<p>⪰ is a strictly convex preference relation on a set X which is the set of all N-tuples of nonnegative real numbers. Further are x,y,z element of this set X and i have the preferences x ≻ y ≻ z given.</p>
<p>How can i show that there exists a unique t ∈ R with y = tx + (1 − t)z?</p>
<p>My approach is the following. I have a convex linear combination for y which lies between x and z, which is also represented in the preferences. I guess this is true for every t ∈ (0,1). So in between this open interval there is a unique solution. But how can i show that in detail?</p>

CC BY-SA 4.0
2023-12-11T14:47:46.343Z
Strictly convex preferences only make sense on a convex set. This rules out the natural numbers. And the result will generally not hold on convex domains either. The lexicographic ordering on $\mathbb{R}^2$ is strictly convex but violates your condition.
Michael Greinecker
CC BY-SA 4.0
2023-12-13T17:24:26.113Z
So that ⪰ is a striclty convex preference relation on X = (R_+)^N does not make sense?
mediation_boy
CC BY-SA 4.0
2023-12-13T17:25:53.550Z
It does, but not on the set of natural numbers.
Michael Greinecker
CC BY-SA 4.0
2023-12-13T17:27:03.030Z
does (R_+)^N mean all positive real numbers or all natural numbers?
mediation_boy
CC BY-SA 4.0
2023-12-13T17:38:40.440Z
Neither. It is the set of all N-tuples of nonnegative real numbers.
Michael Greinecker
CC BY-SA 4.0
2023-12-13T18:23:10.293Z
okay thanks, so i summarized the task wrong. If i deal now with this kind of set, is there a possiblity to show that y = tx + (1-t) z for a unique t element of R?
mediation_boy
CC BY-SA 4.0
2023-12-13T21:55:11.550Z
No, I gave a counterexample in my first comment. You can show it under the assumption that preferences are continuous.
Michael Greinecker
CC BY-SA 4.0
2023-12-13T23:49:57.910Z
Actually, continuity is not enough either.
Michael Greinecker
CC BY-SA 4.0
2023-12-31T22:47:20.757Z
Couldn't you prove it as follows? Let t and t' so that y=tx+(1-t)z and also y=t'x+(1-t')z. It follows that (t-t')x=(t'-t)z which is possible only if t=t'. Hence, t is unique.
Thomas Favory
<microeconomics><preferences>
Proof for convex preference relation
mediation_boy
# Proof for convex preference relation

(Asked by: username_0 on 2023-12-11)

⪰ is a strictly convex preference relation on a set X which is the set of all N-tuples of nonnegative real numbers. Further are x,y,z element of this set X and i have the preferences x ≻ y ≻ z given.

How can i show that there exists a unique t ∈ R with y = tx + (1 − t)z?

My approach is the following. I have a convex linear combination for y which lies between x and z, which is also represented in the preferences. I guess this is true for every t ∈ (0,1). So in between this open interval there is a unique solution. But how can i show that in detail?

**Tags:** microeconomics, preferences
> **Comments:**
> **username_1**: Strictly convex preferences only make sense on a convex set. This rules out the natural numbers. And the result will generally not hold on convex domains either. The lexicographic ordering on $\mathbb{R}^2$ is strictly convex but violates your condition.  
> **username_0**: So that ⪰ is a striclty convex preference relation on X = (R_+)^N does not make sense?  
> **username_1**: It does, but not on the set of natural numbers.  
> **username_0**: does (R_+)^N mean all positive real numbers or all natural numbers?  
> **username_1**: Neither. It is the set of all N-tuples of nonnegative real numbers.  
> **username_0**: okay thanks, so i summarized the task wrong. If i deal now with this kind of set, is there a possiblity to show that y = tx + (1-t) z for a unique t element of R?  
> **username_1**: No, I gave a counterexample in my first comment. You can show it under the assumption that preferences are continuous.  
> **username_1**: Actually, continuity is not enough either.  
> **username_2**: Couldn't you prove it as follows? Let t and t' so that y=tx+(1-t)z and also y=t'x+(1-t')z. It follows that (t-t')x=(t'-t)z which is possible only if t=t'. Hence, t is unique.  

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## 0 answers

