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I understand that Cobb-Douglas preferences represented by $U(x,y)=x^ay^b$ are strictly monotonic, because increasing at least one of the goods in the bundle increases utility.

However, another definition of strict monotonicity says that marginal utility of each good should be strictly positive. This is not the case with the above function at $(0,0)$. Is there a way to resolve this?

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  • $\begingroup$ The definitions are not always consistent (and I usually hear about weak or strong monotonicity), can you please larify what exactly you mean by strict monotonicity? $\endgroup$ – Giskard Jul 9 at 9:59
  • $\begingroup$ I mean strong monotonicity when I say strict monotonicity. I have seen strong and strict being used interchangeably. $\endgroup$ – PGupta Jul 9 at 10:56
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Cobb-Douglas preferences are strongly monotonic over the positive part of the space of baskets, in this case $\mathbb{R}_{++}^2$.

Leontief preferences are the usual example for weakly but not strongly monotonic preferences. The indifference curve passing through (0,0) is L-shaped for both these and for Cobb-Douglas preferences.


Cobb-Douglass preferences and the boundary of the positive quadrant are problematic for other reasons, as another usual utility representation is $$ U(x,y) = a \ln x + b \ln y $$ which is undefined (over real numbers) when either $x$ or $y$ is 0.

However for the Cobb-Douglas case you can prove that the optimal choice of the consumer (assuming positive income) is never on the boundary, as that yields the lowest possible utility. After this, assuming the consumer makes optimal choices, the utility function is strongly monotonic in the local environment of her choice.

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  • $\begingroup$ Thanks, so to check for strong monotonicity, we can exclude the origin from consideration? Is this a general rule or specific to this utility function? $\endgroup$ – PGupta Jul 9 at 10:57
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    $\begingroup$ @PGupta I am not saying this. My exact statement is "Cobb-Douglas preferences are strongly monotonic over the positive part". This the most you can claim. The preferences are not strongly monotonic over the boundary. $\endgroup$ – Giskard Jul 9 at 11:08
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    $\begingroup$ @PGupta I will add that this is enough for some types of results. For the Cobb-Douglas case you can prove that the optimal choice (assuming positive income) is never on the boundary, as that yields the lowest possible utility. After this, assuming the consumer makes optimal choices, the utility function is strongly monotonic in the local environment of her choice. $\endgroup$ – Giskard Jul 9 at 11:37
  • $\begingroup$ Makes sense, thank you. $\endgroup$ – PGupta Jul 9 at 17:39
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They do not satisfy either condition. An increase in the quantity of one good need not increase utility; $U(0,1)=0^a1^b=0=0^a2^b=U(0,2)$.

The condition that all marginal utilities must be positive is inherently problematic because it does depend not just on the underlying preferences. For one, not every utility representation needs to be differentiable. But even that is not enough. Suppose there is a single good and more is better than less. You can represent these preferences by the utility function given by $U(x)=x$. Now $U'(x)=1$ for all $x$, so these preferences are supposedly strictly monotone. But now consider the utility function given by $V(x)=(x-1)^3$. If $x'>x$, then $V(x')>V(x)$, more is better and $V$ represents the same preferences as $U$. But $V'(1)=0$, so the preferences are not strictly monotone according to the definition in terms of marginal utilities.

The problem just mentioned is actually even worse. Let $U$ be a differentiable utility function on $\mathbb{R}_+^2$ with strictly positive partial derivatives everywhere. Let $(x^*,y^*)\gg0$be any commodity bundle. Define $V$ by $$V(x,y)=\big(U(x,y)-U(x^*,y^*)\big)^3.$$ $V$ represents the same preferences as $U$ but the partial derivatives at $(x^*,y^*)$ are zero.

Long story short: Every partial derivative being strictly positive is a sufficient but not necessary condition for a differentiable function to be increasing in every coordinate.

Yet another issue is that it is not entirely clear how one defines the derivative at the boundary of the commodity space. There are different notions of differentiability which need not be equivalent.

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When $\alpha,\beta\in(0,1)$, you cant use the derivative to check for monotonicity - simply because the derivative doesnt exist at 0.

$\frac{\partial u(x,y)}{\partial x} = \alpha\frac{y^{\beta}}{x^{1-\alpha}}\rightarrow\infty$ as $x\rightarrow 0$. Similarly for $MU_y$ doesnt exist when $y\rightarrow 0$.

This is easily seen from @Giskard's answer: as he rightly points out, the log transformation of the CD-utility is not defined when either coordinate is 0. In such situations, you should derive using first principles.

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