<p>Condition B does not imply condition A. Consider an example of the preference relation defined on <span class="math-container">$\mathbb{R}$</span> that is represented by the following utility function:
<span class="math-container">\begin{eqnarray*} u(x)= \begin{cases} 0 &amp;\text{if } x = 0 \\ 1  &amp; \text{if } x \neq 0\end{cases}\end{eqnarray*}</span>
The preference relation <span class="math-container">$R$</span> represented by <span class="math-container">$u$</span> satisfies condition B but not condition A.</p>

CC BY-SA 4.0
2022-09-12T02:16:29.600Z
Thank you so much professor :)
studentp
CC BY-SA 4.0
2022-10-15T08:58:49.880Z
The above statement, in my humble view, is not true. The above utility function also satisfies condition A. Indeed, If $u(x'') > u(x')$,  which can hold if and only if $x' = 0$, you necessarily have that $u(\lambda x'' +(1-\lambda)x') > u(x')$, which implies condition A as well.
Since in any example where you have a strict preference in the above example you need $x=0$, I don't think that the above statement is correct.
A strict preference relation always implies a weak preference relation, if preferences are rational.
Matteo Bulgarelli
CC BY-SA 4.0
2022-10-15T09:32:14.027Z
It does not satisfy condition A because $1 R (-1)$, but for $\lambda = \frac{1}{2}$ we get $\neg 0 R (-1)$.
Amit
CC BY-SA 4.0
2022-10-15T09:32:24.173Z
EDIT: a strict preference relation always implies a weak preference relation, by definition.
Matteo Bulgarelli
CC BY-SA 4.0
2022-10-15T11:41:07.530Z
Ah, ok, you're considering also the negatives. Note, however, that you just assumed that you can represent this preference relation with an utility function, which is not in the text.
Matteo Bulgarelli
CC BY-SA 4.0
2022-09-11T23:21:53.753Z
2022-09-11T23:21:53.753Z
Amit
<p>By definition <span class="math-container">$y \succ x$</span> <em>means</em> that <span class="math-container">$y \succeq x$</span> <em>but not</em> <span class="math-container">$x \succeq y$</span></p>
<p>Thus if <span class="math-container">$y \succ x$</span>, then <span class="math-container">$y \succeq x$</span>, which is the first part of the two conditions. If <span class="math-container">$y \succ x$</span>  implies (for some random reason) that for <em>any</em> <span class="math-container">$\lambda\in(0,1)$</span>, you have that <span class="math-container">$ \lambda y + (1-\lambda)x \succ x$</span>, then, by definition of <span class="math-container">$\succ$</span>, we have that <span class="math-container">$\lambda y + (1-\lambda)x \succeq x $</span>  <em>but not</em> <span class="math-container">$x \succeq \lambda y + (1-\lambda)x$</span>. Read the proof again: I just showed that any time condition B is true, thus any time that <span class="math-container">$y \succ x$</span> and, for some random reason, for any <span class="math-container">$\lambda \in (0,1)$</span>, <span class="math-container">$ \lambda y + (1-\lambda)x \succ x$</span>, then condition A is also true.</p>
<p>Note that: 1) the problem doesn't state that <span class="math-container">$\succeq$</span> is rational (necessary but not sufficient condition for <span class="math-container">$\succeq$</span> to be represented by an utility function), 2) the problem doesn't state that <span class="math-container">$\succeq$</span> can be represented by an utility function.</p>

CC BY-SA 4.0
2022-10-18T14:01:15.567Z
Dear Matreo I have a question on PBE. Someone draw the extensive form game tree but I could not solve it. Can you please help me to do part (b) (c) ? https://economics.stackexchange.com/questions/52557/finding-pbe-in-incomplete-game
studentp
CC BY-SA 4.0
2022-10-18T14:09:41.410Z
I'll try tonight, I need to work, sorry, but I promise I'll have a look.
Matteo Bulgarelli
CC BY-SA 4.0
2022-10-18T14:40:36.663Z
I see, I will be very happy. Many thanks:)
studentp
CC BY-SA 4.0
2022-10-15T09:31:11.117Z
2022-10-15T11:55:30.843Z
2022-10-15T11:55:30.843Z
Matteo Bulgarelli
<p>Condition A:</p>
<p>Given x, y in X such that <span class="math-container">$yRx$</span> then it follows that</p>
<p><span class="math-container">$\lambda y +(1-\lambda)xRx$</span> for all <span class="math-container">$0&lt; \lambda&lt;1$</span></p>
<p>Condition B:</p>
<p>Given x, y in X such that <span class="math-container">$yPx$</span> then it follows that</p>
<p><span class="math-container">$\lambda y +(1-\lambda)xPx$</span> for all <span class="math-container">$0&lt; \lambda&lt;1$</span></p>
<p>Show that the condition B implies the condition A.</p>
<p>R refers a weak preference relation and P is a strict preference relation.</p>
<hr />
<p>I don’t understand how to show this implication.</p>
<p>What do you think? How can I show this? I am very confused.</p>
<p>This question is duplicated. I also asked on math-stack exchange website. But there, I could not get any proper answer. What do you think about my question? Thank you.</p>
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Strict preference relation implies weak preference relation
studentp
# Strict preference relation implies weak preference relation

(Asked by: username_0 on 2022-09-11)

Condition A:

Given x, y in X such that $yRx$ then it follows that

$\lambda y +(1-\lambda)xRx$ for all $0< \lambda<1$

Condition B:

Given x, y in X such that $yPx$ then it follows that

$\lambda y +(1-\lambda)xPx$ for all $0< \lambda<1$

Show that the condition B implies the condition A.

R refers a weak preference relation and P is a strict preference relation.

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I don’t understand how to show this implication.

What do you think? How can I show this? I am very confused.

This question is duplicated. I also asked on math-stack exchange website. But there, I could not get any proper answer. What do you think about my question? Thank you.

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**Tags:** microeconomics, preferences, self-study

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## 2 answers

### Answer by username_1 on 2022-09-11. Score: 2 | Accepted answer

Condition B does not imply condition A. Consider an example of the preference relation defined on $\mathbb{R}$ that is represented by the following utility function:
\begin{eqnarray\*} u(x)= \begin{cases} 0 &\text{if } x = 0 \\ 1 & \text{if } x \neq 0\end{cases}\end{eqnarray\*}
The preference relation $R$ represented by $u$ satisfies condition B but not condition A.

> **Comments:**
> **username_0**: Thank you so much professor :)  
> **username_2**: The above statement, in my humble view, is not true. The above utility function also satisfies condition A. Indeed, If $u(x'') > u(x')$,  which can hold if and only if $x' = 0$, you necessarily have that $u(\lambda x'' +(1-\lambda)x') > u(x')$, which implies condition A as well. Since in any example where you have a strict preference in the above example you need $x=0$, I don't think that the above statement is correct. A strict preference relation always implies a weak preference relation, if preferences are rational.  
> **username_1**: It does not satisfy condition A because $1 R (-1)$, but for $\lambda = \frac{1}{2}$ we get $\neg 0 R (-1)$.  
> **username_2**: EDIT: a strict preference relation always implies a weak preference relation, by definition.  
> **username_2**: Ah, ok, you're considering also the negatives. Note, however, that you just assumed that you can represent this preference relation with an utility function, which is not in the text.  

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### Answer by username_2 on 2022-10-15. Score: 1

By definition $y \succ x$ *means* that $y \succeq x$ *but not* $x \succeq y$

Thus if $y \succ x$, then $y \succeq x$, which is the first part of the two conditions. If $y \succ x$ implies (for some random reason) that for *any* $\lambda\in(0,1)$, you have that $ \lambda y + (1-\lambda)x \succ x$, then, by definition of $\succ$, we have that $\lambda y + (1-\lambda)x \succeq x $ *but not* $x \succeq \lambda y + (1-\lambda)x$. Read the proof again: I just showed that any time condition B is true, thus any time that $y \succ x$ and, for some random reason, for any $\lambda \in (0,1)$, $ \lambda y + (1-\lambda)x \succ x$, then condition A is also true.

Note that: 1) the problem doesn't state that $\succeq$ is rational (necessary but not sufficient condition for $\succeq$ to be represented by an utility function), 2) the problem doesn't state that $\succeq$ can be represented by an utility function.

> **Comments:**
> **username_0**: Dear Matreo I have a question on PBE. Someone draw the extensive form game tree but I could not solve it. Can you please help me to do part (b) (c) ? https://economics.stackexchange.com/questions/52557/finding-pbe-in-incomplete-game  
> **username_2**: I'll try tonight, I need to work, sorry, but I promise I'll have a look.  
> **username_0**: I see, I will be very happy. Many thanks:)  

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