Static Problem Set 2 Solutions

Static Problem Set Solutions Jonathan Kreamer July, 0 Question (i) Let g, h be two concave functions. Is f = g + h a concave function? Prove it. Yes. Proof: Consider any two points x, x and α [0, ]. Let x = αx + ( α)x. Then we have f(x) = g(x) + h(x). By concavity of g, h, this implies f(x) αg(x ) + ( α)g(x ) + αh(x ) + ( α)h(x ). Rearranging, we get f(x) α [g(x ) + h(x )] + ( α) [g(x ) + h(x )], which implies f(x) αf(x ) + ( α)f(x ). (ii) Let g, h be two quasiconcave functions. Is f = g + h a quasiconcave function? Prove it. x, x False. Proof by counter-example. Consider the two functions defined on [0, ], g(x) = 0, x and 0, x h(x) = x, x x, x. Then observe that their sum is the function f(x) = x, x, which is not quasiconcave. Question Let f : R + R + be defined as f(x, y) = x α y β, α 0, β 0. (i) Use two different ways to prove that f is quasiconcave. Proof : We use the theorem that a function f : R n R is quasiconcave if its largest n leading principle minors of the bordered Hessian alternate in sign, with the smallest of these (the third) positive. Here the bordered Hessian is H = 0 αx α y β βx α y β αx α y β α (α ) x α y β αβx α y β βx α y β αβx α y β β (β ) x α y β The third leading principle minor of H is the determinate of H, which is equal to H = ( αx α y β) αx α y β βx α y β αβx α y β β (β ) x α y β + ( βx α y β ) αx α y β βx α y β α (α ) x α y β αβx α y β H = α βx 3α y 3β + αβ x 3α y 3β So we have H = αβ (α + β) x 3α y 3β. Since we have α, β, x, y 0, it follows that we have H > 0, and therefore we have that f is quasiconcave. Proof : We use the fact that monotonic transformations preserve quasiconcavity. Let h (x, y) = α log (x) + β log (y). Observe that h is concave, because log is concave, and the sum of concave functions is

concave. h is also quasiconcave, because concave functions are quasiconcave. Now observe that f = g h, where g (x) = e x is a monotonic transformation. Therefore f is a monotonic transformation of a quasiconcave function, and is therefore quasiconcave. Note: Why is quasiconcavity preserved under monotonic transformations? One definition of quasiconcavity is that all upper level sets are convex. Monotonic transformations preserve level sets, and therefore they preserve all properties that are defined by upper level sets, including quasiconcavity (but not concavity!). (ii) Provide a condition on α and β such that f is concave. As we showed in problem set, an equivalent condition for f to be concave is that f < 0. The analogous condition in R + is that the Hessian f is negative semi-definite. Here the Hessian is [ ] α (α ) x f (x, y) = α y β αβx α y β αβx α y β β (β ) x α y β H is negative semi-definite if the leading principle minors alternate in signs, and the first is negative. Therefore we need α (α ) x α y β < 0 and αβ [ α β] x α y β > 0. Since we have α, β, x, y 0, the first is true if and only if α. The second is true if and only if α + β. Note that the second condition (together with α, β 0) implies the first, and that it also implies β. So a condition for f concave is α + β. Question 3 (i) Prove the following theorem from class: Suppose f : R n R. Then x solves max x R n f(x) if and only if x solves min x R n f(x)}. Proof by contradiction. Suppose we have x that solves max x R n f(x) but not min x R n f(x)}. Then x R n such that f(x ) < f(x ). But then we have f(x ) > f(x ), which contradicts that x solves max x R n f(x). To prove the other direction, suppose we have x that solves min x R n f(x)} but not max x R n f(x). Then there exists x such that f(x ) > f(x ). But then we have f(x ) < f(x ), which contradicts that x that solves min x R n f(x)}. (ii) Show that if f is C and x arg min x f (x), then f(x ) = 0. Since f is C, we have that the two-sided derivative exists. In particular, for each x i we have f(x ) = f(x lim +he i) f(x ) and also f(x ) f(x = lim ) f(x he i), where e i is the basis vector in the direction of x i. x arg min x f (x) implies that for all x, we have f(x ) f(x ). In particular, we have f(x + he i ) f(x ) and f(x he i ) f(x ), which imply f(x + he i ) f(x ) 0 and f(x ) f(x he i ) 0. Therefore we have f(x ) 0 and f(x ) 0, which together imply f(x ) = 0. Since this is true for all x i, we have f(x) = 0. (iii) Suppose f : R R. Show that if f is C and x is a local minimum of f(x), then f(x ) 0. First, to review a definition: x is a local minimum of f(x) if there exists ɛ > 0 such that for all x such that x x < ɛ, we have f(x ) < f(x ). Now consider such an x. The matrix has ij entry corresponding to the cross derivative with respect to i and j. In other words f(x ) dh i h j We can write this sum as = lim h 0 f(x + he i + he j ) f(x + he i ) f(x + he j ) + f(x ) h

f(x ) [f(x + he i + he j ) f(x )] [f(x + he i ) f(x )] [f(x + he j ) f(x )] = lim dh i h j Now we use the fact that x is a local minimum of f. This implies that f(x + he) f(x ) 0. Therefore every term in the numerator is nonnegative, and since we started with generic ij, this holds for every term of the Hessian. Therefore we have f(x ) 0. Question 4 Let f : R R be defined as f(x) = sin(x). Find all the local minima and maxima of f and identify which is which. Do the same for g(x) = x + sin(x). I start with f(x) = sin(x). Since sin(x) is C and defined everywhere on R, the necessary condition for a maximum or a minimum is f(x) = 0. Here, this condition is cos(x) = 0, which is true for x = π + nπ, n Z. To check which of these points are local maxima or minima, we check the second-order sufficient condition f(x) < 0 for a maximum and f(x) > 0 for a minimum. Here we have f(x) = sin(x), which is negative at x = π, 5π, 9π,..., and positive for x = 3π, 7π,... Therefore x = π + nπ is a minimum for n odd and a maximum for n even. Now turning to g(x) = x + sin(x), I find the points that meet the necessary condition g(x) = 0. Here these are those x that satisfy cos(x) =. This is true for x = π, 3π, 5π..., or x = (n + )π for n Z. Turning to the sufficient condition, we compute g(x) = sin(x), which at x = (n + )π is equal to 0. This indicates that these points are neither maxima nor minima, but inflection points. We can see this from the graph as well. Question 5 A function g : R R is said to satisfy the single-crossing property if there is x R such that g(x ) = 0, g(x) 0 for all x x and g(x) 0 for all x x (or if the inequalities are reversed). (i) Suppose f : R R is C. Show that if f is concave, then f (x) is decreasing. Show that if f has a local extremum, then f (x) satisfies the single-crossing property. Note: I interpret the second statement to mean that if f has a local extremum and f is concave, then it satisfies the single-crossing property. Proof. Consider arbitrary points x, x such that x < x. Let m be the slope of the line connecting the f(x ) and f(x ). That is, m = f(x) f(x) x x. Since f is is C, the two-sided limit of f exists at both x and x. Therefore we have f f(x (x ) = lim +h) f(x ) for h smaller than x x. Therefore we have f f(x (x ) lim )+mh f(x ) h 0. Since f is concave, we have f(x + h) f(x ) + mh h = m. We likewise have f f(x (x ) = lim ) f(x h). Again by concavity, we have f(x h) f(x ) mh. Therefore we have f f(x (x ) lim ) f(x )+mh h 0 = m. Together this implies f (x ) f (x ). h Proof. Suppose f has a local extremum. Suppose this occured at x. Then as we showed in Question 3, we have f (x ) = 0. Since f is concave, it follows that f is decreasing. Therefore for all x > x we have f (x) < 0, and for all x < x we have f (x) > 0. (ii) Suppose f : R R is C. Show that if f is quasi-concave and has a global extremum, then f (x) satisfies the single-crossing property. Construct an example such that f is quasi-concave, but f (x) is not decreasing. Let f : R R be C, quasi-concave, and have a global maximum at x. Suppose there exists x > x such that f (x) > 0. Then lim f(x) f(x h) h 0 + h > 0. This implies that there exists x < x such that f(x ) < f(x) (just let x = x h in the expression for the derivative at x). Now consider the level set of f associated with the value h = f(x)+f(x ). We have f(x) > h > f(x ). We also have f(x ) > h. Therefore x and x are 3

in the upper-level set of h, but x is not. Since x lies between x and x, this contradicts the fact that f is quasi-concave. Therefore we have shown that for all x > x, we have f (x) 0. The same argument can be used to show that for x < x we have f (x) 0. Therefore f (x) satisfies the single-crossing property. An example of a quasi-concave function with a local maximum that is not concave is f(x) = e x (a non-normalized normal pdf). Question 6 Let f : R 3 R be given by f(x, y, z) = xyz. Compute the Jacobian and Hessian of f. Answer: Recall that the Jacobian is the vector of first-order partial derivatives and the Hessian is the matrix of second-order partial derivatives. Assuming the order (x, y, z), the Jacobian is and the Hessian is Question 7 J = H = yz xz yz 0 z y z 0 x y x 0 (i) Let f : R N R, and let g : R R be any strictly monotonic function. Show that the level sets of f and the level sets of h = g f are the same. The level sets of f and h are the same if f(x) = f(y) g(x) = g(y). To show the first direction, suppose that f(x) = f(y). Then h(x) = g(f(x)) = g(f(y)) = h(y). To show the other direction, we use proof by contradiction. Suppose not. Then there exists x, y such that h(x) = h(y), but f(x) f(y). Then we must either have f(x) > f(y) or f(x) < f(y). Suppose the first. Then it follows by strict monotonicity of g that we either have g(f(x)) > g(f(y)) or g(f(x)) < g(f(y)), which contradicts that h(x) = h(y). Then we must have f(x) < f(y). But this implies, again by monotonicity of g, that either g(f(x)) > g(f(y)) or g(f(x)) < g(f(y)), which contradicts that h(x) = h(y). Therefore we have contradicted f(x) f(y), and we must have f(x) = f(y). (ii) Show that the function ( log ( (x 3 + y 3 + z 3) 3 )) 4 has the same level sets as the function ( x 3 + y 3 + z 3) 3. This is false. Proof by counter-example. Call the first function f and the second function g. Consider v = (x, y, z ) such that g(v ) = e and v = (x, y, z ) such that g(v ) = e. Then v and v are not in the same level set of g. However, f(v ) = (log (g(v ))) 4 = and f(v ) = (log (g(v ))) 4 =. So v and v are in the same level set of f. Note: The functions have the same level sets on the plane x 3 + y 3 + z 3, because on this plane all transformations are monotonic. (iii) Solve for the level set of f(x, y) = ln x + ln y containing the point (5, 5). By the properties of logarithms, f(x, y) = ln xy. The level set containing (5, 5) is the set of points such that ln xy = ln 5. Exponentiating both sides, we get that this is the set of points such that xy = 5. 4

Question 8 Let f : R R and g : R R be defined as follows x, x < 0.5 f(x) = 0, x 0.5 g(x) = x, x < 0.5, x 0.5 (i) Find the value of x which maximizes f if it exists. If not, explain what the problem is. Such an x does not exist. We can show this by contradiction. Suppose there was an x that maximized f. If x 0.5, then we can choose y = 0.5 for which f(y) > f(x), contradicting that x is a maximum of f. If x < 0.5, we can choose y = x+0.5. Then f(y) > f(x), again contradicting that x is a maximum of f. The reason there does not exist a maximum is that f is not upper semi-continuous. (ii) Find the value of x which maximizes g if it exists. If not, explain what the problem is. Any x [0.5, ) is a global maximum of g. A unique global maximum does not exist because g is not strictly concave. Question 9 Prove that the product of any two continuous functions is continuous. Lemma: If x n x and y n y for x n, y n R, then x n y n xy. (SB.3) Proof: Let x n x and y n y for x n, y n R. Take an arbitrary ɛ > 0. Now choose ɛ = min ( ɛ, ɛ ( x + y ) ). By convergence of x n and y n, there exist numbers N and N such that, for any n > N and n > N, we have x x n < ɛ and y y n < ɛ. Now pick N = max (N, N ). Choose any n > N and consider the distance xy x n y n. We can write this as xy x n y + x n y x n y n. By the triangle inequality, we have xy x n y + x n y x n y n xy x n y + x n y x n y n We pull out the constants y and x n to get y x x n + x n y y n. By convergence of x n and y n we have y x x n + x n y y n < y ɛ + x n ɛ Now we just want to get rid of the x n. To do this we write y ɛ + x n ɛ = y ɛ + x n x + x ɛ. By the triangle inequality again, we get y ɛ + x n x + x ɛ y ɛ + x n x ɛ + x ɛ < ( x + y ) ɛ + ɛ ɛ Since we have ɛ ( x + y ), it follows that ( x + y ) ɛ + ɛ ɛ + ɛ. Since we have ɛ ɛ, it follows that ɛ ɛ and so ɛ + ɛ ɛ + ɛ = ɛ. So we have shown that for any ɛ > 0 we can find an integer N such that for any n > N, we have x n y n xy < ɛ. In other words, x n y n xy. Armed with this lemma, we proceed as follows. Suppose that we have two continuous functions f and g, and suppose we have h = f g, meaning that h(x) = f(x)g(x). Suppose that x n x. Then by continuity of f and g, we have f(x n ) f(x) and g(x n ) g(x). By the lemma, we then have that f(x n )g(x n ) f(x)g(x). Using the definition of h, this implies that h(x n ) h(x). Therefore we have proved that h is continuous. 5

Question 0 Suppose a function f : R R is concave. (i) Show that the set of points in R n+ given by (x, y) : f(x) y} is a convex set. Call this set S. Consider points (x, y ) and (x, y ) in S. Now consider the point (x, y), with x = αx + ( α)x and y = αy + ( α)y for α [0, ]. We are done if we can show that (x, y) S. By the concavity of f, we have f(x) αf(x ) + ( α)f(x ). Since (x, y ) and (x, y ) are in S, we have f(x ) y and f(x ) y, and so αf(x ) + ( α)f(x ) αy + ( α)y = y So we have shown that f(x) y, and so (x, y) S. (ii) Show that the set of points in R n given by x : f(x) b 0} is a convex set. Call this set S. Consider points x, x S. Now consider the point x = αx + ( α)x. We are done if we can show that x S. By the concavity of f, we have f(x) αf(x ) + ( α)f(x ), which we can equivalently write f(x) b α (f(x ) b)+( α) (f(x ) b). Since x and x are S, we have f(x ) b 0 and f(x ) b 0. This implies that f(x) b 0, and so x S. Question Suppose a function f : R n R is both concave and convex. Show that f(x) = a x + b, where a is an n-dimensional vector and b is a real number. (That is, show that any function that is both convex and concave must define a hyperplane). Lemma : For any x, x and α [0, ], we have f(ax + ( α)x ) = αf(x ) + ( α)f(x ). Proof: Consider such x, x and α. Let x = αx + ( α)x. Since f is concave, we have f(x) αf(x ) + ( α)f(x ). Since f is convex, we have f(x) αf(x ) + ( α)f(x ). Therefore we have f(x) = αf(x ) + ( α)f(x ). Lemma : For any x, x and α >, we have f(ax + ( α)x ) = αf(x ) + ( α)f(x ). Proof: Let x = αx + ( α)x. Rearranging, we get x = α x + ( α) x. Since α >, we have α [0, ]. By Lemma, we have f(x ) = α f(x) + ( α) f(x ). Rearranging again, we obtain f(x) = αf(x ) + ( α)f(x ). Lemma 3: For any x, x and α < 0, we have f(ax + ( α)x ) = αf(x ) + ( α)f(x ). ( ) ( ) Proof: Let x = αx + ( α)x. Rearranging, we get x = α x + α x. Since α < 0, it ( ) ( ) follows that α [0, ], and so by Lemma we have f(x ) = α f(x) + α f(x ). Rearranging again, we obtain f(x) = αf(x ) + ( α)f(x ). Corollary : If t R, then f(tx) = tf(x) + ( t)f(0). Proof: Follows immediately from Lemmas - 3, with x = x, x = 0, and α = t. We can now prove the theorem. For any x R n, we can write x = n n i= nx ie i, where x i is the ith element of the vector x, and e i is the ith elementary vector of R n, that is the vector of all zeros with a in the ith place. Therefore we have f(x) = f ( n n i= nx ) ie i. By repeated application of Lemma, we can write f(x) = n n i= f(nx ie i ). Observing that nx i is just a real number, we can apply Corollary to the terms f(nx i e i ) to obtain f(x) = n n i= [nx if(e i ) + ( nx i )f(0)]. We can write this as f(x) = n i= f(0), which can be simplified to f(x) = n i= x i [f(e i ) f(0)] + f(0). We n n i= nx i [f(e i ) f(0)] + n can equivalently write this as f(x) = a x + b, with a i = f(e i ) f(0) and b = f(0). 6