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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.

Short Answer

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Question: Prove that a concave function is also a quasi-concave function. Answer: To prove that a concave function is also a quasi-concave function, we need to show that if function \(f\) is concave, it also satisfies the definition of a quasi-concave function. Since, for any two points \(x\) and \(y\) and for any scalar \(\alpha \in [0,1]\), a concave function \(f\) satisfies: $$ f(\alpha x + (1-\alpha) y) \ge \alpha f(x) + (1-\alpha) f(y) $$ We can see that for any \(x, y\) and \(\alpha \in [0,1]\), if \(f(x) \le f(y)\), then $$ f(\alpha x + (1-\alpha) y) \ge f(x) $$ and if \(f(x) > f(y)\), then $$ f(\alpha x + (1-\alpha) y) \ge f(y) $$ In both cases, we have that $$ f(\alpha x + (1-\alpha) y) \ge \min\{f(x), f(y)\} $$ meaning that \(f\) is also quasi-concave.

Step by step solution

01

Definition of Concave Function

A function \(f\left(x_{1}, x_{2}\right)\) is said to be concave if, for any two points \(x\) and \(y\) and for any scalar \(\alpha \in [0,1]\): $$ f(\alpha x + (1-\alpha) y) \ge \alpha f(x) + (1-\alpha) f(y) $$
02

Definition of Quasi-Concave Function

A function \(f\left(x_{1}, x_{2}\right)\) is said to be quasi-concave if, for any two points \(x\) and \(y\) and for any scalar \(\alpha \in [0,1]\): $$ f(\alpha x + (1-\alpha) y) \ge \min\{f(x), f(y)\} $$ Now, let's prove the given statement.
03

Comparing Concave with Quasi-Concave Definition

Suppose \(f\) is a concave function, then for any \(x, y\) and \(\alpha \in [0,1]\), $$ f(\alpha x + (1-\alpha) y) \ge \alpha f(x) + (1-\alpha) f(y) $$ Now, we want to show that \(f(\alpha x + (1-\alpha) y)\) is at least as big as the minimum of \(f(x)\) and \(f(y)\). Consider the two cases below: 1. If \(f(x) \le f(y)\), then $$ \alpha f(x) + (1-\alpha) f(y) \ge \alpha f(x) + (1-\alpha) f(x) = f(x) \Rightarrow f(\alpha x + (1-\alpha) y) \ge f(x) $$ 2. If \(f(x) > f(y)\), then $$ \alpha f(x) + (1-\alpha) f(y) \ge \alpha f(y) + (1-\alpha) f(y) = f(y) \Rightarrow f(\alpha x + (1-\alpha) y) \ge f(y) $$ In both cases, we have that \(f(\alpha x + (1-\alpha) y) \ge \min\{f(x), f(y)\}\), meaning that \(f\) is also quasi-concave.
04

Intuitive Reason

Intuitively, this result suggests that concave functions are a special case of quasi-concave functions where not only do they fulfill the quasi-concavity requirement, but are also "more concave" due to following the stricter condition posed by concavity.
05

Converse Statement

The converse is not always true. Quasi-concave functions are not necessarily concave. A counterexample is the function \(f(x) = x^3\), which we illustrate in the following steps: 1. \(f(x) = x^3\) is quasi-concave because for any \(x, y\) and \(\alpha \in [0,1]\): $$ f(\alpha x + (1-\alpha) y) = (\alpha x + (1-\alpha) y)^3 \ge \min\{x^3, y^3\} = \min\{f(x), f(y)\} $$ 2. However, \(f(x) = x^3\) is not concave because its second derivative \(f''(x) = 6x\) is positive for \(x>0\), which violates the condition for concavity (a concave function should have a non-positive second derivative). In conclusion, a concave function is always quasi-concave, but a quasi-concave function is not necessarily concave.

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Key Concepts

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

Concave Functions
Concave functions are an important concept in microeconomic theory. They help to describe how outputs change with respect to inputs in various economic models. A function is said to be concave if, for any two points within the function's domain, the line segment connecting these points lies entirely below or on the graph of the function. Mathematically, this is defined as:
  • For function \(f(x_1, x_2)\), and any points \(x\) and \(y\), with scalar \(\alpha\) such that \(0 \leq \alpha \leq 1\), the inequality \(f(\alpha x + (1-\alpha) y) \geq \alpha f(x) + (1-\alpha) f(y)\) holds true.
This property indicates that the average of the function is greater than or equal to the function at the average point, showcasing diminishing increases or even declines in value as inputs increase. In business and economics, this can model scenarios where returns or benefits decrease after a certain level of input is reached. A classic example is the utility function in consumer theory, where satisfaction from consumption experiences diminishing returns.
Quasi-Concave Functions
Quasi-concave functions are a broader category than concave functions. They are particularly useful in economic optimization problems where we deal with preferences, production, or utility functions. A function is quasi-concave if
  • For any points \(x\) and \(y\), and any value of \(\alpha\) where \(0 \leq \alpha \leq 1\), \(f(\alpha x + (1-\alpha) y) \geq \min\{f(x), f(y)\}\).
This condition implies that the function retains at least the minimal value of the two points it's evaluated at, ensuring the level sets of the function are convex sets. Quasi-concavity is often related to preferences in economics which are not strictly diminishing like concave functions. For instance, while concave utility functions always show risk-averse behavior, quasi-concave utility functions can model more nuanced preferences where the consumer could have varying levels of risk tolerance.
Mathematical Proof
In mathematical terms, proving properties like these involves rigorously showing that one condition implies another. To prove that a concave function is also quasi-concave, you'd compare their respective defining inequalities. Given:
  • Concavity: \(f(\alpha x + (1-\alpha) y) \geq \alpha f(x) + (1-\alpha) f(y)\)
  • Quasi-concavity: \(f(\alpha x + (1-\alpha) y) \geq \min\{f(x), f(y)\}\)
You start by assuming the stronger condition, concavity, and show it satisfies the weaker condition, quasi-concavity, through logical steps. Intuitively, once the average in concave functions is greater than individual values, it stretches to cover minimums in quasi-concave functions. A counterexample like \(f(x) = x^3\) shows that the reverse is not true; it demonstrates that a function can be quasi-concave but not concave, because its second derivative changes sign, which violates the necessary conditions for concavity. This highlights the nuanced differences in these mathematical constructs critical for economic theory.

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Most popular questions from this chapter

The height of a ball that is thrown straight up with a certain force is a function of the time ( \(t\) ) from which it is released given by \(f(t)=-0.5 g t^{2}+40 t\) (where \(g\) is a constant determined by gravity). a. How does the value of \(t\) at which the height of the ball is at a maximum depend on the parameter \(g\) ? b. Use your answer to part (a) to describe how maximum height changes as the parameter \(g\) changes. c. Use the envelope theorem to answer part (b) directly. d. On the Earth \(g=32\), but this value varies somewhat around the globe. If two locations had gravitational constants that differed by \(0.1,\) what would be the difference in the maximum height of a ball tossed in the two places?

Suppose a firm's total revenues depend on the amount produced ( \(q\) ) according to the function \\[ R=70 q-q^{2} \\] Total costs also depend on \(q\) \\[ C=q^{2}+30 q+100 \\] a. What level of output should the firm produce to maximize profits \((R-C) ?\) What will profits be? b. Show that the second-order conditions for a maximum are satisfied at the output level found in part (a). c. Does the solution calculated here obey the "marginal revenue equals marginal cost" rule? Explain.

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?

The definition of the variance of a random variable can be used to show a number of additional results. a. Show that \(\operatorname{Var}(x)=E\left(x^{2}\right)-[E(x)]^{2}\) b. Use Markov's inequality (Problem \(2.14 \mathrm{d}\) ) to show that if \(x\) can take on only non-negative values, \\[ P\left[\left(x-\mu_{x}\right) \geq k\right] \leq \frac{\sigma_{x}^{2}}{k^{2}} \\] This result shows that there are limits on how often a random variable can be far from its expected value. If \(k=h \sigma\) this result also says that \\[ P\left[\left(x-\mu_{x}\right) \geq h \sigma\right] \leq \frac{1}{h^{2}} \\]. Therefore, for example, the probability that a random variable can be more than two standard deviations from its expected value is always less than \(0.25 .\) The theoretical result is called Chebyshev's inequality. c. Equation 2.197 showed that if two (or more) random variables are independent, the variance of their sum is equal to the sum of their variances. Use this result to show that the sum of \(n\) independent random variables, each of which has expected value \(\mu\) and variance \(\sigma^{2},\) has expected value \(m \mu\) and variance \(n \sigma^{2}\). Show also that the average of these \(n\) random variables (which is also a random variable) will have expected value \(\mu\) and variance \(\sigma^{2} / n\). This is sometimes called the law of large numbers-that is, the variance of an average shrinks down as more independent variables are included. d. Use the result from part (c) to show that if \(x_{1}\) and \(x_{2}\) are independent random variables each with the same expected value and variance, the variance of a weighted average of the two \(X=k x_{1}+(1-k) x_{2}, 0 \leq k \leq 1\) is minimized when \(k=0.5\) How much is the variance of this sum reduced by setting \(k\) properly relative to other possible values of \(k\) ? e. How would the result from part (d) change if the two variables had unequal variances?

Consider the following constrained maximization problem: \\[ \begin{array}{ll} \text { maximize } & y=x_{1}+5 \ln x_{2} \\ \text { subject to } & k-x_{1}-x_{2}=0 \end{array} \\] where \(k\) is a constant that can be assigned any specific value. a. Show that if \(k=10\), this problem can be solved as one involving only equality constraints. b. Show that solving this problem for \(k=4\) requires that \(x_{1}=-1\) c. If the \(x^{\prime}\) s in this problem must be non-negative, what is the optimal solution when \(k=4 ?\) (This problem may be solved either intuitively or using the methods outlined in the chapter.) d. What is the solution for this problem when \(k=20 ?\) What do you conclude by comparing this solution with the solution for part (a)? Note: This problem involves what is called a quasi-linear function. Such functions provide important examples of some types of behavior in consumer theory-as we shall see.

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