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According to Varian's Macroeconomic Analysis (page 15), when returns to scale are constant, production function will be homogeneous of degree 1.

When he discusses increasing and decreasing returns to scale he does not mention homogeneity of the production function again, but I wonder if we could prove that for increasing returns to scale production function will be homogeneous of degree $k>1$ and for decreasing returns to scale homogeneous of degree $0<k<1$?

Intuitively it feels like this is just corollary to the definition of increasing (decreasing) returns to scale $f(kx)>kf(x)$ ($f(kx)<kf(x)$), but I am not sure. Any clarification on this would be welcome.

HerrMaximus
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2 Answers2

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It is possible to find a relationship between both concepts, but not in the way that you mention.
Let us consider a production function that is homogeneous of degree $k$ (hdk for short), then $$f(tx)=t^kf(x)$$ for all $x$ and all $t>0.$ For a degree of homogeneity greater than one, $k>1$, we have $$f(tx)=t^kf(x)>tf(x),$$ for all $t>1$, which corresponds to the definition of IRTS given by Varian (1992, p.16). Similarly for a production function hdk with $k<1$. In conclusion: $$Hdk, k>1 \implies IRTS, $$ $$Hdk, k<1 \implies DRTS. $$ The converse is not true.

Bertrand
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    I see your point and I agree with what you wrote, but you should write it more clearly:

    "It is true that every function that is homogenous of degree $k>1$ ($k \in (0,1)$ exhibits increasing (decreasing) returns to scale), but the reverse does not necessarily hold, namely there are functions that exhibit increasing (decreasing) returns to scale but are not homogenous of degree k > 1 (k between 0 and 1). Then, the example made in the answer above this one proves this point.

     – Matteo Bulgarelli Jan 11 '24 at 13:39
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    @Matteo Bulgarelli: thanks for the clarification, it is useful. – Bertrand Jan 11 '24 at 14:08
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No. Take $f$ given by $f(x)=x+x^2$. Suppose $f$ would be homogenous of degree $k$. By Euler's homogeneous function theorem, we must have $k f(x)=x f'(x)$. Here, this means $k(x+x^2)=x+2x^2$, or

$$k=\frac{x+2x^2}{x+x^2}=\frac{1+2x}{1+x}.$$

Since the right side is strictly increasing and the left is a constant, we get a contradiction.

Michael Greinecker
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    $f(x) = \ln x$ might be an even simpler example, as there $kf(x) = xf'(x)$ is $k\ln x = x/x = 1$. – Giskard Jan 11 '24 at 08:08
  • "Suppose would be homogenous of degree ". So, you suppose something false by assumption? The function you took is not homogenous! – Matteo Bulgarelli Jan 11 '24 at 13:20
  • Even though I see your point (you are showing that a non-homogenous function can exhibit increasing returns to scale), the wording needs to be improved. – Matteo Bulgarelli Jan 11 '24 at 13:38
  • You might want to look up how proofs by contradiction work,. – Michael Greinecker Jan 11 '24 at 14:14
  • Ok, but you need to contradict a true statement and show that this results in a false statement. Here you are contradicting a FALSE statement and showing that indeed it is true that the statement is false, so you are basically doing nothing. – Matteo Bulgarelli Jan 11 '24 at 14:57
  • Or better, you are showing that f() is not homogenous, not that f() is not homogenous AND can exhibit increasing (or decreasing) returns to scale (which is the point!). – Matteo Bulgarelli Jan 11 '24 at 15:03
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    The negation of being homogenous is not being homogenous. The negation of that is being homogenous. So I use a contradiction argument to show that this specific function is not homogenous. Since this specific function is clearly (maybe not for you) satisfying increasing returns to scale, this shows that I have actually given a counterexample. – Michael Greinecker Jan 11 '24 at 15:15