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Another function we will encounter often in this book is the power function: \\[ y=x^{\delta} \\] the form \(y=x^{8} / 8\) to ensure that the derivatives have the proper sign). a. Show that this function is concave (and therefore also, by the result of Problem \(2.9,\) quasi-concave). Notice that the \(\delta=1\) is a special case and that the function is "strictly" concave only for \(\delta<1\) b. Show that the multivariate form of the power function \\[ y=f\left(x_{1}, x_{2}\right)=\left(x_{1}\right)^{8}+\left(x_{2}\right)^{8} \\] is also concave (and quasi-concave). Explain why, in this case, the fact that \(f_{12}=f_{21}=0\) makes the determination of concavity especially simple. c. One way to incorporate "scale" effects into the function described in part (b) is to use the monotonic transformation \\[ g\left(x_{1}, x_{2}\right)=y^{y}=\left[\left(x_{1}\right)^{8}+\left(x_{2}\right)^{8}\right]^{y} \\] where \(\gamma\) is a positive constant. Does this transformation preserve the concavity of the function? Is \(g\) quasi-concave?

Short Answer

Expert verified
Analyzing the concavity and quasi-concavity of various power functions, we found that: 1. The univariate power function \(y = x^{\delta}\) is concave if \(\delta \le 1\). Since concave functions are also quasi-concave, this result holds for quasi-concavity as well. 2. The multivariate power function \(y=f(x_{1}, x_{2})=(x_{1})^{8}+(x_{2})^{8}\) is concave and thus quasi-concave as well. 3. A monotonic transformation \(g(x_{1}, x_{2}) = [\left(x_{1}\right)^{8}+\left(x_{2}\right)^{8}]^{\gamma}\) preserves concavity only if \(1 \le \gamma\). However, quasi-concavity may still hold regardless of the \(\gamma\) value.

Step by step solution

01

Univariate Power Function Concavity

To demonstrate the concavity of the univariate power function \(y = x^{\delta}\), we need to find the second derivative and check if it is nonpositive. Let's compute the derivatives: 1. First derivative: \\[ \frac{dy}{dx} = \delta x^{\delta - 1} \\] 2. Second derivative: \\[ \frac{d^2 y}{dx^2} = \delta(\delta - 1) x^{\delta - 2} \\] For \(y = x^{\delta}\) to be concave, the second derivative must be nonpositive. Given that \(x\) is positive, we can analyze the sign of the constant term, \(\delta(\delta - 1)\): - For \(\delta < 1\), the second derivative is negative, implying strict concavity. - For \(\delta = 1\), it is a special case where the second derivative is zero, which means the function is linear but not strictly concave. However, it is still concave since nonpositive second derivative still holds. Since the function is concave, it is also quasi-concave by the result of Problem 2.9.
02

Multivariate Power Function Concavity

To show that the multivariate power function \(y=f(x_{1}, x_{2})=(x_{1})^{8}+(x_{2})^{8}\) is concave, we will find its Hessian matrix and check if it is negative semidefinite. The Hessian matrix is given by: \\[ H = \begin{bmatrix} \frac{\partial^2 y}{\partial x_1^2} & \frac{\partial^2 y}{\partial x_1 \partial x_2} \\ \frac{\partial^2 y}{\partial x_2 \partial x_1} & \frac{\partial^2 y}{\partial x_2^2} \end{bmatrix} \\] Finding partial derivatives for the matrix elements: 1. \(\frac{\partial^2 y}{\partial x_1^2} = 56x_1^6\) 2. \(\frac{\partial^2 y}{\partial x_2^2} = 56x_2^6\) 3. \(\frac{\partial^2 y}{\partial x_1 \partial x_2} = \frac{\partial^2 y}{\partial x_2 \partial x_1} = 0\) Our Hessian matrix looks like this: \\[ H = \begin{bmatrix}56x_1^6 & 0 \\ 0 & 56x_2^6\end{bmatrix} \\] Since \(H\) is a diagonal matrix with nonnegative elements, it is negative semidefinite. Therefore, the multivariate power function \(y=f(x_{1}, x_{2})=(x_{1})^{8}+(x_{2})^{8}\) is concave and by Problem 2.9, it's also quasi-concave. The fact that \(f_{12} = f_{21} = 0\) makes the determination of concavity simple because there are no mixed partial derivative terms, which simplifies the condition for a function to be concave.
03

Monotonic Transformation and Concavity

To analyze the effect of the monotonic transformation on concavity, let's find the Hessian matrix of the transformed function \(g(x_{1}, x_{2}) = [\left(x_{1}\right)^{8}+\left(x_{2}\right)^{8}]^{\gamma}\). Applying the chain rule to compute partial derivatives: 1. \(\frac{\partial^2 g}{\partial x_1^2} = \gamma(\gamma - 1)\left[x_1^{8} + x_2^{8}\right]^{\gamma - 2}(56x_1^6)\), 2. \(\frac{\partial^2 g}{\partial x_2^2} = \gamma(\gamma - 1)\left[x_1^{8} + x_2^{8}\right]^{\gamma - 2}(56x_2^6)\), 3. \(\frac{\partial^2 g}{\partial x_1 \partial x_2} = \frac{\partial^2 g}{\partial x_2 \partial x_1} = 0\). The transformed Hessian matrix is: \\[ H_g = \begin{bmatrix} \gamma(\gamma - 1)\left[x_1^8 + x_2^8\right]^{\gamma - 2}(56x_1^6) & 0 \\ 0 & \gamma(\gamma - 1)\left[x_1^8 + x_2^8\right]^{\gamma - 2}(56x_2^6) \end{bmatrix} \\] To preserve concavity, the Hessian matrix of \(g\) must be negative semidefinite. Considering all elements of the Hessian matrix, it is negative semidefinite only if \(\gamma(\gamma - 1) \le 0\), which implies either \(\gamma \le 0\) or \(1\le\gamma\). However, we were given that \(\gamma\) is a positive constant, so the only case preserving concavity is when \(1\le\gamma\). For quasi-concavity, the Hessian matrix's signs are not strictly determined, so the transformed function may still be quasi-concave regardless of the value of \(\gamma\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Function
A power function is an expression of the form \( y = x^\delta \), where \(\delta\) is a constant exponent. These functions are prevalent in economics and other disciplines because they capture relationships where one variable is proportional to a power of another variable. Many natural phenomena, including economic output and resource consumption, can be modeled with power functions.

Understanding the concavity of a power function is critical when analyzing economic functions and their behaviors. A function is concave if its second derivative is less than or equal to zero throughout its domain. For the power function \( y = x^\delta \), the concavity depends on the exponent \(\delta\):
  • If \( \delta < 1 \), the function is strictly concave as the second derivative is negative.
  • If \( \delta = 1 \), the function is linear and hence, concave but not strictly.
  • For \( \delta > 1 \), the function is convex.
The concavity reveals insights about diminishing returns, an essential concept in economics that describes how the additional output from an extra unit of input decreases as the quantity of the input increases.
Quasi-Concavity
Quasi-concavity is a property of functions that allows them to maintain certain levels of output despite changes in input combinations. A function is quasi-concave if, for any points \(x\) and \(y\) in its domain, the function value at the convex combination of these points \((\theta x + (1-\theta) y)\) is at least as large as the minimum of the function values at \(x\) and \(y\) for all \(0 \leq \theta \leq 1\).

In economic terms, quasi-concavity is often associated with preference and production functions, capturing the idea of trade-offs or compromises. A quasi-concave function does not necessarily have a nonpositive second derivative, as required for concavity, but still exhibits the essential property of having no local maxima unless they are global maxima.
  • Quasi-concavity is a weaker condition than concavity; thus, a concave function is always quasi-concave.
  • In the exercise discussed, verifying concavity helped us assert quasi-concavity because concavity implies quasi-concavity.
This concept is crucial in utility functions, where an individual's satisfaction may increase but at a decreasing rate, demonstrating quasi-concavity.
Univariate and Multivariate Analysis
In economic modeling, functions frequently involve more than one variable, leading us to distinguish between univariate and multivariate functions. Univariate functions involve a single variable, while multivariate functions involve multiple variables.

Univariate analysis is typically simpler, allowing for direct application of calculus techniques like differentiation to explore properties such as concavity. Multivariate analysis, on the other hand, involves more complexity since interactions between variables can occur.
  • Univariate functions allow straightforward computation of derivatives for evaluating concavity or convexity.
  • Multivariate functions require partial derivatives to ascertain the same properties.
Making sense of multivariate functions involves tools like Hessian matrices, which systematically organize second derivatives to check concavity or convexity. It's evident from the exercise that moving from a univariate to a multivariate perspective requires understanding how multiple inputs change the outcome, providing a more comprehensive view of economic scenarios.
Hessian Matrix
The Hessian matrix is a mathematical tool used in multivariate calculus to determine the concavity or convexity of a function involving multiple variables. It is a square matrix of second-order partial derivatives of a scalar-valued function.

In the exercise, the Hessian was used to check the concavity of the multivariate function \( y = f(x_{1}, x_{2}) = (x_{1})^8 + (x_{2})^8 \). The elements of the Hessian matrix represent the second derivatives concerning each pair of variables. For such a function:
  • The diagonal elements \((\frac{\partial^2 y}{\partial x_1^2}, \frac{\partial^2 y}{\partial x_2^2})\) signify the curvature influenced by each variable alone.
  • The off-diagonal elements \((\frac{\partial^2 y}{\partial x_1 \partial x_2})\) reflect how the variables jointly affect the curvature.
A diagonal Hessian matrix, like in this case, where the off-diagonal elements are zero, simplifies the determination of concavity. The function is concave if the Hessian is negative semidefinite, which means all its eigenvalues are non-positive. This matrix is vital in economics when examining second-order conditions for optimizing multivariate economic functions.

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Most popular questions from this chapter

Because the expected value concept plays an important role in many economic theories, it may be useful to summarize a few more properties of this statistical measure. Throughout this problem, \(x\) is assumed to be a continuous random variable with PDF \(f(x)\). a. (Jensen's inequality) Suppose that \(g(x)\) is a concave function. Show that \(E[g(x)] \leq g[E(x)] .\) Hint: Construct the tangent to \(g(x)\) at the point \(E(x)\). This tangent will have the form \(c+d x \geq g(x)\) for all values of \(x\) and \(c+d E(x)=g[E(x)]\) where \(c\) and \(d\) are constants. b. Use the procedure from part (a) to show that if \(g(x)\) is a convex function then \(E[g(x)] \geq g[E(x)]\) c. Suppose \(x\) takes on only non-negative values-that is, \(0 \leq x \leq \infty\). Use integration by parts to show that \\[ E(x)=\int_{0}^{\infty}[1-F(x)] d x \\] where \(\left.F(x) \text { is the cumulative distribution function for } x \text { [that is, } F(x)=\int_{0}^{x} f(t) d t\right]\) d. (Markov's inequality) Show that if \(x\) takes on only positive values then the following inequality holds: \\[ \begin{array}{r} P(x \geq t) \leq \frac{E(x)}{t} \\ \text {Hint: } E(x)=\int_{0}^{\infty} x f(x) d x=\int_{0}^{t} x f(x) d x+\int_{t}^{\infty} x f(x) d x \end{array} \\] e. Consider the PDF \(f(x)=2 x^{-3}\) for \(x \geq 1\) 1\. Show that this is a proper PDF. 2\. Calculate \(F(x)\) for this PDF. 3\. Use the results of part (c) to calculate \(E(x)\) for this PDF. 4\. Show that Markov's inequality holds for this function. f. The concept of conditional expected value is useful in some economic problems. We denote the expected value of \(x\) conditional on the occurrence of some event, \(A,\) as \(E(x | A) .\) To compute this value we need to know the PDF for \(x\) given that \(A\) has occurred [denoted by \(f(x | A)]\). With this notation, \(E(x | A)=\int_{-\infty}^{+\infty} x f(x | A) d x\). Perhaps the easiest way to understand these relationships is with an example. Let $$f(x)=\frac{x^{2}}{3} \quad \text { for } \quad-1 \leq x \leq 2$$ 1\. Show that this is a proper PDF. 2\. Calculate \(E(x)\) 3\. Calculate the probability that \(-1 \leq x \leq 0\) 4\. Consider the event \(0 \leq x \leq 2,\) and call this event \(A\). What is \(f(x | A) ?\) 5\. Calculate \(E(x | A)\) 6\. Explain your results intuitively.

Show that if \(f\left(x_{1}, x_{2}\right)\) is a concave function then it is also a quasi-concave function. Do this by comparing Equation 2.114 (defining quasi-concavity) with Equation 2.98 (defining concavity). Can you give an intuitive reason for this result? Is the converse of the statement true? Are quasi-concave functions necessarily concave? If not, give a counterexample.

Here are a few useful relationships related to the covariance of two random variables, \(x_{1}\) and \(x_{2}\) a show that \(\operatorname{Cov}\left(x_{1}, x_{2}\right)=E\left(x_{1} x_{2}\right)-E\left(x_{1}\right) E\left(x_{2}\right) .\) An important implication of this is that if \(\operatorname{Cov}\left(x_{1}, x_{2}\right)=0, E\left(x_{1} x_{2}\right)\) \(=E\left(x_{1}\right) E\left(x_{2}\right) .\) That is, the expected value of a product of two random variables is the product of these variables' expected values. b. Show that \(\operatorname{Var}\left(a x_{1}+b x_{2}\right)=a^{2} \operatorname{Var}\left(x_{1}\right)+b^{2} \operatorname{Var}\left(x_{2}\right)+2 a b \operatorname{Cov}\left(x_{1}, x_{2}\right)\) c. In Problem \(2.15 \mathrm{d}\) we looked at the variance of \(X=k x_{1}+(1-k) x_{2} \quad 0 \leq k \leq 1 .\) Is the conclusion that this variance is minimized for \(k=0.5\) changed by considering cases where \(\operatorname{Cov}\left(x_{1}, x_{2}\right) \neq 0 ?\) d. The correlation coefficient between two random variables is defined as \\[ \operatorname{Corr}\left(x_{1}, x_{2}\right)=\frac{\operatorname{Cov}\left(x_{1}, x_{2}\right)}{\sqrt{\operatorname{Var}\left(x_{1}\right) \operatorname{Var}\left(x_{2}\right)}} \\] Explain why \(-1 \leq \operatorname{Corr}\left(x_{1}, x_{2}\right) \leq 1\) and provide some intuition for this result. e. Suppose that the random variable \(y\) is related to the random variable \(x\) by the linear equation \(y=\alpha+\beta x\). Show that \\[ \beta=\frac{\operatorname{Cov}(y, x)}{\operatorname{Var}(x)} \\] Here \(\beta\) is sometimes called the (theoretical) regression coefficient of \(y\) on \(x\). With actual data, the sample analog of this expression is the ordinary least squares (OLS) regression coefficient.

Suppose that \(f(x, y)=x y .\) Find the maximum value for \(f\) if \(x\) and \(y\) are constrained to sum to \(1 .\) Solve this problem in two ways: by substitution and by using the Lagrange multiplier method.

One of the most important functions we will encounter in this book is the Cobb-Douglas function: \\[ y=\left(x_{1}\right)^{\alpha}\left(x_{2}\right)^{\beta} \\] where \(\alpha\) and \(\beta\) are positive constants that are each less than 1 a. Show that this function is quasi-concave using a "brute force" method by applying Equation 2.114 b. Show that the Cobb-Douglas function is quasi-concave by showing that 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 \(\alpha+\beta>1\) then the Cobb-Douglas function is not concave (thereby illustrating again that not all quasiconcave functions are concave). Note: The Cobb-Douglas function is discussed further in the Extensions to this chapter.

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