Question

One of the most important functions we will encounter in this book is the Cobb-Douglas function: \\[ y=(x \\] where a and \(P\) are positive constants that are each less than one. a. Show that this function is quasi-concave using a "brute force" method by applying Equation 2.107 b. Show that the Cobb-Douglas function is quasi-concave by showing that the any contour line of the form \(y=c\) (where \(c\) is any positive constant) is convex and therefore that the set of points for which \(y>c\) is a convex set. c. Show that if \(a+y 8>1\) then the Cobb-Douglas function is not concave (thereby illus trating that not all quasi-concave functions are concave). (Note: The Cobb-Douglas function is discussed further in the Extensions to this chapter.)

Short answer

In summary, the Cobb-Douglas function \(y = x_1^{\alpha}x_2^{\beta}\) is shown to be quasi-concave by (a) using the Hessian matrix method and (b) showing that its contour lines are convex. However, if \(\alpha + \beta > 1\), the function is not concave, as demonstrated in (c) by finding a specific violation of the concavity inequality.

Step-by-step solution

a. Quasi-concavity using Hessian Matrix

To check for quasi-concavity using the Hessian Matrix, first, we need to find the matrix of second derivatives for the given function. Start by finding the first partial derivatives with respect to \(x_1\) and \(x_2\). First derivative w.r.t. \(x_1\): \\[f_{x_1} = \alpha x_1^{\alpha-1} x_2^{\beta}\\] First derivative w.r.t. \(x_2\): \\[f_{x_2} = \beta x_1^{\alpha} x_2^{\beta-1}\\] Next, find the second partial derivatives: Second derivative w.r.t. \(x_1\): \\[f_{x_1x_1} = \alpha (\alpha - 1) x_1^{\alpha-2} x_2^{\beta}\\] Second derivative w.r.t. \(x_2\): \\[f_{x_2x_2} = \beta (\beta - 1) x_1^{\alpha} x_2^{\beta-2}\\] Second derivative w.r.t. both \(x_1\) and \(x_2\): \\[f_{x_1x_2} = f_{x_2x_1} = \alpha\beta x_1^{\alpha-1} x_2^{\beta-1}\\] Now, form the Hessian matrix \\[ H = \begin{pmatrix} f_{x_1x_1} & f_{x_1x_2} \\ f_{x_2x_1} & f_{x_2x_2} \end{pmatrix} \\] A function is quasi-concave if its Hessian matrix is negative semi-definite. To check negative semi-definiteness, we will check whether the determinant of Hessian is non-negative and if at least one of the diagonal entries is non-positive. Determinant: \\[\text{det}(H) = f_{x_1x_1}f_{x_2x_2} - f_{x_1x_2}^2 = \alpha(\alpha-1)\beta(\beta-1)x_1^{\alpha-2}x_2^{\beta-2}-\alpha^2\beta^2x_1^{\alpha+\beta-2}x_2^{\alpha+\beta-2}\\] Notice that since \(\alpha\) and \(\beta\) are positive constants less than 1, then \(\alpha-1 < 0\) and \(\beta-1 < 0\). This means that \(\alpha(\alpha-1) < 0\) and \(\beta(\beta-1) < 0\) and therefore, the determinant is non-negative. Taking a look at the diagonal elements, for example \(f_{x_1x_1}\), since \(\alpha(\alpha - 1)\) is non-positive, we have negative semi-definite Hessian, which implies that the Cobb-Douglas function is quasi-concave.

b. Quasi-concavity using contour lines

We now want to show that any contour line of the form \(y = c\), where \(c\) is a positive constant, is convex. Rewrite the Cobb-Douglas function \(y\) in terms of contour lines: \\[ c = x_1^{\alpha}x_2^{\beta} \\] Now, consider two points \((x_1^1, x_2^1)\) and \((x_1^2, x_2^2)\), let \(\lambda \in (0,1)\) and form the convex combination of these points, with \(x_1^* = \lambda x_1^1 + (1-\lambda) x_1^2\) and \(x_2^* = \lambda x_2^1 + (1-\lambda) x_2^2\). Our goal is to show that \(c^* \ge c\), where \(c^* = (x_1^*)^{\alpha}(x_2^*)^{\beta}\), because this means that the contour lines are convex. Using the convex combination property, calculate \(c^*\): \\[ c^* = [\lambda x_1^1 + (1-\lambda) x_1^2]^{\alpha}[\lambda x_2^1 + (1-\lambda) x_2^2]^{\beta} \\] Use Jensen's inequality which states that if $$\phi$$ is a concave function, then \(\phi(\lambda x_1^1 + (1-\lambda) x_1^2) \ge \lambda\phi(x_1^1) +(1-\lambda) \phi(x_1^2)\). Since \(\alpha\) and \(\beta\) are less than 1, we know that the power functions \(x^{\alpha}\) and \(x^{\beta}\) are concave. Apply Jensen's inequality separately to \(x^{\alpha}\) and \(x^{\beta}\) terms: \\[ c^* \ge [\lambda x_1^{\alpha} + (1-\lambda) x_1^{\alpha}][\lambda x_2^{\beta} + (1-\lambda) x_2^{\beta}] = c \\] Since \(c^* \ge c\), the contour lines are convex, and thus the Cobb-Douglas function is quasi-concave.

c. Not concave if \(\alpha + \beta > 1\)

To prove that the Cobb-Douglas function is not concave when \(\alpha + \beta > 1\), we have to show that for some \(x_1^1, x_2^1, x_1^2, x_2^2\) and \(\lambda \in (0,1)\), the inequality \(f(\lambda x_1^1 + (1-\lambda) x_1^2, \lambda x_2^1 + (1-\lambda) x_2^2) < \lambda f(x_1^1, x_2^1) + (1-\lambda) f(x_1^2, x_2^2)\) holds. Consider two points: \((x_1^1, x_2^1) = (1,1)\) and \((x_1^2, x_2^2) = (0,0)\), and let \(\lambda = 1/2\). Then: \\[ f\left(\frac{1}{2}, \frac{1}{2}\right) = \left(\frac{1}{2}\right)^{\alpha}\left(\frac{1}{2}\right)^{\beta} = \left(\frac{1}{2}\right)^{\alpha + \beta} \\] \\[ \frac{1}{2}f(1, 1) + \frac{1}{2}f(0, 0) = \frac{1}{2} \\] If \(\alpha + \beta > 1\), then \(\left(\frac{1}{2}\right)^{\alpha + \beta} < \frac{1}{2}\). As a result, the function is not concave in this case.

Key concepts

Quasi-concavity

Quasi-concavity is an important property in economics relating to the shape of functions. A function is quasi-concave if its upper-level sets (regions where the function takes values larger than a certain constant) are convex. In simpler terms, if you look at a map of the function, the areas where the function is relatively high form shapes without any dents.

For the Cobb-Douglas function, quasi-concavity implies that combining two different input bundles leads to at least as high of an output as taking the average output of these bundles on their own. To apply quasi-concavity precisely, we use mathematical tools like the Hessian matrix, often involving derivatives that analyze how different input changes affect output.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a scalar-valued function. In the context of determining concavity and quasi-concavity, the Hessian serves as a powerful tool.

Specifically, for the Cobb-Douglas function, this matrix helps in examining the behavior by involving the second partial derivatives with respect to each variable. Here's a brief step-by-step of how it's used:

• Compute first-order partial derivatives with respect to each variable (e.g., labor, capital).
• Calculate second-order partial derivatives to fill the Hessian matrix.
• Examine the quasi-concavity by checking the determinant of the Hessian matrix.
• If the determinant is non-negative, and at least one leading principal minor is non-positive, the function is quasi-concave.

In summary, the Hessian matrix provides a structured way to identify the character of the function's curvature and establish the quasi-concavity property.

Jensen's Inequality

Jensen's inequality is an essential inequality in mathematics and economics, often used in the context of proving quasi-concavity or convexity of functions.

This inequality states that for any convex function \( \phi \), the function value at the weighted average of points is always less than or equal to the weighted average of the function values at those points.

• For the Cobb-Douglas function, power terms are often involved, which are concave or convex based on the parameter values.
• This means when the exponents are between 0 and 1, these power functions exhibit concave behavior.
• Jensen's inequality aids by affirming that combining different inputs (or wealth, resources) will not reduce the function output below certain thresholds by ensuring convexity of 'contour lines'.

Utilizing Jensen's inequality provides assurance on how the Cobb-Douglas function behaves between different procedural inputs, confirming stability in quasi-concavity.