That this utility function represents monotonic preferences, I think it's clear. Both goods have positive and constant marginal utilities. What I think is less clear is if this preference relation is transitive. How can I assess it?
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3$\begingroup$ How about applying the definitions of "utility function represents preferences" and of "transitive preferences"? $\endgroup$– GiskardOct 4, 2018 at 21:06
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$\begingroup$ A preference relation is transitive if $A \succ B, B \succ C \implies A \succ C$. So if there's $p,q, r \in \mathbb{R}^2$ and $U(p) > U(q), U(q) > U(r) \implies U(p) > U(r)$ then $U()$ represents a transitive preference relation. Hmm... Yeah, positive and constant derivatives should do the trick to guarantee it. Thanks, man. $\endgroup$– Pedro CavalcanteOct 4, 2018 at 21:41
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1$\begingroup$ $U(x,y) = (x+y)^2$ is a monotonic transformation of $U^{'}(x,y) = x+y$.(the transformation is $f(z) = z^2$) $\endgroup$– superhulkOct 5, 2018 at 1:56
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3$\begingroup$ @PedroCavalcanteOliveira Why do you need anything about derivaties? Why is $$ U(A) > U(B) \text{ and } U(B) > U(C) \Rightarrow U(A) > U(C) $$ not enough? $\endgroup$– GiskardOct 5, 2018 at 4:21
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2 Answers
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Transitivity.
Definition. If $A \succsim B$ and $B \succsim C$, then $A \succsim C$.
Proof. Suppose $A \succsim B$ and $B \succsim C$. Then by definition of a utility representation, $U(A) \geq U(B)$ and $U(B) \geq U(C)$. By the transitivity of $\geq$, we have $U(A) \geq U(C)$. And so again by definition of a utility representation, $A \succsim C$, so that $\succsim$ is transitive.
Monotonicity.
Definition. If $x_1\geq x_2$ and $y_1\geq y_2$, then $\left(x_1,y_1\right)\succsim\left(x_2,y_2\right)$.
Proof. Suppose $x_1\geq x_2 \left(\geq 0\right)$ and $y_1\geq y_2\left(\geq 0\right)$. Then $$U\left(x_1,y_1\right) = x_1^2 +y_1^2 +2x_1y_1 \geq x_2^2 +y_2^2 +2x_2y_2 = U\left(x_2,y_2\right),$$ so that definition of a utility representation, $\succsim$ is monotonic.
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Any preference relation represented by an utility function is transtive.
Suppose $x \succsim y$ and $y \succsim z$ and $\succsim$ is represented by $U$. Then $U(x) \geq U(y) $ and $U(y) \geq U(z)$, so $U(x) \geq U(z)$ and then $x \succsim z$.
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$\begingroup$ This is just a consequence of the transitivity in the real numbers to which the utility function maps, right? $\endgroup$– DaveDec 26, 2022 at 4:44