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Suppose that $F(\cdot)$ has CRS in $K$ and $L$, the elasticity of Substitution is $\sigma_{K L} \equiv F_{L} F_{K} / F F_{L K}$.

I once derived this equation but I remember that it takes me quite long time and the process is quite tedious. I wonder is there any easy way to derive this or remember it intuitively?

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I am not sure if it is intuitive but this is because because CRS function is homogenous of degree 1.

Full derivation:

First, general formula for any arbitrary elasticity of substitution between $L$ and $K$* is given by (see Sydsæter et al. EMEA pp 430):

$$\sigma_{L,K} = \frac{-F_K'F_L'(xF_K'+ yF_L')}{xy \left( (F_L')^2F_{KK}^{''} - 2 F_K'F_L'F_{KL}'' + (F_K')^2F_{LL}''\right)}, \text{ for: } F(K,L)=c \tag{1}$$

Where c is arbitrary constant. This is our starting point.

Now by definition property of CRS production function is that they are homogenous of degree 1 (since by definition we have CRS when $F(tK,tL)= tF(K,L)$).

If $F$ is homogenous of degree 1, then the numerator of (1) will be $= −F_K'F_L'F$.

This is because of the Euler's theorem which tells us that if:

$$f(x,y) \text{ is homogenous of degree k} \implies xf_x'(x,y) + yf_y'(x,y)= kf(x,y)$$

Next Euler theorem also implies that (assuming $f$ is twice continuously differentiable) that:

$$xf_{xx}''(x,y) + yf_{yx}''(x,y)= (k-1)f_x'(x,y)$$ $$xf_{yx}''(x,y) + yf_{yy}''(x,y)= (k-1)f_y'(x,y)$$

The above implies that in our case:

$$K F_{KK}'' = - LF_{KL}'' $$ and that $$L F_{LL}'' = - K F_{LK}'' = - K F_{KL}'' $$. Hence the denominator will be given by:

$$ -F{KL}''\left( L^2 (F_L')^2 + 2KLF_K'F_L'+ K^2(F_K')^2\right)= - F_{KL}''(KF_K'+ L F_L')^2 = - F_{KL}''F^2$$

We use Euler theorem above again.

Now finally we are done:

$$\sigma_{LK} = \frac{−F_K'F_L'F}{- F_{KL}''F^2} =\frac{F_K'F_L'}{ F_{KL}''F} $$

I personally do not think the result above is very intuitive (if it is so its intuition eludes me), but it is a consequence of CRS function being homogenous of degree 1 and this result actually applies for any elasticity of substitution between two variables of arbitrary function that is homogenous of degree 1 and twice continuously differentiable.


* or $y$ and $x$ for that matter this generalizes to any elasticity of substitution problem

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    $\begingroup$ Yes, what confuse me is that this result should be so general but neither its derivation nor the result itself quite intuitive. Anyway, thanks for your answer and I think it's nice to have an answer that contains the full derivation in this site. $\endgroup$ – Alalalalaki May 16 at 21:17
  • $\begingroup$ @Alalalalaki yea I don't really think there is an easier derivation, not all results in economics need to be intuitive, I mean this is why we model stuff, sometimes we learn something non-intuitive yet tremendously useful. $\endgroup$ – 1muflon1 May 16 at 21:19
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1muflon1's answer is entirely correct.

Let me give another alternative derivation, which might be a little bit easier to remember, although probably not more intuitive.

Let $k = K/L$ be the capital to labour ratio. Then we can define $f(k) = F(K/L, 1)$ to be the output per unit of labour. then by the CRS assumption we have: $$ F(K,L) = L f(K/L) = L f(k). $$ Taking derivatives with respect to $K$ and $L$ gives: $$ \begin{align*} &F_K = L f'(k) \frac{1}{L} = f'(k),\\ &F_L = f(k) + L f'(k) \left(-\frac{K}{L^2}\right) = f(k) - k f'(k). \end{align*} $$ Finally: $$ \begin{align*} &F_{K,L} = f''(k)\left(-\frac{K}{L^2}\right) = -kf''(k) \frac{1}{L},\\ \to &F_{K,L} F = -k f''(k) f(k). \end{align*} $$ Let's now have a look at the IES: $$ \sigma_{L,K} = -\frac{\partial \ln(F_K/F_L)}{\partial \ln (K/L)} = \frac{\partial \left(\frac{f'(k)}{f(k) - k f'(k)}\right) }{\partial \ln k} $$ Now take derivative of both numerator and denominator with respect to $k$: $$ \begin{align*} \sigma_{L,K} &= -k \frac{f - k f'}{f'}\frac{f'' \left(f - k f'\right) - f'\left(f' - f'- k f''\right)}{(f - k f)^2}\\ &= \frac{-k f'' f}{f'(f - kf)},\\ &= \frac{F_{K,L} F}{F_K F_L} \end{align*} $$

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