# Strictly Convex and Strictly Monotonic Preferences

On the assumptions for a well behaved preference, I read recently that a preference must be complete, transitive, continuous, strictly monotonous (if the vectors x≥y and x≠y, then x is strictly preferable to y) or sometimes just monotonous (if x>>y, then x is strictly preferable to y) and strictly convex. But my doubt now follows: if you assume strict convexity (any linear combination with a different bundle in the same indifference curve will be strictly preferable) for a preference, it seems to me that any degree of monotonicity will be strict -- right? Am I missing something?

• What you call "strictly monotonous" should instead read "strongly monotonic." Preferences that are strongly monotonic state that when $x \geq y$ and $x \neq y$, then $x \succeq y$. Preferences that are strictly monotonic when $x \geq y$, then $x \succeq y$ and when $x >> y$, $x \succ y$. – jmbejara Mar 18 '14 at 22:38
• Also, just to note, a "preference" relation is usually defined as a binary relation that is complete and transitive. The other properties are assumed so that we can derive "demand functions." We want the consumer's constrained maximization problem to produce a unique solution that is continuous in prices (among other properties). – jmbejara Mar 19 '14 at 1:18
• @jmbejara: From what I have seen, there is unfortunately no real convention in the literature on the definition of the different variants of monotonicity axioms (weak,strong,strict,...). For instance, Mas-Collel's textbook defines "strongly monotonicity" as [$x \geq y$ and $x\neq y$ implies $x \succ y$], not $x \succeq y$. Kreps' textbook calls the same property "strict monotonicity". Jehle and Reny on the other hand agree with you definition of strict monotonicity. Finally I have seen papers referring to what you call "strict monotonicity" as "monotonicity". Quite a mess... – Martin Van der Linden Mar 19 '14 at 15:10
• You're right about strong monotonicity. It is correct, however, in my solution below. However, I think there is definitely agreement between strict and strong, as the same concepts appear elsewhere and more generally in mathematics. (Check me if I'm wrong.) See wikipedia. I believe the idea is that strongly convex means that strictly increasing the quantity of at least one good strictly increases utility. Strict convexity means all goods must be increased to guarantee strictly more utility. en.wikipedia.org/wiki/Convex_function#Strongly_convex_functions – jmbejara Mar 19 '14 at 15:27
• I am not sure I follow you on "strongly convex means that strictly increasing the quantity of at least one good strictly increases utility". Do you mean "strongly monotonic"? I know nothing about the conventional use of the "strong" and "strict" qualifiers in math, so I would not dare checking you on that ;). But I do know that there is alas no agreement in the economic literature or profession at least... I have checked some other references quickly to get a better sample and there is a real confusion, strong and strict being used interchangeably to define the same axiom by different authors. – Martin Van der Linden Mar 19 '14 at 16:47

### Definitions

To clarify, preferences that are strongly monotonic state that when $x \geq y$ and $x \neq y$, then $x \succ y$. Preferences are strictly monotonic when $x \geq y$ imply $x \succeq y$ and $x>>y$ imply $x \succ y$. The difference is that strong monotonicity requires only one more of any good to be strictly preferred.

Strict convexity (of preferences) means that when $x \neq y$ and $x \succeq y$, then $t x + (1-t) y \succ y$ for all $t \in (0,1)$.

### Example with no (positive) monotonicity

Consider preferences like those depicted below. The top indifference curve (U1) is strictly less preferred than the bottom (U2). Preferences over these goods (which might better be called "bads") are strictly convex. Whenever $x \neq y$ and $x \succeq y$, then $t x + (1-t) y \succ y$ for all $t \in (0,1)$.

However, these preferences are in no degree monotonic. (In a sense, they are monotonically decreasing, but we usually take "monotonic preferences" to be increasing as in the definition I gave above.) For the right domain, preferences of this kind are indeed complete and transitive.

So, strictly convex preferences do not imply monotonicity (at least in the direction that you want).

### Monotonicity and strict convexity imply strong monotonicity

Suppose we use a definition of monotonicity where $x \geq y$ imply $x \succeq y$. Then, it is true that this form of monotonicity together with strict convexity (as defined above) imply strong monotonicity (as defined above).

To show this, let $x$ and $y$ be given such that $x \geq y$ and $x \neq y$. Then by this form of monotonicity, $x \succeq y$. This allows us to use strict convexity to say that $\forall t \in (0,1)$ we have $t x + (1-t) y \succ y$. Now, either $x \succ y$ or $x \sim y$. Suppose by way of contradiction that $x \sim y$. Then $\forall t \in (0,1)$ we have $$t x + (1-t) y \succ y \sim x.$$ But it is clear that $t x + (1-t) y \leq x$, so by monotonicity $t x + (1-t) y \preceq x$. This is a contradiction. So it must be that $x \succ y$. Thus, the preferences are strongly monotonic.

• But in your proof you're not deliberately raising every dimension of the vector? And by strict convexity tx+(1−t)y≤x is actually tx+(1−t)y<x. Right? Now I'm thinking my assumption maybe wrong. Another thing: I was thinking with the definition of Mas-Collel for monotonocity, but the document in which I was reading the definition of monotonicity uses this definition: if x>>y, then x⪰y. edit: I apologize for my lack of rigorousness in this question (and other questions before). – John Doe Mar 20 '14 at 22:11
• I am not raising every dimension of the vector. The expression $t x + (1-t) y \leq x$ is correct as it stands. $<$ would not make sense because $x$ and $y$ are vectors. $<<$ would make conform to the fact that these are vectors, but like you mentioned, I am not raising every dimension. So $\leq$ is appropriate. As for the definition, Martin has pointed out that the definitions vary a lot. So, my proof is careful to provide the definition that it relies. on. The proof would work just as well with the alternative definition that you gave, after adjusting the assumptions. – jmbejara Mar 21 '14 at 3:33

Here is an attempt to make up for my ugly mistake on the strict convexity of Leontieff preferences ;).

If I read you right, the guess you want to check is whether for complete, transitive and continuous preferences, strict convexity implies that any monotonic preference relation is also strictly monotonic. I believe your guess is right and this can be shown by contradiction (the construction of the different bundle in the proof are illustrated in the picture).

• Assume that there exists $x,y$ such that $x \geq y, x\neq y$ and $y$ is not strictly preferred to $x$.
• By monotonicity, $x_l = y_l$ for some good $l$ (otherwise $x \succ y$).

• By completeness we must have $y \succeq x$.
• As $x \neq y$ and preferences are continuous, there must exist a bundle $k$, a convex combination of $x$ and $y$, which leaves the agent indifferent with $x$, that is for some $\alpha \in [0,1)$ we have $k = \alpha x + (1-\alpha) y$ and $x \sim k$. (This step is not really needed but helps sticking to your definition of strict convexity).
• By strict convexity, there exists a convex combination $z$ of $k$ and $x$ such that $z$ is strictly preferred to $x$, that is for some $\lambda \in (0,1)$ we have $z = \lambda x + (1-\lambda) y$ and $z \succ x$.
• By continuity, strict upper contour sets are open, so there exists a neighborhood of $z$, $N_\epsilon(z)$ such that for every $m \in N_\epsilon(z)$, $m \succ x$.
• Take any $w \in N_\epsilon(z)$ such that $w << z$. By construction $w << x$, so by monotonicity we must have $x \succ w$. But by $w \in N_\epsilon(z)$, we must have $w \succ x$, a contradiction.