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We say $\succsim$ represents weak monotonic preferences if $$x,y \in X, \,\, y >> x \implies y \succ x $$

where $y >> x$ means that every element of $y$ is greater than every element of $x$.

And we say $\succsim$ represents strong monotonic preferences if $$x,y \in X, \,\, y \geq x, y \neq x \implies y \succ x$$

where $y \geq x$ means that at least one element of $y$ is greater than an element of $x$ and all others are equal.

My question is: do strong monotonic preferences imply weak monotonic preferences?

My answer: yes. The reasoning is as follows:

Since the set $A = \{ x,y \in X: y >> x \}$ is a subset of $B = \{x,y \in X: y \geq x, y \neq x \}$, then if $B \implies y \succ x$ it must also be true that $A \implies y \succ x$ and thus strong monotonic preferences imply weak monotonic preferences.

In plain English, if a bundle with more of one commodity and the same of all others is preferred, then a bundle with more of every commodity must also be preferred.

Is my reasoning correct?

Thanks!

Pedro Cunha
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  • Thanks for the input, @1muflon1. I don't think I made myself very clear with regards to the math. I think we can agree that the "plain English" explanation is correct, right? What I meant to say with the math is exactly the same. If $B \implies y \succ x$, that is, if having more of one commodity and the same of others implies that $y \succ x$, then it must also be true that having more of all commodities, that is, $A$, also implies $y \succ x$. I'm not saying $B \implies A$, I'm saying that if $B \implies y \succ x$ than it must also be true that $A \implies y \succ x$, by their design. – Pedro Cunha May 13 '20 at 11:34
  • When you say "if you say that strong monotonic preferences imply weak then all results for weak monotonic preferences should apply to strong ". That's not true, because weak monotonicity requires a stronger condition than strong monotonicity, that is for all elements of a bundle to be greater than the elements of another bundle. Thus, if strong monotonicity happens, there's no guarantee weak monotonicity will also happen. Take the sets $A = (2,3)$ and $B = (3,3)$ and $C=(3,4)$. Then if s.m is valid, $B \succ A, C \succ A, C \succ B$. However, if only w.m is valid $C \succ A$ is all you can say – Pedro Cunha May 13 '20 at 11:48
  • But that is precisely what I'm trying to prove. If strong monotonicity is valid, then weak monotonicity must also be valid. Again, referring to the "plain English" explanation: if having more of one of the goods implies that a bundle is preferred to another, then it must also be true that having more of all goods also implies that a bundle if preferred to another. If having more of something is better, then having more of all things is also better, considering, of course, that all commodities are "goods". What do you think? Again, I really appreciate the back and forth! – Pedro Cunha May 13 '20 at 12:09
  • Let us continue this discussion in chat. – Pedro Cunha May 13 '20 at 12:13
  • "If strong monotonicity is valid then bundle A is unambiguously better than bundle B and hence any rational decision maker would not hesitate picking A, but in weak monotonicity the bundle A is better only in some way than bundle B so it is no longer unambiguously the best choice and hence rational decision maker might hesitate here.". This depends on the bundles A and B. If s.m. is valid, then $A \succ B$ only if A has more of at least one of the goods. In much the same way, if w.m. is valid, then $A \succ B$ only if A has more of all of the goods. If s.m. is valid and we have a bundle with.. – Pedro Cunha May 13 '20 at 12:22
  • more of all goods, that this bundle will also be preferred, because it also has more of one of the goods (it has more of all goods). And thus, if a bundle with more of all goods is preferred, then the s.m. is also valid. – Pedro Cunha May 13 '20 at 12:24
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    The proof seems fine, albeit could be cleaner. – Walrasian Auctioneer May 13 '20 at 17:41
  • I'd appreciate any tips on how to improve it, @WalrasianAuctioneer – Pedro Cunha May 13 '20 at 19:07
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    @1muflon1: "Does it make sense to say that a natural number implies real number? I don’t think so in my mind those are two different concepts, but I might be wrong." Yes, if $x$ is a natural number, then $x$ is a real number. Yes you are wrong. –  May 14 '20 at 02:57
  • @KennyLJ thanks for clarifying – 1muflon1 May 14 '20 at 03:01
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    @PedroCunha Kenny's answer was the cleaner more direct proof I would have written! – Walrasian Auctioneer May 14 '20 at 04:24

1 Answers1

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Let $x,y\in X$. Suppose $y \gg x$ (so that in particular, $y\geq x$ and $y\neq x$). By assumption, $\succsim$ satisfies strong monotonicity; and so, $y\succ x$.

(We've just shown that for any $x,y\in X$ such that $y \gg x$, we have $y\succ x$. Hence, $\succsim$ satisfies weak monotonicity.)

The argument you've given above is correct and roughly the same as the proof I've just given. The only problem is that your argument is somewhat indirect, convoluted, and unclear.

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    I'm not sure what your math/econ background is, but it is probably worthwhile reviewing the basics of proof techniques. –  May 14 '20 at 02:51
  • I'm working on it. I'm taking my first 'proof-based' course now. And also it doesn't help that I'm not a native english speaker. Thank you very much for the proof. – Pedro Cunha May 14 '20 at 03:56