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I am going through some micro concepts and I am confused, is there a difference between deriving preferences through indifference curves and actual preferences of the consumers?

The question I was going through was:

Jim's utility function is $U(x; y) = xy$. Jerry's utility function is $U(x; y) = 1,000xy + 2, 000.$ Tammy's utility function is $U(x; y) = xy(1-􀀀xy)$. Oral's utility function is 􀀀$-1/(10+xy)$. Billy's utility function is $U(x; y) = x/y$. Pat's utility function is $U(x; y) = -􀀀xy$.

(a) No two of these people have the same preferences.

(b) They all have the same preferences except for Billy.

(c) Jim, Jerry, and Pat all have the same indiff erence curves, but Jerry and Oral are the only ones with the same preferences as Jim.

(d) Jim, Tammy, and Oral all have the same preferences.

(e) There is no truth in any of the above statements.

The answer apparently is C, which implies that even though Jim, Jerry and Pat have the same indifference curves, The preferences of Pat and Jim are different, while those of Oral and Jim are the same even though they have different indifference curves. How is that possible?

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  • $\begingroup$ I am confused about "same indifference curves but different preferences" point though. This all seems like a very legalistic interpretation (so the stated answer in the question might be right on this point, maybe they're going for a map v curve distinction), and I am having a hard time finding any precise definition for indifference curves (101 definitions are necessarily imprecise). I don't think it's a useful distinction. $\endgroup$
    – Pburg
    Commented Apr 21, 2015 at 13:50
  • $\begingroup$ They must be using a definition that two utility functions give identical indifference curves if and only if they produce the same partition of bundles. An indifference curve is a set $S_\alpha = \{x\in\mathbb{R}^n : U(x)=\alpha \}$. Let $U(x)$ give sets $\{S_\alpha \}_{\alpha \in \mathbb{R}}$ and $V(x)$ give sets $\{S'_\alpha \}_{\alpha \in \mathbb{R}}$. $U,V$ identical $\iff$ $\forall \alpha \in \mathbb{R}, \exists \beta \in \mathbb{R} \; \text{s.t.} S_\alpha = S'_\beta $. $\endgroup$
    – Pburg
    Commented Apr 21, 2015 at 14:05
  • $\begingroup$ There is a problem with your notation, it doesn't render correctly. Please fix $\endgroup$
    – Alecos Papadopoulos
    Commented Apr 21, 2015 at 14:26

3 Answers 3

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A nerd loves both mathematics and physics equally, and hates sports and drinking beer equally.

A football player hates both mathematics and physics equally, and loves sports and drinking beer equally.

The nerd and the football player have the same indifference curves, but not the same preferences.

The point is this: Both agree on the "grouping" of things. But they disagree in the ranking of the groups.

Your Example

If you look at Jim and Pat, they have exactly inverse utility functions and hence inverse ranking in their preferences.

That is, if for Jim $A > B = C = D > E$, Pat will have $A < B = C = D < E$. Both agree on the three groups, but disagree on the order. Jerry's just a monotone transformation (poor Jerry).

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Perhaps you overlooked something, as I think Oral and Jim do not have different indifference curves.

Pat and Jim do have identical indifference curves, but whereas Jim prefers curves that are 'higher' or 'more to the right' in the x,y coordinate system Pat prefers curves that are 'lower or more to the left'.

An analogy: If you would be looking at an altitude map you cannot tell if one line is higher or lower than the other without additional information. In micro this additional information is the utility value (corresponding to altitude) of the indifference curve.

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  • $\begingroup$ I agree on Oral and Jim and Jerry having the same indifference curves. Simple differentiation shows these preferences give the same marginal rate of substitution. $\endgroup$
    – Pburg
    Commented Apr 21, 2015 at 13:41
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Everyone who has the same Marginal rate of substitution will have the same preferences.

Everyone who has the same equation of indifference curves will have same ICs.

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