# The factor elasticities from a Cobb-Douglas function in Romer's macroeconomy book

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.

• 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 $F(K,L)=KL$. Oct 25, 2016 at 16:19

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

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