<p>The power family of production functions satisfies Inada's conditions over a set of admissible parameter values. 
In the case with two inputs, the functional form is given by: </p>

<p><span class="math-container">$$ y = \alpha_{11} x_1^{\gamma_1} + \alpha_{22} x_2^{\gamma_2} + \alpha_{12} x_1^{\gamma_1/2}x_2^{\gamma_2/2}, $$</span></p>

<p>The power production function <span class="math-container">$f$</span> satisfies <span class="math-container">$f(0)=0$</span>, <span class="math-container">$f(x) \geq 0$</span> and is increasing in each <span class="math-container">$x_i$</span> if all <span class="math-container">$\alpha_{ij}&gt;0$</span> and <span class="math-container">$ \gamma_{i}&gt;0 $</span>.
It is concave in <span class="math-container">$x$</span> if in addition all <span class="math-container">$\gamma_{i}&lt;1$</span>, as a sum of concave functions.</p>

<p>The partial derivatives are:
<span class="math-container">$$ \frac{\partial f}{\partial x_i}( x ) = \frac{ \alpha_{ii} \gamma_i }{ x_i^{1-\gamma_i} } + \frac{ \alpha_{ij} \gamma_i/2 }{ x_i^{1-\gamma_i/2} }{x_j^{\gamma_j/2}} $$</span>
The partial derivatives' limiting behavior are compatible with Inada's conditions:
<span class="math-container">$$ \lim_{x_{i}\rightarrow 0}\frac{\partial f}{\partial x_{i}}\left( x\right)
=+\infty ,\qquad \lim_{x_{i}\rightarrow +\infty }\frac{\partial f}{\partial
x_{i}}\left( x\right) =0. $$</span></p>

<p>This power production function can be generalized to the case of <span class="math-container">$J$</span> inputs as follows:
<span class="math-container">$$ y = h( \sum_{i=1}^J \sum_{j=1}^J \alpha_{ij} x_i^{\gamma_i/2}x_j^{\gamma_j/2} ), $$</span>
where function <span class="math-container">$h$</span> satisfies <span class="math-container">$h(0)=0$</span>, and is increasing and concave.
See for instance Diewert (1971) for a contribution with <span class="math-container">$\gamma_i=\gamma_j=1$</span>.
A similar argumentation should apply to the Box-Cox family of production functions.</p>

<p>Diewert, W. E., 1971, "An Application of the Shephard Duality Theorem: A Generalized Leontief Production Function," <em>Journal of Political Economy</em>, 79, 481-507. </p>

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2020-02-06T21:11:55.690Z
2020-02-06T21:11:55.690Z
Bertrand
<p>I believe this question is too broad.  As discussion in the comments indicates, even functional forms that are commonly used for their ability to satisfy the Inada conditions can fail to satisfy them for allowed values of their parameters.  On the other hand, as also noted, one can always define a piecewise-continuous function that satisfies them.</p>

<p>With that said, it's been shown that every Inada-satisfying function is asymptotically Cobb-Douglas.  <a href="https://www.sciencedirect.com/science/article/abs/pii/S0165176503002180" rel="nofollow noreferrer">https://www.sciencedirect.com/science/article/abs/pii/S0165176503002180</a></p>

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2020-02-04T15:39:04.787Z
2020-02-04T15:39:04.787Z
heh
<p>The Wikipedia statement should come with a citation or proof, as it is not entirely accurate: Cobb-Douglas functions with increasing returns to scale do not necessarily satisfy the Inada conditions, e.g., <span class="math-container">$f(x_1,x_2) = x_1^2x_2^2$</span> does not.</p>

<hr>

<p>There are many non Cobb-Douglas functions that satisfy the Inada conditions, albeit they may not have nice compact formulas. E.g., 
<span class="math-container">$$
f(x) = \left\{\begin{array}{ll}
x^{1/2} &amp; \mbox{ if } x \leq 1 \\
2x^{1/4} - 1 &amp; \mbox{ if } 1 &lt; x 
\end{array}\right.
$$</span>
This satisfies the conditions:</p>

<ol>
<li><span class="math-container">$f(0) = 0$</span></li>
<li>The Hessian is negative semidefinite as <span class="math-container">$f$</span> is strictly concave.</li>
<li><span class="math-container">$$\lim_{x \to 0}\frac{\text{d} f(x)}{\text{d} x} = \infty$$</span></li>
<li><span class="math-container">$$\lim_{x \to \infty} \frac{\text{d} f(x)}{\text{d} x} = 0$$</span></li>
</ol>

<p>You can easily manufacture similar functions. I just combined two strictly concave functions at a point of tangency and made sure that the piece going to zero would cross the origin.</p>

<hr>

<p><a href="https://economics.stackexchange.com/questions/33859/combining-functions-that-satisfy-the-inada-conditions">I am not sure</a>, but it seems to me that if you have an Inada satisfying function <span class="math-container">$h(\mathbf{x})$</span> then the function <span class="math-container">$f(h(\mathbf{x}))$</span> will also satisfy all the conditions.</p>

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2020-02-04T09:40:54.290Z
(+1) Thanks. Yes, I forgot to add that the Cobb-Douglas production function must be CRS to satisfy the Inada conditions. Do you know of any nice functions that are used in the literature and that satisfy the Inada conditions? It should be possible to construct functions satisfying Inada conditions by taking linear combinations of CRS Cobb-Douglas production funtions too.
Elias
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2020-02-04T09:42:33.350Z
Sorry, I am not a macroeconomist :)
Giskard
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2020-02-04T09:42:49.670Z
P.s. It occurs the me that the Hessian of $f$ is not defined at $x=1$. One could also "smooth" this over, but I do not want to tinker with the formula further. The idea is just to draw a strictly concave function that satisfies 1,3, an 4.
Giskard
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2020-02-04T09:43:48.493Z
Exactly. One can simply fill in the discontinuities.
Elias
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2020-02-04T20:23:16.650Z
Wouldn't the Hessian be negative *definite* if $f$ is strictly concave and be negative semidefinite if $f$ is concave?
Brennan
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2020-02-04T20:24:03.457Z
@Brennan It is my understanding that negative definite is a subset of negative semidefinite? My point was that Inada needs the Hessian to be negative semidefinite, which is fulfilled due to $f$ being strictly concave and hence having a negative definite Hessian.
Giskard
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2020-02-04T20:26:42.607Z
Yes that is true, I just figured maybe to be more specific about the relationship between negative definiteness and strict/non-strict concavity in case other readers are not too familiar. Ah ok that makes much more sense, the less strict requirement is sufficient
Brennan
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2020-02-04T09:25:21.073Z
2020-02-04T09:25:21.073Z
Giskard
<p><a href="https://en.wikipedia.org/wiki/Inada_conditions" rel="nofollow noreferrer">It is known that in the class of CES production functions, only the Cobb-Douglas production function satisfies the Inada conditions.</a> Which other functions satisfy the Inada conditions? </p>

<macroeconomics><production-function>
List of production functions that satisfy the Inada conditions
Elias
# List of production functions that satisfy the Inada conditions

(Asked by: username_0 on 2020-02-01)

[It is known that in the class of CES production functions, only the Cobb-Douglas production function satisfies the Inada conditions.](https://en.wikipedia.org/wiki/Inada_conditions) Which other functions satisfy the Inada conditions?

**Tags:** macroeconomics, production-function


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## 3 answers

### Answer by username_3 on 2020-02-04. Score: 5 | Accepted answer

The Wikipedia statement should come with a citation or proof, as it is not entirely accurate: Cobb-Douglas functions with increasing returns to scale do not necessarily satisfy the Inada conditions, e.g., $f(x\_1,x\_2) = x\_1^2x\_2^2$ does not.

---

There are many non Cobb-Douglas functions that satisfy the Inada conditions, albeit they may not have nice compact formulas. E.g.,
$$
f(x) = \left\{\begin{array}{ll}
x^{1/2} & \mbox{ if } x \leq 1 \\
2x^{1/4} - 1 & \mbox{ if } 1 < x
\end{array}\right.
$$
This satisfies the conditions:

1. $f(0) = 0$
2. The Hessian is negative semidefinite as $f$ is strictly concave.
3. $$\lim\_{x \to 0}\frac{\text{d} f(x)}{\text{d} x} = \infty$$
4. $$\lim\_{x \to \infty} \frac{\text{d} f(x)}{\text{d} x} = 0$$

You can easily manufacture similar functions. I just combined two strictly concave functions at a point of tangency and made sure that the piece going to zero would cross the origin.

---

[I am not sure](https://economics.stackexchange.com/questions/33859/combining-functions-that-satisfy-the-inada-conditions), but it seems to me that if you have an Inada satisfying function $h(\mathbf{x})$ then the function $f(h(\mathbf{x}))$ will also satisfy all the conditions.

> **Comments:**
> **username_0**: (+1) Thanks. Yes, I forgot to add that the Cobb-Douglas production function must be CRS to satisfy the Inada conditions. Do you know of any nice functions that are used in the literature and that satisfy the Inada conditions? It should be possible to construct functions satisfying Inada conditions by taking linear combinations of CRS Cobb-Douglas production funtions too.  
> **username_3**: Sorry, I am not a macroeconomist :)  
> **username_3**: P.s. It occurs the me that the Hessian of $f$ is not defined at $x=1$. One could also "smooth" this over, but I do not want to tinker with the formula further. The idea is just to draw a strictly concave function that satisfies 1,3, an 4.  
> **username_0**: Exactly. One can simply fill in the discontinuities.  
> **username_4**: Wouldn't the Hessian be negative *definite* if $f$ is strictly concave and be negative semidefinite if $f$ is concave?  
> **username_3**: @username_4 It is my understanding that negative definite is a subset of negative semidefinite? My point was that Inada needs the Hessian to be negative semidefinite, which is fulfilled due to $f$ being strictly concave and hence having a negative definite Hessian.  
> **username_4**: Yes that is true, I just figured maybe to be more specific about the relationship between negative definiteness and strict/non-strict concavity in case other readers are not too familiar. Ah ok that makes much more sense, the less strict requirement is sufficient  

---

### Answer by username_1 on 2020-02-06. Score: 2

The power family of production functions satisfies Inada's conditions over a set of admissible parameter values.
In the case with two inputs, the functional form is given by:

$$ y = \alpha\_{11} x\_1^{\gamma\_1} + \alpha\_{22} x\_2^{\gamma\_2} + \alpha\_{12} x\_1^{\gamma\_1/2}x\_2^{\gamma\_2/2}, $$

The power production function $f$ satisfies $f(0)=0$, $f(x) \geq 0$ and is increasing in each $x\_i$ if all $\alpha\_{ij}>0$ and $ \gamma\_{i}>0 $.
It is concave in $x$ if in addition all $\gamma\_{i}<1$, as a sum of concave functions.

The partial derivatives are:
$$ \frac{\partial f}{\partial x\_i}( x ) = \frac{ \alpha\_{ii} \gamma\_i }{ x\_i^{1-\gamma\_i} } + \frac{ \alpha\_{ij} \gamma\_i/2 }{ x\_i^{1-\gamma\_i/2} }{x\_j^{\gamma\_j/2}} $$
The partial derivatives' limiting behavior are compatible with Inada's conditions:
$$ \lim\_{x\_{i}\rightarrow 0}\frac{\partial f}{\partial x\_{i}}\left( x\right)
=+\infty ,\qquad \lim\_{x\_{i}\rightarrow +\infty }\frac{\partial f}{\partial
x\_{i}}\left( x\right) =0. $$

This power production function can be generalized to the case of $J$ inputs as follows:
$$ y = h( \sum\_{i=1}^J \sum\_{j=1}^J \alpha\_{ij} x\_i^{\gamma\_i/2}x\_j^{\gamma\_j/2} ), $$
where function $h$ satisfies $h(0)=0$, and is increasing and concave.
See for instance Diewert (1971) for a contribution with $\gamma\_i=\gamma\_j=1$.
A similar argumentation should apply to the Box-Cox family of production functions.

Diewert, W. E., 1971, "An Application of the Shephard Duality Theorem: A Generalized Leontief Production Function," *Journal of Political Economy*, 79, 481-507.


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### Answer by username_2 on 2020-02-04. Score: 1

I believe this question is too broad. As discussion in the comments indicates, even functional forms that are commonly used for their ability to satisfy the Inada conditions can fail to satisfy them for allowed values of their parameters. On the other hand, as also noted, one can always define a piecewise-continuous function that satisfies them.

With that said, it's been shown that every Inada-satisfying function is asymptotically Cobb-Douglas. <https://www.sciencedirect.com/science/article/abs/pii/S0165176503002180>


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