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In two dimensions, we have on an indifference curve that $dU=0$.

Does this apply to indifference objects in higher dimensions?

I was thinking that if $dU > 0$, then one is moving to a higher indifference curve/object

My Macroeconomics professor, in the first week of class, gave a microeconomics review. I asked if $dU = 0$ in higher dimensions. He said he would answer the question next time, and during the next class, he said it may not hold true in higher dimensions because of the problem of integrability.

I asked another economics professor in the department who said that $dU=0$ holds by definition of indifference curves/objects.

What is this problem of integrability, and what does it have to do with indifference curves?

Is it this one?

I've taken 2 courses of real analysis, 2 on probability theory and 2 on stochastic calculus (and 4 on statistics), but the mathematics I've encountered in economics is only up to basic linear algebra and separable ordinary differential equations (Actually the linear algebra's mainly just matrices. I hardly recall learning concepts like 'kernel' or 'eigenvalue' in economics, though I have learned uses for both in finance).

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    $\begingroup$ $dU$ seems ill-defined? $\endgroup$
    – Giskard
    Oct 17 '16 at 9:59
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    $\begingroup$ @denesp ? $\endgroup$
    – BCLC
    Oct 21 '16 at 0:05
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    $\begingroup$ Okay, so its the full differential. $\endgroup$
    – Giskard
    Oct 21 '16 at 4:32
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    $\begingroup$ It's not clear what your professor means by the problem of integrability. Usually, when we talk about "the integrability problem" we mean the problem of recovering preferences from a demand function (MWG have a section on this, starting on p75). But this is not obviously related to the question of whether utility is differentiable (I would agree with you that, if $U$ is differentiable, then $dU=0$ by definition of an indifference hypersurface). $\endgroup$
    – Ubiquitous
    Oct 21 '16 at 9:09
  • $\begingroup$ @Ubiquitous Thanks! Maybe there are some conditions where Not integrable --> Not differentiable or something? $\endgroup$
    – BCLC
    Oct 21 '16 at 11:52
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Seems to me $dU = 0$ is true by definition unless $dU$ is not defined

  1. because the function $U$ is not differentiable in some variable. E.g. $$U(x,y) = |x| + y$$
  2. because there are a countably infinite number of dimensions but the partial differentials are not always absolute convergent and hence cannot be summed up. E.g.: $$U(x_1,x_2,x_3,x_4...) = x_1 - x_2 + x_3 -x_4...$$
  3. because the dimensions have a larger than countably infinite cardinality. I have never seen this in micro but it is theoretically possible. E.g. expected utility over continuum states.

Edit: This seems to offer an extremely thorough walkthrough of the integrability problem in $n$ dimensions.

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  • $\begingroup$ 2 and 3 --> thanks! 1 --> this case is probably excluded because we are supposed to have $dU=0$ for 2D. So I guess we must assume U is differentiable in all variables. Any idea what any of this has to do with integrability? $\endgroup$
    – BCLC
    Oct 21 '16 at 5:44
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    $\begingroup$ $U$ not being differentiable would make it undefined. E.g. $|x|$ is not differentiable w.r.t $x$ but you can calculate the integral just fine. An integrability problem may arise if you try to calculate the indifference curve from its slope, i.e. $MRS(x,y)$. The slope is the derivative of the function, integrating returns the function ($+c$) from the slope. If the function was not differentiable to begin with $MRS(x,y)$ does not exist. $\endgroup$
    – Giskard
    Oct 21 '16 at 6:31
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    $\begingroup$ On continuum states: The consumption bundle may be a stock, this will have different values in different world states $i$. The actual good is not the stock though, it is unit value in the different world states. As you say these may be thought of as options. $\endgroup$
    – Giskard
    Oct 21 '16 at 6:38
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    $\begingroup$ @BCLC for detailed definition of the problem and conclusions please see my edit. $\endgroup$
    – Giskard
    Oct 21 '16 at 12:49
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    $\begingroup$ Thanks but TL;DR, denesp. Where's non-integrability there? $\endgroup$
    – BCLC
    Oct 22 '16 at 8:28

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