<p>If the function <span class="math-container">$ f(x,y) = (x-\alpha)^\beta \cdot y^{1-\beta} $</span> is homogeneous of degree <span class="math-container">$ k $</span>, then it must be true that
 <span class="math-container">$ f(\lambda x, \lambda y) = \lambda^k \cdot f(x,y) $</span> for <span class="math-container">$ \lambda \in \mathbb{R} $</span> and some constant <span class="math-container">$ k $</span>. But see that:</p>

<p><span class="math-container">$ f(\lambda x, \lambda y) = (\lambda x-\alpha)^\beta \cdot (\lambda y)^{1-\beta} \neq \lambda^k \cdot (x-\alpha)^\beta \cdot y^{1-\beta} $</span> </p>

<p>unless it should happen that <span class="math-container">$ \alpha = 0 $</span>. In that case, the function will be homogeneous of degree one.</p>

<p>Kind regards,
Pedro.</p>

CC BY-SA 4.0
2019-06-03T00:06:27.050Z
2019-06-03T00:06:27.050Z
Pedro Cunha
<p>I dont know how to demonstrate homogeneity degree of this function</p>

<p><span class="math-container">$(X-\alpha)^{\beta}(Y)^{1-\beta}$</span></p>

<p>Any idea?</p>

<p>Thanks.</p>

CC BY-SA 4.0
2019-06-03T16:37:59.920Z
Probably you will only be able to put some bounds and state whether it shows decreasing/constant/increasing returns to scale.  This type of function is sometimes called quasi-homothetic, so maybe you can find something on that.
Regio
<consumer-theory><producer-theory>
How can i determine the homogeneity degree of Stone Gaery function?
ESC
# How can i determine the homogeneity degree of Stone Gaery function?

(Asked by: username_0 on 2019-06-02)

I dont know how to demonstrate homogeneity degree of this function

$(X-\alpha)^{\beta}(Y)^{1-\beta}$

Any idea?

Thanks.

**Tags:** consumer-theory, producer-theory
> **Comments:**
> **username_1**: Probably you will only be able to put some bounds and state whether it shows decreasing/constant/increasing returns to scale.  This type of function is sometimes called quasi-homothetic, so maybe you can find something on that.  

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## 1 answers

### Answer by username_2 on 2019-06-03. Score: 1 | Accepted answer

If the function $ f(x,y) = (x-\alpha)^\beta \cdot y^{1-\beta} $ is homogeneous of degree $ k $, then it must be true that
$ f(\lambda x, \lambda y) = \lambda^k \cdot f(x,y) $ for $ \lambda \in \mathbb{R} $ and some constant $ k $. But see that:

$ f(\lambda x, \lambda y) = (\lambda x-\alpha)^\beta \cdot (\lambda y)^{1-\beta} \neq \lambda^k \cdot (x-\alpha)^\beta \cdot y^{1-\beta} $

unless it should happen that $ \alpha = 0 $. In that case, the function will be homogeneous of degree one.

Kind regards,
Pedro.


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