# Cobb-Douglas nested in CES model

Assume, using the following equation $Y=[((A_1L)^\alpha K^\beta)^\sigma+ (A_2X^\gamma)^\sigma]^{1/\sigma}$, we back out (obtain) the evolution of $A_1$ & $A_2$. Can we interpret $A_1$ as labour-saving technical change index since it augments labour in the model?

• Say, there is a limited substitutability between the C-D composite and X. – london Sep 4 '17 at 10:50

$$Y=[A_1Q^\sigma+ A_2X^\sigma]^{1/\sigma}$$

where $Q=L^\alpha K^\beta$ (I removed unnecessary bits like $\gamma$ and exponent of $A_i$). This is a standard CES, where the optimal use of $Q$ and $X$ depend on $\frac{A_1}{A_2}$ and $\sigma$.

For a relatively low substitution ($\sigma<0$), an increase in $A_1$ (holding factor prices fixed) means firms make use of that extra increase in productivity of factor $Q$ by using less of it and more of $X$. Since $A_1$ corresponds to a Hick neutral change regarding $Q$, you will use the bundle $Q$ with the same initial $K/L$ ratio than before the change in $A_1$. Therefore, you will reduce both $K$ and $L$. In consequence, an increase in $A_1$ could be thought as labour-saving technical change with respect to $X$, and in absolute value (although notice that $K/L$ remains unchanged).

The reverse is true if $0<\sigma \leq 1$, namely an increase in $A_1$ leads to more use of $Q$ relative to $X$, thereby expanding the use of $L$ and $K$.

Meanwhile, in the inbetween case ($\sigma=0$), the function collapses to a Codd-Douglas:

$$Y = \left(L^\alpha K^\beta\right)^{\phi} X^{1-\phi}$$

where $\phi = \frac{A_1}{A_1 + A_2}$. Here a change in $A_1$ does not affect the K/L ratio, albeit it does affect the optimal ratio with respect to $X$. But commonly technical change is though as a change in some $A$ rather than in the elasticities of factors, so this case does not nest well with the above. I think this might be because the original function is not properly normalised. This is, the terms $A_1$ and $A_2$ do not add up to one, which is necessary given your original assumption of constant returns to scale.

• Once I get the time I will add the mathematical proof. – luchonacho Sep 4 '17 at 12:17
• First, the in-between case is $\sigma=1$ not $\sigma=0$. $\sigma=1$ yields directly a linear $Y=A_1Q+A_2X$. In the case of $\sigma=0$ you get of course $Y=Q^\alpha X^{1-\alpha}$ where $\alpha=A_1/(A_1+A_2)$. – Fato Sep 5 '17 at 6:41
• @Fato You are right. Confused $\sigma$ with $\rho$. Updated. – luchonacho Sep 5 '17 at 6:53