When our Isoquant map exhibits constant MRTS along a ray from the origin making. Why do we make specific reference to.

  1. Constant returns to scale
  2. Homothetic Functions

I'm asking because it seems to me that any homogenous function is satisfactory i.e. according to my text, for a homogenous function of degree $d>0$ $MRTS$ will depend only on the ratio of it's input. E.g. Looking at our standard cobb-duglas we get $MRTS_{l,k} = \frac{βk}{αl}$ This seems to depend only on the ratio of $k$ & $l$, for whatever α & β chosen, which depending on choice can give DRS, CRS, or ICS.


My book "intermediate Microeconomic Extensions" - Walter Nicholson (A great book btw which i really love). Is emphasising the role of homogenous function of degree one (CRS), as giving us Isoquant maps where MRTS are constant along a ray.

I'm confused why CRS and not just homogeneity greater than 0 are being emphasised here. For example i don't get the significance of the following:

Because A CRS production function is homothetic, the RTS depends only on the ratio of $k$ to $l4 not on the scale of production.

Why doesn't it just read. 'Because the function is homogenous to a degree greater than 0, RTS depends only on the ratio of $k$ to $l$ not on the scale of production'

My understanding is as follows:

  1. Any functions homogenous to degree $d > 0$ will be Homothetic.
  2. Any function $f$ that satisfies the following is homothetic.
  • It can be written as $f(x,y) = q(r(x,y))$
  • Where $r$ is a function that is homogenous of degree 1
  • $q$ is strictly increasing.
  • This makes sense, and i can prove it.
  1. The ratio of the partial derivatives e.g. the MRTS, of any homogenous function can be shown to depend only on the ratio of inputs and not their absolute value. I'm happy with the proof of this.

Question recap: For our constant MRTS along a ray from the origin why do we care about:

  1. CRS
  2. Homothetic functions
  3. I think Im generally missing the added value of a homothetic function vs homogenous function.
  • $\begingroup$ not any function can be written as in your point 2. This is only possible for homothetic functions. $\endgroup$
    – Bertrand
    May 20, 2023 at 12:06
  • $\begingroup$ Hahaha no no I meant. Any function that can be written like this…is homothetic (: exactly as you have said. My grammer must be off let me try and adjust, thanks for pointing it out $\endgroup$
    – CormJack
    May 20, 2023 at 12:17

1 Answer 1

  1. I think I’m generally missing the added value of a homothetic function vs homogenous function.

The added value of homothetic functions is that they are a wider class with respect to homogeneous functions.

So that one can work under less restrictive assumptions, preserving some important properties.

Indeed, homothetic functions can be considered a generalization of homogeneous functions.

It can be proved that homogeneity $\implies$ homotheticity.

But the converse is not true, homothetic functions constitute a wider class, that is there are many homothetic functions that are not homogeneous.

Consider for example the function

$$f(x,y)= \frac {1}{2} \ln(x)+ \frac {1}{2} \ln (y)= \ln(x^{\frac {1}{2}}y^{\frac {1}{2}})\qquad (1)$$

It is evidently written as $f(x,y) = q(r(x,y))$ where $r(x,y)=x^{\frac {1}{2}}y^{\frac {1}{2}}$ is homogenous of degree $1$ , and $q$ (the logaritm) is strictly increasing.

But calculations show that it is not homogeneous. Multiplying $x$ and $y$ by $t>0$ we get

$$\ln((tx)^{\frac {1}{2}}(ty)^{\frac {1}{2}}) = \ln (t x^{\frac {1}{2}} y^{\frac {1}{2}})=\ln(t) + \ln (x^{\frac {1}{2}}y^{\frac {1}{2}})\neq t \ln (x^{\frac {1}{2}}y^{\frac {1}{2}}) \qquad (2)$$


Homothetic functions inherit some features of homogeneous functions, in particular the property that along the lines passing through the origin the marginal rate of substitution is constant. And it has been proved that the converse is also true, that is homothetic functions are the unique class of functions that have this property.

Ide and Takayama, On homothetic functions discusses the issue and gives a proof:

In economic analysis, the importance of the homotheticity of production functions (or utility functions), which is due to Shepard (1953), has been well recognized. Its important feature lies in the fact that every expansion path is a ray from the origin if and only if the underlying production (or utility) function is homothetic. Although the proof of the “if” part of this statement is easy, the proof of the “only if” part, at least as it appears in the literature, is not easy […] It would thus be desirable to obtain an elementary, short, alternative proof of the above important proposition. […]In this paper we prove both the “only if” and the “if” part simultaneously.(emphasis mine)

But homothetic functions lose some other properties of homogeneous functions.

For instance, they lose property exemplified in $(2)$, and lose the Euler's Theorem for homogeneous functions:

Let ${\displaystyle f\colon A\rightarrow \mathbb {R^n} }$ a differentiable function defined on an open cone $A\subset \mathbb{R} ^{n}$. Then $f$ is homogeneous of degree $k$ on $A$ if and only if the following identity holds (Euler's identity):

$${\displaystyle \sum _{i=1}^{n}{\frac {\partial f(x)}{\partial x_{i}}}\ x_{i}=k\ f(x),\quad \forall x\in A,}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.