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There are a number of different definitions of Homothetic functions i have come across. I have used each of them to prove that a function $f(x, y) = x^a y^b$ with $a+b > 0$ is homothetic. But i have some questions.

  • Question 1): Compared to definition one, why the stricter conditions on definition two and three, if definition one is so simple?
  • Question 2): Why are Definitions two and three equivalent despite subtle differences?
  • Question3): An exam question used $f$ as a production function i.e. $x = k$ Capital, $y = l$ Labour. The answer then used definition 3 to prove it was homothetic. While not complicated, it seems the most complicated out of the alternatives, can anyone give insight into why this approach might have been used? is there anything special about the function as a production function? Is defintion 3 a tighter definition?

Definition 1 Source: My Mathematical Economics course pack

  • Any homogeneous function of degree $d>0$ is homothetic .

$f(x,y) = x^a y^b$ which is homogenous to degree $a+b > 0$ and is therefore homothetic.

Definition 2 Source:

  • A function $f$ is homothetic if:
  • $f(x,y) = q(r(x,y))$
  1. $r$ is a homogenous function of any degree
  2. $q$ is monotonically increasing (Which i believe means non-decreasing)

$f(x,y) = h(r(x,y))= (x^{a/2}y^{b/2})^2$ Where i have assumed $x,y >0$ and so $h^2$ is monotonically increasing $r$ is homogenous to degree

Definition 3 Source: My Mathematical Economics course pack

  • A function $f: \mathbb{R^{n+}} \to \mathbb{R}$ is homotheic if it has the form:
  • $f(x,y) = q(r(x,y))$
  1. Where $r$ is a function that is homogenous of degree 1
  2. $q$ is strictly increasing

$f(x,y) = h(r(x,y))= (x^qy^p)^{a+b}$ Where $q = \frac{a}{a+b}$ $p+\frac{b}{a+b}$ Hence we have written $f$ in terms of a monotonic transformation of a homogenous to degree $d = 1$ function $r$.

Question 3): This was the technique used in an exam answer scheme, where $x = k$ Capital, $y = l$ Labour, therefore we can assume $x\ge0$ and $y\ge0$ I don't understand why they would use all this when surely just the simple first definition would suffice? Noting that the $f$ is homogenous of degree d > 0. Or at least definition 4 which is super simple? Is there any reason one of these other approaches would have been invalid?

Definition 4 Source:

A function $f: \mathbb{R}^n_+ \to \mathbb{R}$ is homothetic if its marginal rate of substitution (MRS) between any two variables is constant along rays from the origin, i.e., the MRS depends only on the ratio of the variables. I.e. MRS is homogeneous of degree 0.

Using $x = k$ Capital, $y = l$ Labour

$MRTS(k,l) = \frac{al}{bk}$ which is clearly homogenous of degree 0.

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  • $\begingroup$ @BakerStreet you should have left it as homeopathic functions hahah $\endgroup$
    – CormJack
    Commented Apr 23, 2023 at 13:29
  • $\begingroup$ Yes, it was an original, interesting notion! $\endgroup$ Commented Apr 23, 2023 at 13:36
  • $\begingroup$ No worries, @solowsupremacy the idea of my post becoming a messaging board sounds fun!. I to am often in a constant battle to steal BakerStreets time hahah lol $\endgroup$
    – CormJack
    Commented Apr 23, 2023 at 13:50
  • $\begingroup$ @solowsupremacy pretty sure him and a few others are like the guardians of the Econ Stack exchange. They should start a Patreon hahahah $\endgroup$
    – CormJack
    Commented Apr 23, 2023 at 13:52
  • $\begingroup$ @ solow supremacy I would be glad to check your question, but I don't know if I can during these days, I'm very busy and I'm away from home. I will read it as soon as possible, if I've time. $\endgroup$ Commented Apr 23, 2023 at 13:53

1 Answer 1

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Question 1.

Definition 1 is not exactly a definition. $u(x) = \ln(x) + 3$ is homothetic but not homogenous of degree $1$. Verify using the second definition.

Question 2.

Definition 2 requires strict monotonicity. Otherwise, if $r(x,y) = x^\alpha y^{1-\alpha}$ and $q(x) = 5$, you lose both the utility function and the ordering.

Here's an image of Varian's definition:

enter image description here

If you see, he has defined a monotonic function function as a positive and strictly increasing function.

Question 3.

Because the third is the correct definition.

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  • $\begingroup$ Thanks for this Solow! Although I'm unclear on some of your points: R.e. definition one? Are you saying it's wrong, I believe the "definition" is trying to say that a homogenous function of degree d > 0 is nomothetic, but not necessarily the other way around? I'm also not sure why you have referenced homogenous of degree one, in relation to definition one? Apologies for my lack of understating here. $\endgroup$
    – CormJack
    Commented Apr 23, 2023 at 14:09
  • $\begingroup$ Regarding definition 2. I am also confused because the source I sited clearly states any degree and monotonicty? I have also found a 5th definition which is a hybrid between definition 2 and 3. Written by Varian, see this post: economics.stackexchange.com/questions/55179/… Edit: I have just seen you have posted the same definition. Let me double check it. $\endgroup$
    – CormJack
    Commented Apr 23, 2023 at 14:13
  • $\begingroup$ okay so this is what confused me in his definition. He talks about positive monotonic transformation, where g is strictly increasing. But then at the bottom he says g is monotonic. And monotonic doesn't necessarily mean strictly increasing? Is he just assuming g is strictly increasing or not? And it's still not clear to me why my course pack has also that super simple definition one that I mentioned (as well as this more intricate definition). $\endgroup$
    – CormJack
    Commented Apr 23, 2023 at 14:17
  • $\begingroup$ @CormJack Comment 1: The statement in "definition 1" isn't wrong, but it isn't a definition. Yes, the converse doesn't hold (which is necessary for a definition). Comment 3: In economics, it is assumed that monotone refers to positive and strictly monotone functions. You might have even studied in the initial years that strict monotonicity is required to preserve the order. Look at my answer to question 2. You'll see that $q(r)$ no more describes the same preferences as $r$. $\endgroup$
    – user43302
    Commented Apr 23, 2023 at 15:57

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