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The convexity of an indifference curve results from the fact that the absolute value of its (negative) derivative, which is the marginal rate of substitution is decreasing. But why do we say that it's convex to the origin?

What is an implicit function that is convex and concave to the origin?

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I think what people mean when they say "convex to the origin" (or to any point $p$) is that the function is convex when looked at in a new basis, namely the basis resulting from a rotation such that the new x axis (call it x') is, up to a constant, tangent to the IC and the distance $|p-IC|$ is minimized by that point of tangency ($w$).

enter image description here

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  • $\begingroup$ This seems like the only answer that addresses the actual question. Can you please back up your definition with a reference? $\endgroup$ – Giskard Oct 30 '18 at 17:22
  • $\begingroup$ @denesp no because I made it up, that’s how I understand it but there’s no reason why this should be true $\endgroup$ – John Oct 30 '18 at 20:13
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The expression tries to convey a visual notion of convexity. It is "convex to the origin" in the sense that if we "stand" at the origin, the point $(0,0)$, and "look towards" the graph, we will perceive it as convex.

In contrast, if we stand "above" such a graph looking towards it, taking "our" position as being the origin, it will be concave. This "relativity" I suspect comes from physics and the arbitrariness of the origin of the coordinate system there.

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    $\begingroup$ What does it mean to be convex respective to one point? $\endgroup$ – John Oct 30 '18 at 9:10
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Note: this answer is more mathematically complicated than an undergraduate level, but I will try to give intuition as well.

I think it's easier to think about this problem in terms of preferences. A preference relation $\succsim$ over a set of alternatives $X$ is convex if $$x \succsim y \Rightarrow \lambda x + (1-\lambda) y \succsim y$$ for every $\lambda \in (0,1]$.

Basically, what this is saying is that if I like x more than y, I also like any bundle which is a mix between them better than y. So essentially, if I'm at y, then I'd like to get as close to x as possible with respect to y.

To visualize this, consider being at some value on an arbitrary indifference curve for convex preferences (e.g. Cobb-Douglass utility function). Pick another point which is on a strictly higher indifference curve. Then I can draw a straight line between the two bundles, and everything on that line is better than where I started. Graphically, this should be obvious, as everything between the two points should also be on a strictly higher indifference curve. Mathematically speaking, this is the same as saying that the upper contour set for x, $\{y : y \succsim x\}$ is a convex set for every x.

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  • $\begingroup$ I like water better than fanta but fanta mixed with water is a disgrace and I'd rather drink just fanta. But I see your point $\endgroup$ – John Oct 30 '18 at 9:15
  • $\begingroup$ "So essentially, if I'm at y, then I'd like to get as close to x as possible with respect to y." This is not true. Taking symmetric Cobb-Douglass preferences and $y = (0,1)$, $x = (1,0.01)$ I would much rather stay somewhere close to $\lambda = 1/2$ than get as close to $x$ as possible. $\endgroup$ – Giskard Oct 30 '18 at 17:19
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    $\begingroup$ Also the question is not about convexity, which is explained well in its first line, but about the concept of "convex to origin". $\endgroup$ – Giskard Oct 30 '18 at 17:20
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In case of any convex function, the value of the Arithmetic mean of any two points on the function is not less than the value of midpoint of the function in that interval.

Look at this function below: enter image description here

This function is convex to the point $(0,0)$ and concave to the point $(10,10)$.

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  • $\begingroup$ What you describe in your first sentence is the convex property, not the convex to origin property. $\endgroup$ – Giskard Oct 30 '18 at 17:25
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I too have pondered this question.

I found a neat definition from the book 'Existence and Optimality of Competitive Equilibria' by Charalambos D. Aliprantis, Donald J. Brown, Owen (see page 9). I quote it here:

"Mathematically, a 'convex to the origin' curve is described by saying that if $A$ and $B$ are any two points on the curve, then a ray passing through the origin $O$ and any point $X$ on the line segment $AB$ will meet the curve at most at one point $D$ between $O$ and $X$". EDIT - this should read 'at exactly one point $D$ between $0$ and $X$ as opposed to at most one; the author appears to have made an error.

Taken from the same source

Under this definition, a curve shaped like the left-half of a U-shaped parabola, but never reaching a point at which its derivative is 0, is the type of nice convex to the origin indifference curve we dream about as economists. If we were to allow the slope of the indifference curve to become positive after some point then it will be possible to find draw a line from the origin that crosses two points of the curve, in which case the curve would not be convex to the origin.

When a utility function is a function of two variables x and y, an indifference curve is convex to the origin if the derivative of the indifference curves are always negative and the second derivatives are positive. That is, the indifference curves slope downwards always but the slope of the slope increases (from a negative value to a less negative value) as we move to the right. That is, the marginal rate of substitution (the negative of the slope of the indifference curve or equivalently, the magnitude of the slope of the indifference curve) decreases as we move to the right along an indifference curve.

Best,

Dutchman

Edit in response to Giskard's comment:

Giskard is right, the definition would need to be amended to exactly one.

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  • $\begingroup$ I don't get it. "a ray passing through the origin $O$ and any point $X$ on the line segment $AB$ will meet the curve at most at one point $D$ between $O$ and $X$". Seems to me that this works even if the utility function is strictly concave. E.g. $U(x,y) = x^2 + y^2$ has this property. Perhaps if the definition was "exactly at one" and not "a most at one". $\endgroup$ – Giskard Jun 2 '19 at 9:43
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    $\begingroup$ @Giskard, I think that you are right. Thanks. $\endgroup$ – Dutchman Jun 2 '19 at 10:29

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