.
From Cobb-Douglas simplest function $Y=K^\alpha L^{1-\alpha}$ we have, for $Y=const$, $K=(YL^{\alpha-1})^{1/\alpha}$ then
$$\frac {d K}{d L}=\frac{ \alpha - 1}{\alpha}\left(\frac{Y}{L}\right)^{1/\alpha}$$
now replacing in the formula:
$$El_L(K)=\frac{ \alpha - 1}{\alpha}\left(\frac{Y}{L}\right)^{1/\alpha}\times\frac{L}{K}$$ operating we have
$$El_L(K)=\frac{\alpha-1}{\alpha}$$
like $El_L(K)=\frac{1}{El_K (L)}$ then $El_K (L)=\frac{\alpha}{\alpha-1}$.
The point is that this values are unequal to one.</p>

<p>Am I interpreting wrong the affirmation? Thanks for your help.</p>

CC BY-SA 3.0
2016-10-25T16:19:24.670Z
Well, viewing other definition about the factor elasticity based on TSR (Technical Substitution Rate) the result is $El_K(L)=\alpha/(\alpha -1)$ then the Romer's affirmation isn't correct.  For this affirmation would be true, the production function have to be $F(K,L)=KL$.
Luis Salazar
<macroeconomics><production-function><cobb-douglas>
The factor elasticities from a Cobb-Douglas function in Romer's macroeconomy book
Luis Salazar
# The factor elasticities from a Cobb-Douglas function in Romer's macroeconomy book

(Asked by: username_0 on 2016-10-25)

Good night, I'm reading the Romer's macroeconomy book in the page 42, "A complication" section title. The begin of third paragraph say:

> This is not a general property of production functions, however. **With Cobb–Douglas production, the elasticity of substitution between inputs is 1**. ...

I understand that the factors elasticity of sustitution is defined like $El\_x(y)=\frac{d\,\ln(y)}{d\,\ln(x)}$ in this case will be $El\_K(L)=\frac{d\,\ln(L)}{d\,\ln(K)}=\frac{d L}{d K}\frac{K}{L}$.
From Cobb-Douglas simplest function $Y=K^\alpha L^{1-\alpha}$ we have, for $Y=const$, $K=(YL^{\alpha-1})^{1/\alpha}$ then
$$\frac {d K}{d L}=\frac{ \alpha - 1}{\alpha}\left(\frac{Y}{L}\right)^{1/\alpha}$$
now replacing in the formula:
$$El\_L(K)=\frac{ \alpha - 1}{\alpha}\left(\frac{Y}{L}\right)^{1/\alpha}\times\frac{L}{K}$$ operating we have
$$El\_L(K)=\frac{\alpha-1}{\alpha}$$
like $El\_L(K)=\frac{1}{El\_K (L)}$ then $El\_K (L)=\frac{\alpha}{\alpha-1}$.
The point is that this values are unequal to one.

Am I interpreting wrong the affirmation? Thanks for your help.

**Tags:** macroeconomics, production-function, cobb-douglas

> **Comments:**
> **username_0**: Well, viewing other definition about the factor elasticity based on TSR (Technical Substitution Rate) the result is $El_K(L)=\alpha/(\alpha -1)$ then the Romer's affirmation isn't correct.  For this affirmation would be true, the production function have to be $F(K,L)=KL$.  

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## 1 answers

### Answer by username_1 on 2018-09-07. Score: 1

This is not the elasticity of substitution formula. The correct one is, between factors, $K,L$

$$\sigma = \frac{\mathrm{d}\log \left(\frac{K}{L}\right)}{\mathrm{d}\log \left(\frac{F\_L}{F\_K}\right)}$$

This expression can conveniently be worked as follows:

$$\frac{\mathrm{d}\log \left(\frac{K}{L}\right)}{\mathrm{d}\log \left(\frac{F\_L}{F\_K}\right)} = \frac {(L/K)[d(K/L)]}{(F\_K/F\_L)[d(F\_{L}/F\_{K})]}$$

$$=\frac {LF\_L}{KF\_K} \cdot \frac{d(K/L)}{d(F\_{L}/F\_{K})}$$

In the Cobb-Douglas case we have

$$ \frac {F\_L}{F\_L} = \frac {1-a}{a}\frac {K}{L} $$

so

$$d\left(\frac{F\_{L}}{F\_{K}}\right) = \frac {1-a}{a}d\left(\frac{K}{L}\right)$$

Substituting both expressions we have

$$\sigma = \frac {L}{K}\frac {1-a}{a}\frac {K}{L} \cdot \frac{d(K/L)}{[(1-a)/a]d(K/L)}=1.$$


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