<p>I am not sure if it is intuitive but this is because because CRS function is homogenous of degree 1.</p>
<p><strong>Full derivation:</strong></p>
<p>First, general formula for any arbitrary elasticity of substitution between <span class="math-container">$L$</span> and <span class="math-container">$K$</span>* is given by (see Sydsæter et al. EMEA pp 430):</p>
<p><span class="math-container">$$\sigma_{L,K}  = \frac{-F_K'F_L'(xF_K'+ yF_L')}{xy \left( (F_L')^2F_{KK}^{''} - 2 F_K'F_L'F_{KL}'' + (F_K')^2F_{LL}''\right)}, \text{ for: } F(K,L)=c \tag{1}$$</span></p>
<p>Where c is arbitrary constant. This is our starting point.</p>
<p>Now by definition property of CRS production function is that they are homogenous of degree 1 (since by definition we have CRS when <span class="math-container">$F(tK,tL)= tF(K,L)$</span>).</p>
<p>If <span class="math-container">$F$</span> is homogenous of degree 1, then the numerator of (1) will be <span class="math-container">$= −F_K'F_L'F$</span>.</p>
<p>This is because of the Euler's theorem which tells us that if:</p>
<p><span class="math-container">$$f(x,y) \text{ is homogenous of degree k} \implies xf_x'(x,y) +  yf_y'(x,y)= kf(x,y)$$</span></p>
<p>Next Euler theorem also implies that (assuming <span class="math-container">$f$</span> is twice continuously differentiable) that:</p>
<p><span class="math-container">$$xf_{xx}''(x,y) +  yf_{yx}''(x,y)= (k-1)f_x'(x,y)$$</span>
<span class="math-container">$$xf_{yx}''(x,y) +  yf_{yy}''(x,y)= (k-1)f_y'(x,y)$$</span></p>
<p>The above implies that in our case:</p>
<p><span class="math-container">$$K F_{KK}'' = - LF_{KL}'' $$</span> and that <span class="math-container">$$L F_{LL}'' = - K F_{LK}'' = - K F_{KL}'' $$</span>. Hence the denominator will be given by:</p>
<p><span class="math-container">$$ -F{KL}''\left( L^2 (F_L')^2 + 2KLF_K'F_L'+ K^2(F_K')^2\right)= - F_{KL}''(KF_K'+ L F_L')^2 = - F_{KL}''F^2$$</span></p>
<p>We use Euler theorem above again.</p>
<p>Now finally we are done:</p>
<p><span class="math-container">$$\sigma_{LK} = \frac{−F_K'F_L'F}{- F_{KL}''F^2} =\frac{F_K'F_L'}{ F_{KL}''F}  $$</span></p>
<p>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.</p>
<hr />
<p><sub> * or <span class="math-container">$y$</span> and <span class="math-container">$x$</span> for that matter this generalizes to any elasticity of substitution problem </sub></p>

CC BY-SA 4.0
2021-05-16T21:17:39.247Z
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.
Alalalalaki
CC BY-SA 4.0
2021-05-16T21:19:45.470Z
@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.
1muflon1
CC BY-SA 4.0
2021-05-16T21:11:31.060Z
2021-05-16T21:11:31.060Z
1muflon1
<p><strong>1muflon1</strong>'s answer is entirely correct.</p>
<p>Let me give another alternative derivation, which might be a little bit easier to remember, although probably not more intuitive.</p>
<p>Let <span class="math-container">$k = K/L$</span> be the capital to labour ratio. Then we can define  <span class="math-container">$f(k) = F(K/L, 1)$</span> to be the output per unit of labour. then by the CRS assumption we have:
<span class="math-container">$$
F(K,L) = L f(K/L) = L f(k).
$$</span>
Taking derivatives with respect to <span class="math-container">$K$</span> and <span class="math-container">$L$</span> gives:
<span class="math-container">$$
\begin{align*}
&amp;F_K = L f'(k) \frac{1}{L} = f'(k),\\
&amp;F_L = f(k) + L f'(k) \left(-\frac{K}{L^2}\right) = f(k) - k f'(k).
\end{align*}
$$</span>
Finally:
<span class="math-container">$$
\begin{align*}
&amp;F_{K,L} = f''(k)\left(-\frac{K}{L^2}\right) = -kf''(k) \frac{1}{L},\\
\to &amp;F_{K,L} F = -k f''(k) f(k).
\end{align*}
$$</span>
Let's now have a look at the IES:
<span class="math-container">$$
\sigma_{L,K} = -\frac{\partial \ln(F_K/F_L)}{\partial \ln (K/L)} = \frac{\partial \ln \left(\frac{f'(k)}{f(k) - k f'(k)}\right) }{\partial \ln k}
$$</span>
Now take derivative of both numerator and denominator with respect to <span class="math-container">$k$</span>:
<span class="math-container">$$
\begin{align*}
\sigma_{L,K} &amp;= -k \frac{f - k f'}{f'}\frac{f'' \left(f - k f'\right) - f'\left(f' - f'- k f''\right)}{(f - k f)^2}\\
&amp;= \frac{-k f'' f}{f'(f - kf)},\\
&amp;= \frac{F_{K,L} F}{F_K F_L}
\end{align*}
$$</span></p>

CC BY-SA 4.0
2021-05-17T12:37:32.840Z
2021-09-30T17:22:01.200Z
2021-09-30T17:22:01.200Z
tdm
<p>Suppose that <span class="math-container">$F(\cdot)$</span> has CRS in <span class="math-container">$K$</span> and <span class="math-container">$L$</span>, the elasticity of Substitution is  <span class="math-container">$\sigma_{K L} \equiv F_{L} F_{K} / F F_{L K}$</span>.</p>
<p>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?</p>

<production-function><elasticity-of-substitution>
Elasticity of Substitution of CRS Production Function
Alalalalaki
# Elasticity of Substitution of CRS Production Function

(Asked by: username_0 on 2021-05-16)

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?

**Tags:** production-function, elasticity-of-substitution


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## 2 answers

### Answer by username_1 on 2021-05-16. Score: 4

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

> **Comments:**
> **username_0**: 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.  
> **username_1**: @username_0 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.  

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### Answer by username_2 on 2021-05-17. Score: 2

**username_1**'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:
$$
\sigma\_{L,K} = -\frac{\partial \ln(F\_K/F\_L)}{\partial \ln (K/L)} = \frac{\partial \ln \left(\frac{f'(k)}{f(k) - k f'(k)}\right) }{\partial \ln k}
$$
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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