# Thread: Concavity of a multivariate regression function

1. ## Concavity of a multivariate regression function

I am conducting a research in the field of economics. Based on empiric data I have obtained a multivariate regression function as follows:

F = – 65,9569 + 5,08x – 0,83y – 5,27z + 6,65w

where F is the dependent variable and x,y,z and w the independent variables.

Is it possible to determine the concavity or convexity of this function? How?

Thanks

2. ## Re: Concavity of a multivariate regression function

Hey Kohoutek.

For concavity you basically have to show that if you take any two points in the set of the function (call them A and B), then you have to show that all the points in-between them (on a straight line path) also lie in the function.

Something is convex if f(a + (b-a)t) where to goes from 0 to 1 and a and b are in the domain.

So to start off with, pick two general points corresponding to f(a) and f(b) where a = -65,9569 + 5,08x1 - 0,83y1 - 5,27z + 6,65w1 and b = -65,9569 + 5,08x2 - 0,83y2 - 5,27z2 + 6,65w2. Now if you can show that f(a + (b-a)t) is always in the possible function values. If you show that it fails then its concave.

So basically you have f(a + (b-a)t) = f(-65,9569 + 5,08x1 - 0,83y1 - 5,27z + 6,65w1 + (-65,9569 + 5,08x2 - 0,83y2 - 5,27z2 + 6,65w2 - (-65,9569 + 5,08x1 - 0,83y1 - 5,27z + 6,65w1))t)

= f(-65,9569 + 5,08x1 - 0,83y1 - 5,27z + 6,65w1 + (5,08(x2-x1)t - 0,83(y2-y1)t - 5,27(z2-z1)t + 6,65(w2-w1)t) where t is in between 0 and 1.

So for all values of t, do these values lie in the set of potential function values? If they do your function is convex.

Basically the question now becomes what the constraints are on your x,y,z,w parameters and in some cases you will get convexity and in others you won't.

3. ## Re: Concavity of a multivariate regression function

Thanks Chiro.

I'd like to make another approach to evaluate concavity (not sure wether it's right).

In the simplest case of a function y=f(x), the curve representing f(x) is concave if the first derivative f'(x)>0 and the second derivative f''(x)<0. On the other hand, it would be convex if f'(x)>0 and f''(x)>0.

If the function is multivariate like the one indicated in my first post F=F(x,y,z,w) it will be be concave if the Hessian H<0 and convex if H>0.

Therefore, the calculation of the Hessian can assess concavity or convexity of a multivariate function. If H=0 then concavity or convexity cannot be evaluated.

In case I am right with this approach, could you or anyone else help me obtaining the Hessian of the posted function F to evaluate its concavity?

Thanks

4. ## Re: Concavity of a multivariate regression function

I did a quick search to confirm your suspicions and I got this page which goes into a lot of depth:

Math tutorial: concave and convex functions of many variables

5. ## Re: Concavity of a multivariate regression function

Thanks Chiro for the web site. I clarifies a lot on Hessians. Having read it, I am afraid I cannot use this operator to determine concavity/convexity of my multivariate function. Functions need to be twice differentiable and, to the best of my knowledge, mine it is not.

The posted equations is a regression function obtained from a data base giving values for 142 countries of variables x, y, z and w that represent different socio-economic attributes. I used statistical program SPSS to obtain the regression.

Now I think an alternative way to measure concavity would be to transform the posted equation into a Cobb Douglas function. This is a function we use often for different applications in economics. Hopefully I will be able to use Hessians over it and find out wether it is concave or convex.

I still leave the thread open just in case someone can give me a hand...

6. ## Re: Concavity of a multivariate regression function

Do you have a restriction on the values for (x,y,z,w)?

Since this is a linear object, the convexity will depend on these constraints. If you have constraints at all, then it would seem to be convex (by its definition).