I am at my wits end. Maybe one of you can explain it to me. A utility function is given: $U(x,y)=\sqrt{x^2−y^2}$ and we should determine whether the indifference curve is convex.

From the lecture there is two options to tell if a indifference curve is convex:

  1. See if the MRS is decreasing for increases in $x$.
  2. Check if $α[U(x_1, y_1)]+(1-α)[U(x_2, y_2)] ≤ U[(αx_1+(1-α)x_2), (αy_1+(1-α)y_2)]$.

For my function, both these conditions are full-filled. However, when setting utility to a constant level and drawing the indifference curve, one can see that the curve is concave with its opening to the right side. So how do I know and especially show, except drawing it, that the curve is concave?

There were more than one function given to evaluate. For another function, $U(x,y)=\sqrt{xy}$, both of these condiments hold as well. But one can quickly see that this is another form of a Cobb-Douglas utility function and that it is indeed convex. Both functions differ only in that $U(x,y)=\sqrt{xy}$ has a positive MRS and $U(x,y)=\sqrt{x^2−y^2}$ has a negative MRS. For both, the MRS differentiated for $x$ is negative as given by option one above.

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    $\begingroup$ This utility function doesn't make much sense. The more $y$, the less utility, so the $y$-good is really a bad. Moreover, utility is undefined for $y>x$ ...? $\endgroup$
    – VARulle
    Commented Jan 24, 2023 at 12:25

3 Answers 3

  1. $U(x,y) = \sqrt{x^2 - y^2}$ indeed has concave indifference curves. As you pointed out, this can be found by setting utility to a constant level $\overline{U}$.

$\overline{U} = \sqrt{x^2 - y^2} \iff \overline{U}^{2} = x^2 - y^2 \iff y^2 = x^2 - \overline{U}^2 \iff y = \sqrt{x^2 - \overline{U}^2}$,

the last step assuming the quantities can't be negative.

Note that since $\overline{U}$ is constant, then the quantity $\overline{U}^2$ is a constant, which we'll rename as $C$.

We get the family of indifference curves as $y = \sqrt{x^2 - C}$.

Differentiating w.r.t. x,

$\frac{dy}{dx} = \frac{x}{\sqrt{x^2 - C}}$

Differentiating w.r.t. x again,

$\frac{d^2 y}{dx^2} = \frac{\sqrt{x^2 - C} - x \frac{x}{\sqrt{x^2 - C}}}{x^2 - C} = \frac{\frac{x^2 - C - x^2}{\sqrt{x^2 - C}}}{x^2 - C} = \frac{-C}{(x^2 - C)^{\frac{3}{2}}}$

The quantity in the denominator is positive as long as itself and $\frac{d^2 y}{dx^2}$ well-defined and since $C = \overline{U}^2$, it is positive as long as $\frac{d^2 y}{dx^2}$ is well defined.

That well definedness I talk about occurs when $x > y$ as square roots can't be taken from negative numbers. The inside of the square root can be $0$ but the derivate wouldn't exist as the square root is in the denominator.

So we got

$\frac{d^2 y}{dx^2} < 0$ when $x > y$.

Since $x > y \iff \overline{U} > 0,$ this gives us concavity of the indifference curves for $\overline{U} \in (0,\infty)$.

The points where $y = x$ are exactly the indifference curve for $\overline{U} = 0$. Since it is a straight line, it is concave. (Lines are both convex and concave, they're kind of a wild card).

Therefore, the indifference curve for $\overline{U} = 0$ is also concave.

We got that the indifference curves for $\overline{U} \in [0,\infty)$ are concave.

Since the range of $U(x,y) = \sqrt{x^2 - y^2}$ is $[0,\infty)$, we can conclude that all the indifference curves are concave.

As you pointed out, this utility function has a negative MRS, because of the following:

$\frac{\partial U}{\partial y} = \frac{-y}{\sqrt{x^2 - y^2}}$ which is negative for all $y > 0$.

This means that $U$ is decreasing in $y$ which makes $y$ a bad rather than a good, as pointed out in the comment. This is the underlying problem.

Most of the theory is based on both $x, y$ being economic goods, i.e. the utility function is non-decreasing in both $x,y$, so you'd have to be more careful with the rules you're taught when one of the "goods" is actually an economic bad.

Note: This utility function makes no sense when $y > x$ because it would give the square root of a negative number.

On the other hand, for that Cobb-Douglas utility function, $x,y$ are both always economic goods, since

$\frac{\partial U}{\partial x} = \frac{y}{2 \sqrt{xy}}$,

$\frac{\partial U}{\partial y} = \frac{x}{2 \sqrt{xy}}$.

Since both can't be negative,

$MRS = \frac{\frac{\partial U}{\partial x}}{\frac{\partial U}{\partial y}}$ can't be negative, like you pointed out.

Edit after comment from OP:

$MU_x = \frac{\partial U}{\partial x} = \frac{x}{\sqrt{x^2 - y^2}}$,

$MU_y = \frac{\partial U}{\partial y} = \frac{-y}{\sqrt{x^2 - y^2}}$.

From here we get

$MRS_{x,y} = \frac{MU_x}{MU_y} = - \frac{x}{y}$.

The $MRS$ we usually calculate is $MRS_{x,y}$ unless stated otherwise.

Even though this is negative, when $x$ increases, the fraction $\frac{x}{y}$ gets bigger as $x$ is in the numerator and both quantities are positive.

This implies that $MRS_{x,y} = - \frac{x}{y}$ gets more negative as $x$ gets bigger. Getting more negative implies it is decreasing.

  • $\begingroup$ Thank you for this very elaborate answer! I also did check $\frac{dy}{dy}$ and $\frac{d^2 y}{dx^2}$. However, what was given in the exercise was to check if the $MRS$ declines for increases in $x$. Is there a way to see this? After you suggested that $y$ really is a "bad", I checked $MRS_{y,x}$ rather than $MRS_{x,y}$ as previously. This does not make a difference. Both are negative, and both are negative when building the first derivative, same as for the convex function. $\endgroup$
    – Charles
    Commented Jan 25, 2023 at 18:12
  • $\begingroup$ @Charles, edited my answer taking into account your comment. Is everything clear? $\endgroup$ Commented Jan 26, 2023 at 5:10
  • $\begingroup$ I am sorry, but it is still not clear to me. From $\frac{dMRS}{dx}=-\frac{1}{y}$ we already know that $MRS$ is decreasing for increases in $x$. However, in the exercise it was stated that $\frac{dMRS}{dx}<0$ was the condition for a convex curve. In the case of $U(x,x)=\sqrt{x^2-y^2}$ this is also fulfilled and this is what causes my problem. If I was just to check $\frac{dMRS}{dx}$ and nothing else, I would conclude that $U(x,x)=\sqrt{x^2-y^2}$ is also convex. The only difference between the concave and the convex function is that one has a positive and the other a negative $MRS$. $\endgroup$
    – Charles
    Commented Jan 26, 2023 at 10:56
  • $\begingroup$ @NicolasTorres the answer above yours which states the IC curves are convex seems to contradict yours, or at least creates some confusion, do you have any thoughts! Thanks! $\endgroup$
    – CormJack
    Commented Feb 14, 2023 at 18:10
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    $\begingroup$ @CormJack this discussion made me realize that both definitions are equivalent if the x-good and y-good are both “goods”, but when we make good y a “bad”, the upper contour set switches from lying above the IC to lying below the IC. $\endgroup$ Commented Feb 15, 2023 at 1:51

It's hard to be sure without knowing the terminology that your teacher is using, but I think a source of confusion here could be two different definitions of convexity.

Firstly, a function is convex if $\alpha f(x)+(1-\alpha)f(y)> f(\alpha x+(1-\alpha)y)$ (notice this is the opposite inequality to the one in your question, which is what suggests to me that this is not the notion of convexity that your instructor has in mind).

Graphically, for a single-variable function, this means the plot is u-shaped such that connecting any two points on the function yields a line that lies everywhere above the function; a concave function is one where the line lies below.

This is what you are checking when you plot the indifference curves in the $x$-$y$ space and see that they curve up from south to east.

Please excuse the crude drawing, but this shows a concave (non-convex) function because the red line lies everywhere below the blue function.

enter image description here

In set theory, we say that a set is convex if, for any two points inside the set, a weighted average of those points is also inside the set. Graphically, this means that if we draw a straight line between two points in the set then the entire line lies inside the set. A non-convex set is a set where you can connect two points and find that the line passes outside of the set. The first picture below shows a convex set, the second shows a non-convex set.

enter image description here

enter image description here

Now, an indifference curve defines a set of all of the points (i.e., all of the bundles) that are better than those on the indifference curve. We call that set the upper contour set. For a standard textbook Cobb-Douglas indifference curve, that set would look like this (notice the upper contour set is convex):

enter image description here

For the utility function in your question, the set looks like this (is still convex because connecting any two points in the shaded area yields a line entirely inside the shaded area):

enter image description here

This second notion of convexity corresponds to the condition that $α[U(x_1, y_1)]+(1-α)[U(x_2, y_2)] ≤ U[(αx_1+(1-α)x_2), (αy_1+(1-α)y_2)]$.

  • $\begingroup$ So what you are saying is, that $U(x,y)=\sqrt{x^2-y^2}$, or rather its corresponding indifference curve, is also convex? $\endgroup$
    – Charles
    Commented Jan 26, 2023 at 13:25
  • $\begingroup$ @Charles Yes. Or,to be slightly more precise, that the upper contour sets of the indifference curves generated by $U(x,y)$ are convex sets. Normally, if this property is satisfied then we say that the consumer's preferences are convex (check out the Wikipedia article on convex preferences). $\endgroup$
    – Ubiquitous
    Commented Jan 26, 2023 at 13:30
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    $\begingroup$ So you are contradicting what Nicolas Torres is saying? Why then, is the indifference curve looking so concave? And why would Nicolas Torres first suggestion of checking $\frac{d^2y}{dx^2}$ yield different results for $U(x,y)=\sqrt{xy}$ and $U(x,y)=\sqrt{x^2-y^2}$ ? $\endgroup$
    – Charles
    Commented Jan 26, 2023 at 13:48
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    $\begingroup$ @Charles No, I am not contradicting what he is saying. I am saying that there are two (actually more, but let's not go there…) definitions of what it means for an indifference curve to be convex because the language is not standardised. By one definition, an IC is convex if the function y(x) is a convex function—this is the definition used in Nicolas Torres' answer. If that's your definition then his answer is correct. But the second definition (an IC is convex if its upper contour set is convex) matches the conditions you were given by the instructor, so is probably the one s/he had in mind. $\endgroup$
    – Ubiquitous
    Commented Jan 26, 2023 at 17:03
  • $\begingroup$ @Ubiquitous I think the confusion the OP is referring too is that, your post and comments highlight that the indifference curve is Convex. But the answer bellow by NicolasTorres opens by saying "$U(x,y) = \sqrt{x^2 - y^2}$ Indeed has Concave indifference curves." Your answer seems great, could you possibly clarify the confusion for us? $\endgroup$
    – CormJack
    Commented Feb 14, 2023 at 18:04

Just an observation, that is too long for a comment, about a possible source of confusion.

Maybe an origin of the problem is a confusion about the signs of the derivatives.

You wrote:

A utility function is given: $U(x,y)=\sqrt{x^2−y^2}$ and we should determine whether the indifference curve is convex. From the lecture there is two options to tell if a indifference curve is convex

  1. See if the MRS is decreasing for increases in $x$.
  2. Check if $α[U(x_1, y_1)]+(1-α)[U(x_2, y_2)] ≤ U[(αx_1+(1-α)x_2), > (αy_1+(1-α)y_2)]$.

For my function, both these conditions are full-filled. However, when setting utility to a constant level and drawing the indifference curve, one can see that the curve is concave with its opening to the right side.

I think that Point 1) means that the absolute value of the slope of the indifference curve is decreasing, as the MRS is usually defined as the absolute value, that is Slope of the indifference curve= $ - \frac {\frac {\partial U}{\partial x} }{{\frac {\partial U}{\partial y} } }=-MRS$ (if the slope is negative). $^1$

(this sometimes gives rise to confusion).

Think of the usual indifference curves as Cobb-Douglas indifference curves: enter image description here

Indifference curves have a negative slope, which is decreasing in absolute value but, as it is negative, is actually increasing: the second derivative is positive.

On the contrary, the indifference curves of your function $U(x,y)=\sqrt{x^2−y^2}$ are positively sloped, and their slope decreases (in this case it is the MRS): the second derivative is negative, therefore they are concave.

These are, in both cases, the conditions of convexity/concavity that use the second derivatives: a twice differentiable function is convex if and only if $f'' \geq 0$, is concave if and only if the second derivative is $f'' \leq 0$. That's all.

So, I suppose that point 1) of your lecture concerns the more usual cases in which the indifference curves are negatively sloped.

$^1$ See for example Mas Colell, Microeconomic Theory, p. 53.


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