Trouble with book example on concavity vs strict concavity

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hellosoupy

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Hi!

I was just working through an example in my textbook where I obtained the right answer, then they threw me a curveball at the end of their explanation.

The example:
Use the sign of the second-order total differential to show that the function [MATH]y = 5-(x_1 + x_2)^2[/MATH] is concave.

So I found [MATH]f11=f22=f12=f21=-2[/MATH]
therefore

[MATH]d^2y=-2dx_1^2+-4dx_1dx_2+-2dx_2^2[/MATH]
and we can say that y is concave. I get that. But then the textbook mentions the following (I'm copying it nearly verbatim):

"[T]o see that this function is not strictly concave, notice that along the set of [MATH](x_1, x_2)[/MATH] values satisfying [MATH]x_2 =a - x_1[/MATH], where a is any constant, we have [MATH]y = (x_1 + a - x_1)^2 =a^2[/MATH]. These [MATH](x_1, x_2)[/MATH] values generate horizontal linear segments on the graph of this function."

Is there any way to figure out this linear segments bit without looking at the graph of the function? If yes, what would have clued me in to this without looking at the graph? And I still don't understand how that equation indicates that there are linear segments on the graph. Any help would be appreciated!

I actually figured out (by continuing to read through the text) that you can show that this function is concave but not strictly concave using the matrix method, but I'd like to try to wrap my head around the above.

Thanks in advance!
 
This is not something I know about, but I'll just make these observations on what you have said. If this is not helpful, I apologise, and just ignore!

1. The function is [MATH]y = 5-(x_1 + x_2)^2[/MATH] so I don't know why they said:
"[T]o see that this function is not strictly concave, notice that along the set of [MATH](x_1, x_2)[/MATH] values satisfying [MATH]x_2 =a - x_1[/MATH], where a is any constant, we have [MATH]\boldsymbol{y = (x_1 + a - x_1)^2 =a^2}[/MATH]. These [MATH](x_1, x_2)[/MATH] values generate horizontal linear segments on the graph of this function."

It should surely be [MATH]y=5-a^2[/MATH] which is still a constant.

2. I don't think it was necessary to draw the graph to see that there was a straight line segment on the surface. They just used a linear segment in the [MATH]x_1x_2[/MATH] plane, viz. [MATH](x_1,a-x_1)[/MATH] and found that it generated a straight line segment on the surface, viz. [MATH](x_1, a-x_1,5-a^2)[/MATH].
[MATH](x_1, a-x_1,5-a^2) = (0,a,5-a^2)+x_1(1,-1,0)[/MATH] which is a straight line segment (on the surface).
The look at the graph is only to illustrate what is happening.
1617614622600.png
 
Beer soaked query follows.
hellosoupy said:
Hi!

I was just working through an example in my textbook where I obtained the right answer, then they threw me a curveball at the end of their explanation.

The example:
Use the sign of the second-order total differential to show that the function [MATH]y = 5-(x_1 + x_2)^2[/MATH] is concave.

So I found [MATH]f11=f22=f12=f21=-2[/MATH]
therefore

[MATH]d^2y=-2dx_1^2+-4dx_1dx_2+-2dx_2^2[/MATH]
and we can say that y is concave. I get that. But then the textbook mentions the following (I'm copying it nearly verbatim):

"[T]o see that this function is not strictly concave, notice that along the set of [MATH](x_1, x_2)[/MATH] values satisfying [MATH]x_2 =a - x_1[/MATH], where a is any constant, we have [MATH]y = (x_1 + a - x_1)^2 =a^2[/MATH]. These [MATH](x_1, x_2)[/MATH] values generate horizontal linear segments on the graph of this function."

Is there any way to figure out this linear segments bit without looking at the graph of the function? If yes, what would have clued me in to this without looking at the graph? And I still don't understand how that equation indicates that there are linear segments on the graph. Any help would be appreciated!

I actually figured out (by continuing to read through the text) that you can show that this function is concave but not strictly concave using the matrix method, but I'd like to try to wrap my head around the above.

Thanks in advance!
Click to expand...
What textbook are you using?
 
lex said:
This is not something I know about, but I'll just make these observations on what you have said. If this is not helpful, I apologise, and just ignore!

1. The function is [MATH]y = 5-(x_1 + x_2)^2[/MATH] so I don't know why they said:
"[T]o see that this function is not strictly concave, notice that along the set of [MATH](x_1, x_2)[/MATH] values satisfying [MATH]x_2 =a - x_1[/MATH], where a is any constant, we have [MATH]\boldsymbol{y = (x_1 + a - x_1)^2 =a^2}[/MATH]. These [MATH](x_1, x_2)[/MATH] values generate horizontal linear segments on the graph of this function."

It should surely be [MATH]y=5-a^2[/MATH] which is still a constant.

2. I don't think it was necessary to draw the graph to see that there was a straight line segment on the surface. They just used a linear segment in the [MATH]x_1x_2[/MATH] plane, viz. [MATH](x_1,a-x_1)[/MATH] and found that it generated a straight line segment on the surface, viz. [MATH](x_1, a-x_1,5-a^2)[/MATH].
[MATH](x_1, a-x_1,5-a^2) = (0,a,5-a^2)+x_1(1,-1,0)[/MATH] which is a straight line segment (on the surface).
The look at the graph is only to illustrate what is happening.
View attachment 26192
Click to expand...
Thank you, Lex. I think I get it. If you can create an equation "y = a constant" (which is a horizontal line) within the function, then the function can't be strictly concave. I need to review the definitions of concavity and strict concavity to figure out why, but thanks for your help!

Jonah, the textbook is mathematics for economics by Hoy et al
 
hellosoupy said:
If you can create an equation "y = a constant" (which is a horizontal line) within the function, then the function can't be strictly concave. I need to review the definitions of concavity and strict concavity to figure out why, but thanks for your help!
Click to expand...

That is true. My understanding is:
f is concave if the line segment joining any two points on the graph of f is never above the graph. f is strictly concave if the (open) line segment is always below the graph.
In your example, they have created a line segment joining 2 points on the graph, where EVERY point on the line segment is on the graph (not below the graph), so it is not STRICTLY concave.
So a horizonal straight line on the graph or any straight line on the graph would be sufficient to make it not strictly concave. However, all you need is ONE point of the (open) line segment joining two points on the graph to be on/above the graph for it to fail to be strictly concave.
 
That's so helpful! Thanks for saving me from rereading formulas and drawing pictures :)
 
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