Answers to Econ 210A Midterm, October A. The function f is homogeneous of degree 1/2. To see this, note that for all t > 0 and all (x 1, x 2 )

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1 Question. Answers to Econ 20A Midter, October 200 f(x, x 2 ) = ax {x, x 2 } A. The function f is hoogeneous of degree /2. To see this, note that for all t > 0 and all (x, x 2 ) f(tx, x 2 ) = ax {tx, tx 2 } = t ax {x, x 2 } = t /2 ax {x, x 2 } () = t /2 f(x, x 2 ) B. The function f is neither concave, nor quasi-concave. One should be able to see this quickly by drawing an indifference curve and seeing that the at-leastas-good set is not convex. Therefore the function is not quasi-concave. Since a concave function ust be quasi-concave, f cannot be concave either. One can show that the function is not quasi-concave by eans of a single exaple. Let x = (, 0) and x = (0, ). Then f(x) = f(x ) =. Now consider the convex cobination tx + ( t)x = (t, t) where 0 < t <. If f is quasi-concave, then it ust be that f(tx + ( t)x ) f(x ). But f(tx + ( t)x ) = t < f(x ) =. So f cannot be quasi-concave. We also note fro this sae exaple that tf(x) + ( t)f(x ) = > f(tx + ( t)x ) = t. But if f is a concave function tf(x) + ( t)f(x ) f(tx + ( t)x ), so x is not a concave function. C The function f is not a convex function. The square root sign should be a tipoff. Square root is a strictly concave function. To show that f is not a convex function, consider the two points x = (, 0) and x = (4, 0). Then f(x) = and f(x ) = 2. Therefore 2 f(x) + 2 f(x ) = 2.5. But f( 2 x + 2 x ) = f(2.5, 0) = 2.5 < 2 f(x) + 2 f(x ) The function f is a quasi-convex function. A quick way to see that this ust be true is to look at an indifference curve and see that it looks like the worse-than sets ust be convex. To show this forally, suppose f(x, x 2 ) f(x, x 2). Then f(x, x 2 ) = ax x, x 2, x 3, x 4 If 0 < t <, then f(tx + ( t)x ) = ax{tx + ( t)x 2, tx + ( t)x 2} ax x, x 2, x 3, x 4 = f(x, x 2 )

2 Therefore if f(x) f(x ), then f(tx + ( t)x ) f(x ) which eans that f is quasiconcave. Question 2. Ms Rodent s utility function is Wild Thing s utility function is u(x, x 2 ) = (in{x, x 2 }) 2. u(x, x 2 ) = ax{x, x 2 }. A. Ms Rodent s Marshallian deand function is x (p, ) = x 2 (p, ) = Her indirect utility function is ( v(p, ) = ) 2 Her expenditure function is e(p, u) = u( ). Her Hicksian deand function is h (p, u) = u h 2 (p, u) = u B. Wild Thing s deand function is defined as follows: If p < p 2, x (p, ) = p and x 2 (p, ) = 0. If p > p 2, x (p, ) = 0 and x 2 (p, ) = p 2. If p = p 2 = p, then Wild Thing can axiize his utility at either of two possible deand vectors. These are (x, x 2 ) = ( p, 0), and (x, x 2 ) = (0, p ). Wild Thing s indirect utility function is His expenditure function is v(p, ) = in{p, p 2 } e(p, u) = in {p, p 2 }u If p < p 2, his Hicksian deands are h (p, u) = u and h 2 (p, u) = 0. If p > p 2, his Hicksian deands are h (p, u) = 0 and h 2 (p, u) = u. 2

3 If p = p 2 = p, then there are two bundles either of which would iniize cost of achieving utility u. These are (/p, 0) and (0, /p). C. The Slutsky substitution atrix for Ms Rodent is ( 0 0 ) 0 0 The substitution effect of a price change is zero. If prices are changed, she will continue to consue equal aounts of the two goods. If her incoe is adjusted so that she is exactly as well off as before the price change, she will consue exactly as uch of each bundle as before the price change. Question 3. f(x, x 2 ) = x + x 2 + ax x 2 x2 2 x2 2 2 A. The gradient of f is the vector ( + ax 2 x, + ax x 2 ). Evaluated at (x, x 2 ) = (, 2), this gradient is (2a, a ). B. The Hessian of f is ( a ) a C. The function f will be concave if and only if the Hessian is negative seidefinite. This will be the case if the deterinants of the odd-nubered principle inors are nonpositive and the even-nubered are nonnegative. In the case of our atrix this eans that the diagonals ust be non-positive and the 2x2 Hessian ust be non-negative. In out case, both diagonals are negative. The deterinant of the Hessian is a 2 which is nonnegative if and only if a. So f is concave if and only if a. D. The point at which the gradient of f is zero will be a global ax if f is strictly concave. The function f will be strictly concave if < a <. The gradient of f is zero at (x, x 2 ) if + ax 2 x = 0 and + ax x 2 = 0. Solving these two equations, we find x = x 2 = a. Assuing that < a <, at the point of axiization, we have f(x, x 2 ) = 2 a + a ( a) 2 = a. Notice that if a, then f(x, x) = 2x + (a )x 2, which can be ade arbitrarily large by choosing x large enough. So f has no axiu. 3

4 If a <, then f(x, x) = ( a )x 2 where a > 0. Therefore f can be ade arbitrarily large by choosing x large enough. So again f has no axiu. We have one ore case to worry about. What if a =? Here is one way to dispose of this case. If a =, then f is concave, but not strictly concave. If there is a global axiu, it ust be at the point ( ) x 0 = a, = (/2, /2). a Consider the function F (t) = f(x 0 + ty). Consider the Taylor s series expansion of F, which is F (t) = F (0) + tf (0) + t2 2 F (t) + ultiples of higher derivatives of F. Since the Hessian H(x 0 ) of f is negative seidefinite when a =, we know that F (t) = i j H ijy i y j 0 for all y. The second derivative F (t) does not depend on t. So all higher derivatives are zero. Therefore F (t) = F (0) + tf (0) + t2 2 F (t) 0. This eans that for all y and all t, F (x 0 + ty) F ( 0 ) which iplies that f achieves a global axiu at x 0 = (/2, /2). At this point, f(x 0 ) = /2. Here is an alternative way to show the result for a =. This ethod has the advantage of giving us a full description of all the points at which f takes a global axiu. If a =, then f(x, x 2 ) = 2 ( (x + x 2 ) 2) + 2. The first ter of this expression is necessarily non-positive and is axiized when x + x 2 =. Therefore f(x, x 2 ) is axiized whenever x + x 2 = and at any such point takes a axiu value of /2. Question 4. A. Euler s theore on hoogeneous functions is as follows: If f is a real valued function of n variables and if f is hoogenous of degree k, then n f(x,..., x n ) x i = kf(x,..., x n ). B. If f is hoogeneous of degree k, then partial derivative functions f i (x,..., x n ) are hoogeneous of degree k. C. Proof of Euler s theore: If f is hoogeneous of degree k, then f(tx,..., tx n ) = t k f(x,..., x n ) 4

5 for all t > 0 and all x in the doain of f. Therefore the derivative with respect to t of the left side of this identity is equal to the derivative of the right side with respect to t. This iplies that n x i f(tx,..., tx n ) = kt k f(x,..., x n ) for all t > 0. In particular, this equation holds when t =, which eans that n x i f(x,..., x n ) = kf(x,..., x n ). Proof that if f is hoogeneous of degree k, its partial derivatives are hoogeneous of degree k : If f is hoogeneous of degree k, then f(tx,..., tx n ) = t k f(x,..., x n ) for all t > 0 and all x in the doain of f. equation with respect to x i. This gives us Differentiate both sides of this t f(tx,..., tx n ) = t k f(x,..., x n ). Dividing both sides of this equation by t, we have f(tx,..., tx n ) = t k f(x,..., x n ) which eans that the function f(x,...,xn) is hoogeneous of degree k. 5