2
Jan 18

Solution to exercise 6.1: how to use homogeneity

Solution to exercise 6.1: how to use homogeneity

Suppose a firm produces just one output y using three inputs x_1, x_2, and x_{3} according to the production function: y=x_1(x_2+x_3).  The prices of goods y,x_1,x_2,x_3 are p,w_1,w_2,w_3>0, respectively. We assume that the firm can neither produce negative quantities of y nor use negative quantities of the inputs, so that y,x_1,x_2,x_3\geq 0.

This is a good opportunity to learn using the homogeneity notion. Assuming, for simplicity, that a function f has positive arguments, we say that it is homogeneous of degree \alpha if f(tx_1,...,tx_n)=t^{\alpha}f(x_1,...,x_n) for all t>0. For example, our production function f(x_1,x_2,x_3)=x_1(x_2+x_3) is homogeneous of degree 2 and the cost function g(x_1,x_2,x_3)=w_1x_1+w_2x_2+w_3x_3 is homogeneous of degree 1. Let's see how this affects the properties of the profit function

\pi(x_1,x_2,x_3)=px_1(x_2+x_3)-(w_1x_1+w_2x_2+w_3x_3).

Suppose we scale the inputs by t>0, then

\pi(tx_1,tx_2,tx_3)=t^2px_1(x_2+x_3)-t(w_1x_1+w_2x_2+w_3x_3)

=t^2[px_1(x_2+x_3)-\frac{1}{t}(w_1x_1+w_2x_2+w_3x_3)].

This means the following: if we start with any bundle of positive inputs (x_1,x_2,x_3) and move along the ray (tx_1,tx_2,tx_3) to infinity, the value of the output px_1(x_2+x_3) stays fixed (positive), while the value of the inputs \frac{1}{t}(w_1x_1+w_2x_2+w_3x_3) tends to zero. For t sufficiently large, the value in the brackets [px_1(x_2+x_3)-\frac{1}{t}(w_1x_1+w_2x_2+w_3x_3)] becomes close to px_{1}(x_{2}+x_{3}). As there is the factor t^2 in front of the brackets, the profit function tends to infinity along such a ray. Since the initial bundle (x_1,x_2,x_3) is arbitrary, such rays cover the whole quadrant \{(x_1,x_2,x_3):x_1,x_2,x_3>0\}. So the profit tends to infinity along any ray and not only along the "diagonal" (x,x,x), as the guide says.

Homogeneity is a notion whose usefulness contrasts with its simplicity. See homogeneity of means, of varianceof standard deviation, of correlationof conditional variance, and application to the Gauss-Markov theorem. And the Cobb-Douglas function is also in this club.

12
Nov 16

Properties of standard deviation

Properties of standard deviation are divided in two parts. The definitions and consequences are given here. Both variance and standard deviation are used to measure variability of values of a random variable around its mean. Then why use both of them? The why will be explained in another post.

Properties of standard deviation: definitions and consequences

Definition. For a random variable X, the quantity \sigma (X) = \sqrt {Var(X)} is called its standard deviation.

    Digression about square roots and absolute values

In general, there are two square roots of a positive number, one positive and the other negative. The positive one is called an arithmetic square root. The arithmetic root is applied here to Var(X) \ge 0 (see properties of variance), so standard deviation is always nonnegative.
Definition. An absolute value of a real number a is defined by
(1) |a| =a if a is nonnegative and |a| =-a if a is negative.
This two-part definition is a stumbling block for many students, so making them plug in a few numbers is a must. It is introduced to measure the distance from point a to the origin. For example, dist(3,0) = |3| = 3 and dist(-3,0) = |-3| = 3. More generally, for any points a,b on the real line the distance between them is given by dist(a,b) = |a - b|.

By squaring both sides in Eq. (1) we obtain |a|^2={a^2}. Application of the arithmetic square root gives

(2) |a|=\sqrt {a^2}.

This is the equation we need right now.

Back to standard deviation

Property 1. Standard deviation is homogeneous of degree 1. Indeed, using homogeneity of variance and equation (2), we have

\sigma (aX) =\sqrt{Var(aX)}=\sqrt{{a^2}Var(X)}=|a|\sigma(X).

Unlike homogeneity of expected values, here we have an absolute value of the scaling coefficient a.

Property 2. Cauchy-Schwarz inequality. (Part 1) For any random variables X,Y one has

(3) |Cov(X,Y)|\le\sigma(X)\sigma(Y).

(Part 2) If the inequality sign in (3) turns into equality, |Cov(X,Y)|=\sigma (X)\sigma (Y), then Y is a linear function of X: Y = aX + b, with some constants a,b.
Proof. (Part 1) If at least one of the variables is constant, both sides of the inequality are 0 and there is nothing to prove. To exclude the trivial case, let X,Y be non-constant and, therefore, Var(X),\ Var(Y) are positive. Consider a real-valued function of a real number t defined by f(t) = Var(tX + Y). Here we have variance of a linear combination

f(t)=t^2Var(X)+2tCov(X,Y)+Var(Y).

We see that f(t) is a parabola with branches looking upward (because the senior coefficient Var(X) is positive). By nonnegativity of variance, f(t)\ge 0 and the parabola lies above the horizontal axis in the (f,t) plane. Hence, the quadratic equation f(t) = 0 may have at most one real root. This means that the discriminant of the equation is non-positive:

D=Cov(X,Y)^2-Var(X)Var(Y)\le 0.

Applying square roots to both sides of Cov(X,Y)^2\le Var(X)Var(Y) we finish the proof of the first part.

(Part 2) In case of the equality sign the discriminant is 0. Therefore the parabola touches the horizontal axis where f(t)=Var(tX + Y)=0. But we know that this implies tX + Y = constant which is just another way of writing Y = aX + b.

Comment. (3) explains one of the main properties of the correlation:

-1\le\rho(X,Y)=\frac{Cov(X,Y)}{\sigma(X)\sigma(Y)}\le 1.