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<p>This is the solution to your difficulty. By the way, which text are you following? </p>

CC BY-SA 4.0
2018-09-30T16:35:37.557Z
I think OP's confusion is over why average productivity is defined in this manner. To me, average labor productivity (in efficiency units) would just be $\frac{Y}{AL}$. The paper in question is [here](https://www.econstor.eu/bitstream/10419/55055/1/684015625.pdf) (see page 6 of the paper). Maybe it's just a language issue -- it is a working paper.
Kenneth Rios
CC BY-SA 4.0
2018-10-01T11:42:00.883Z
Yes, I'm also trying to understand this. Anyway, it is not a working paper, it has been published on Journal of Macroeconomics (Volume 34, Issue 1, March 2012, Pages 181-198).
Alessandro
CC BY-SA 4.0
2018-10-01T17:53:52.363Z
I think this is a language issue. I see no relevance in raising average productivity to a function of elasticity of substitution and then calling it average productivity. If you find an answer, please do let me know.
DrStrangeLove
CC BY-SA 4.0
2018-10-01T21:02:42.110Z
Agreed that this is likely just semantics. The language preceding the expression given for average productivity is indicative to me.
Kenneth Rios
CC BY-SA 4.0
2018-09-30T10:51:20.097Z
2018-09-30T10:51:20.097Z
DrStrangeLove
<p>Following the work of <em>Raurich et al. (2012)</em> I got stuck trying to derive the average productivity starting from the following CES production function:</p>

<p><span class="math-container">$$Y=A\left [ \alpha K^{\frac{\sigma -1}{\sigma }}+(1-\alpha)(AL)^{\frac{\sigma -1}{\sigma }} \right ]^{\frac{\sigma}{\sigma-1 }}$$</span></p>

<p>The result they get is: 
<span class="math-container">$$\left (\frac{Y}{AL}  \right )^{\frac{1-\sigma }{\sigma }}=\frac{1}{\alpha \left ( \frac{K}{AL} \right )^{\frac{\sigma-1 }{\sigma }}+(1-\alpha )}$$</span></p>

<p>Is there anyone who can help me with this?</p>

CC BY-SA 4.0
2018-09-29T23:52:41.667Z
I checked just to be sure. But actually the exponent of the LHS  is right.
Alessandro
<macroeconomics><microeconomics><mathematical-economics>
Deriving average productivity from a CES production function
Alessandro
# Deriving average productivity from a CES production function

(Asked by: username_0 on 2018-09-29)

Following the work of *Raurich et al. (2012)* I got stuck trying to derive the average productivity starting from the following CES production function:

$$Y=A\left [ \alpha K^{\frac{\sigma -1}{\sigma }}+(1-\alpha)(AL)^{\frac{\sigma -1}{\sigma }} \right ]^{\frac{\sigma}{\sigma-1 }}$$

The result they get is:
$$\left (\frac{Y}{AL} \right )^{\frac{1-\sigma }{\sigma }}=\frac{1}{\alpha \left ( \frac{K}{AL} \right )^{\frac{\sigma-1 }{\sigma }}+(1-\alpha )}$$

Is there anyone who can help me with this?

**Tags:** macroeconomics, microeconomics, mathematical-economics
> **Comments:**
> **username_0**: I checked just to be sure. But actually the exponent of the LHS  is right.  

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## 1 answers

### Answer by username_1 on 2018-09-30. Score: 2 | Accepted answer

[![Solution](https://i.sstatic.net/3zNON.jpg)](https://i.sstatic.net/3zNON.jpg)

This is the solution to your difficulty. By the way, which text are you following?

> **Comments:**
> **username_2**: I think OP's confusion is over why average productivity is defined in this manner. To me, average labor productivity (in efficiency units) would just be $\frac{Y}{AL}$. The paper in question is [here](https://www.econstor.eu/bitstream/10419/55055/1/684015625.pdf) (see page 6 of the paper). Maybe it's just a language issue -- it is a working paper.  
> **username_0**: Yes, I'm also trying to understand this. Anyway, it is not a working paper, it has been published on Journal of Macroeconomics (Volume 34, Issue 1, March 2012, Pages 181-198).  
> **username_1**: I think this is a language issue. I see no relevance in raising average productivity to a function of elasticity of substitution and then calling it average productivity. If you find an answer, please do let me know.  
> **username_2**: Agreed that this is likely just semantics. The language preceding the expression given for average productivity is indicative to me.  

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