<p>Contrarily to utility functions, production functions are cardinal, and so they are not arbitrarily normalized. In the case of your CES production function, however, if we consider</p>
<p><span class="math-container">$$
F(x; A, a, \rho, \nu) = A \left( \sum_{i=1}^n a_i \, x_i^\rho \right)^{\nu/\rho},
$$</span>
with <span class="math-container">$\sum_{i=1}^n a_i \neq 1$</span> it is easy via a simple reparameterization of <span class="math-container">$(A,a_i)$</span> to rewrite it as
<span class="math-container">$$
G(x; B, b, \rho, \nu) = B \left( \sum_{i=1}^n b_i \, x_i^\rho \right)^{\nu/\rho},
$$</span>
with <span class="math-container">$\sum_{i=1}^n b_i = 1,$</span> and such that for any <span class="math-container">$x$</span> we have
<span class="math-container">$$ F(x; A, a, \rho, \nu) = G(x; B, b, \rho, \nu). $$</span></p>

CC BY-SA 4.0
2021-09-30T06:28:34.590Z
2021-09-30T06:43:28.933Z
2021-09-30T06:43:28.933Z
Bertrand
<p>Can we normalize a production function in the same way we can normalize a utility function? For example, consider the CES function
<span class="math-container">$$
F(x; A, a, \rho, \nu) = A \left( \sum_{i=1}^n a_i \, x_i^\rho \right)^{\nu/\rho}
$$</span>
where <span class="math-container">$x$</span> and <span class="math-container">$a$</span> are vectors of length <span class="math-container">$n$</span>; and <span class="math-container">$A$</span>, <span class="math-container">$\rho$</span>, and <span class="math-container">$\nu$</span> are scalars. <span class="math-container">$x$</span> represents inputs and <span class="math-container">$A, a, \rho, \nu$</span> are parameters.
If <span class="math-container">$F$</span> is a utility function, then we can make the normalization <span class="math-container">$\sum_{i=1}^n a_i = 1$</span>, while keeping the same preferences. Can we do the same if <span class="math-container">$F$</span> is a production function or in that case the normalization matters?</p>

CC BY-SA 4.0
2021-09-30T05:30:38.340Z
Yes, this can be done wlg if you reparameterize $A$. Your function is homogeneous in $(A,a_i)$.
Bertrand
CC BY-SA 4.0
2021-09-30T06:10:41.220Z
@Bertrand Please post answers as answers, and elaborate on this; it may not be clear to some why reparameterizing $A$ is necessary.
Giskard
CC BY-SA 4.0
2021-09-30T06:30:30.907Z
@Giskard: I was reluctant as it looks a bit like an exercise...
Bertrand
<production-function>
Normalization of production function
amrods
# Normalization of production function

(Asked by: username_0 on 2021-09-30)

Can we normalize a production function in the same way we can normalize a utility function? For example, consider the CES function
$$
F(x; A, a, \rho, \nu) = A \left( \sum\_{i=1}^n a\_i \, x\_i^\rho \right)^{\nu/\rho}
$$
where $x$ and $a$ are vectors of length $n$; and $A$, $\rho$, and $\nu$ are scalars. $x$ represents inputs and $A, a, \rho, \nu$ are parameters.
If $F$ is a utility function, then we can make the normalization $\sum\_{i=1}^n a\_i = 1$, while keeping the same preferences. Can we do the same if $F$ is a production function or in that case the normalization matters?

**Tags:** production-function

> **Comments:**
> **username_1**: Yes, this can be done wlg if you reparameterize $A$. Your function is homogeneous in $(A,a_i)$.  
> **username_2**: @username_1 Please post answers as answers, and elaborate on this; it may not be clear to some why reparameterizing $A$ is necessary.  
> **username_1**: @username_2: I was reluctant as it looks a bit like an exercise...  

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## 1 answers

### Answer by username_1 on 2021-09-30. Score: 2 | Accepted answer

Contrarily to utility functions, production functions are cardinal, and so they are not arbitrarily normalized. In the case of your CES production function, however, if we consider

$$
F(x; A, a, \rho, \nu) = A \left( \sum\_{i=1}^n a\_i \, x\_i^\rho \right)^{\nu/\rho},
$$
with $\sum\_{i=1}^n a\_i \neq 1$ it is easy via a simple reparameterization of $(A,a\_i)$ to rewrite it as
$$
G(x; B, b, \rho, \nu) = B \left( \sum\_{i=1}^n b\_i \, x\_i^\rho \right)^{\nu/\rho},
$$
with $\sum\_{i=1}^n b\_i = 1,$ and such that for any $x$ we have
$$ F(x; A, a, \rho, \nu) = G(x; B, b, \rho, \nu). $$


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