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.

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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2 Answers 2


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.

  • $\begingroup$ Thank you so much professor :) $\endgroup$
    – studentp
    Sep 12, 2022 at 2:16
  • $\begingroup$ 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. $\endgroup$ Oct 15, 2022 at 8:58
  • $\begingroup$ It does not satisfy condition A because $1 R (-1)$, but for $\lambda = \frac{1}{2}$ we get $\neg 0 R (-1)$. $\endgroup$
    – Amit
    Oct 15, 2022 at 9:32
  • $\begingroup$ EDIT: a strict preference relation always implies a weak preference relation, by definition. $\endgroup$ Oct 15, 2022 at 9:32
  • $\begingroup$ 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. $\endgroup$ Oct 15, 2022 at 11:41

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.

  • $\begingroup$ 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) ? economics.stackexchange.com/questions/52557/… $\endgroup$
    – studentp
    Oct 18, 2022 at 14:01
  • $\begingroup$ I'll try tonight, I need to work, sorry, but I promise I'll have a look. $\endgroup$ Oct 18, 2022 at 14:09
  • $\begingroup$ I see, I will be very happy. Many thanks:) $\endgroup$
    – studentp
    Oct 18, 2022 at 14:40

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