Unformatted text preview: Fall 2004 F INAL E XAM  A NSWER
KEY ARE211 Problem 1 (25 points). Let f : R2 R and g : R2 R be two concave functions and fix R2 . Let Y = {(y1 , y2 ) R2 : x1 , x2 R2 s.t. y1 f (x1 ), y2 g(x2 ) and x1 + x2 }. To give you some intuition, functions f and g might be interpreted as production functions, x1 and x2 as input vectors and Y as a production set. A) (3 points) Give a formal definition of a convex set.
A set S is convex if x, y S and all [0, 1], x + (1  )y S. B) (3 points) Give a formal definition of a concave function.
A function f : X Y is concave if the set of points below the graph of the function, i.e., {(x, y) X Y : y f (x)}, is a convex set. C) (19 points) Using your answers to parts a) and b), prove formally that Y is convex.
Pick (y1 , y2 ) Y and [0, 1]. We need to prove that y = (1  )y1 + y2 Y . For i = A, B, since yi Y , there exists xi1 , xi2 R2 such that yi f (xi1 ), yi g(xi2 ) and xi1 + xi2 . For j = 1, 2, de2 1
fine x j = (1  )xA j + xB j . Since f and g are concave, we have, f (x1 ) (1  ) f (xA1 ) + f (xB1 ) and g(x2 ) (1  )g(xB2 ) + g(xB2 ). Therefore y 1 y 2 = = (1  )yA + yB 1 1 (1  )yA + yB 2 2 (1  ) f (xA1 ) + f (xB1 ) (1  )g(xA2 ) + g(xB2 ) f (x1 ) g(x2 ) Clearly, x1 + x2 . Therefore, x1 and x2 satisfy the conditions required for y = (y , y ) to belong 1 2 to Y . Problem 2 (25 points). Let f : Rn R and fix x Rn . A) (9 points) Suppose that at x, all of the partial derivatives of f exist. a) Give a mathematical definition of a partial derivative. 2 For i = 1, ..., n, let ei Rn be defined by, for j = 1, ..., n: eij = derivative of f at x0 is now defined by 1 if j = i otherwise . The i'th partial 0 lim f (x0 + ei /k)  f (x0 ) 1/k f (x0 ) xi = k b) Explain intuitively what this definition means in terms of the graph of f .
It's the slope at x0 of the crosssection of the graph of f obtained by "slicing" the graph through the point vertically above x0 and in the direction of ei . c) Can you say that f is differentiable at x?
No, existence of partials does not imply differentiability. See for example Figure 3 in the lecture notes CALCULUS2. d) Does your answer to the previous question change if you are told that f has directional derivatives in all directions at x?
No, existence of all directional derivatives does not imply differentiability. An example was provided in 4.3.7 of the lecture notes CALCULUS2 B) (16 points) Now suppose that f is differentiable at x. and denote by d f x the differential of f at x. a) Does f have directional derivatives in all directions at x?
Yes, differentiability implies the existence of all directional derivatives. b) If so, for h Rn , specify the directional derivative of f at x in the direction h in terms of the differential. f (x) h/h. c) Can you say that the partial derivatives of f , if they exist, are continuous at x?
No. Existence of continuous partials is a sufficient but not necessary condition for differentiability. f (x) xi 's, d) Can you say that the partial derivatives of f at x, i.e., the
No. f (x) xi are continuous? is a point, and it doesn't make sense to say that a point is continuous. Problem 3 (25 points). 2/3 1/3 x1 x2 A) (9 points) Consider the function f : R2 R2 defined by f (x) = . Use the differential to 1/3 2/3 x1 x2 estimate the value of f () at (997, 29). 3 1/3 1/3 2/3 2/3 1/3x1 x2 2/3x1 x2 J f (x) = . Now (997, 29) = (1000, 27) + (3, 2) so we'll evaluate 2/3 1/3 2/3 1/3 1/3x1 x2 2/3x1 x2 the differential of f at x = (1000, 27) at the point dx = (3, 2). Now 200 100 3 1 27  3 306.8074 300 + 27  5 300 5 f (x) + J f (x)dx = = = + 20 9 3 94.3544 2 90 + 40  100 90 100 9 9
So that the B) (8 points) Consider the function f : R R defined by f (x) = x4 . Use a secondorder Taylor approximation of f about x = 2 to estimate the value of f (3).
The second order Taylor expansion of f about x is 4x3 dx+6x2 dx2 . Evaluating at x = 2 and dx = 3 2 = 1, we obtain T2 ( f , 2, 1) = 56. So the second order Taylor approximation to f (3) is f (2) + 56 = 72. C) (8 points) At what point must one compute f () so that the second order Taylor expansion of f about x = 2 (including the remainder term) delivers exactly the difference between f (3) and f (2)?
Since f (3) = 81, the second order Taylor approximation to f (3) is 9 short of its target. We need to make this up by choosing y so that the third order Taylor term T f3 (y, dx) = 4ydx = 9. Since dx = 1, we have y = 9/4. That is, we need to evaluate f () at 9/4. 4
x2 {x : g(x) = 0} F IGURE 1. Quasiconvex constraint set Problem 4 (50 points). Let f : R2 R and g : R2 R be two differentiable functions. Consider the following NPP problem, which we shall refer to as (4). max f (x) s.t. g(x) 0
x0 where (C0) g is a quasiconvex function. A) (4 points). Draw a constraint set that is consistent with assumption (C0).
See Fig. 1 B) (12 points). Assume in this part that f is not quasiconcave. (a) A local solution to (4) is a point x R2 such that for some neighborhood U of x, f (x) f (x ), for all x = x U such that g(x ) 0. Say that a local solution is a strict local solution if f (x) > f (x ), for all x = x U such that g(x ) 0. Show graphically that a point x may exist that is a strict local solution to (4) but not a solution.
See Fig. 2. x is a local solution, but x belongs to the feasible set and f (x ) > f (x). (b) Show graphically that a point x may exist that satisfies the KT conditions but is not a local solution to (4).
See Fig. 3. C) (12 points). Fix x R2 and suppose that (C1) at x the KT conditions are satisfied. (C2) f is quasiconcave (a) Give a condition of f () which, differs from M.K.9 g x1 (4) D) (12 points). An alternative to conditions (C2) and (C3) is the one referred to in the lecture notes as condition M.K.9. If f satisfies M.K.9 and x satisfies (C1), then f is a solution to (4). (b) Provide a graphical example demonstrating that if (C1) and (C2) are satisfied but (C3) is not, F IGURE 2. A local solution that is not a solution x2 x x x2 x level set of f (a) write down the assumption M.K.9 on the constraint set at this point. Denote this condition by (C3). functions f and g that map R to R. then x may fail to be a solution to (4). If you prefer, your example can be constructed using See Fig. 4. The gradient of f vanishes at x, the KT conditions are satisfied but f is not maximized (C3) The gradient of f never vanishes. That is, for all x in the domain of f , f (x) = 0 together with (C1) and (C2), guarantees that x is a solution to (4). F IGURE 3. A point satisfying KT that is not a local solution x, x X, if f (x ) > f (x) then f (x) (x  x) > 0. g g x1 x1 5 6
x2 x f (x)
x1 Constraint set F IGURE 4. A point satisfying (C1) and (C2) that is not a solution
x2 x
Constraint set
x1 f (x) F IGURE 5. A point satisfying M.K.9 but not (C3) (b) explain why it is preferable to assume M.K.9 than to assume both (C2) and (C3). Your explanation should include an example. If you prefer, your example can be constructed using functions f and g that map R to R.
In general, one wants to impose the weakest possible assumption which accomplishes the task at hand. M.K.9 is much weaker than (C3), and hence preferable on general grounds. More specifically, if (C3) is imposed, then one cannot obtain an unconstrained solution to the NPP. Fig. 5 provides an example of an NPP that one would not want to exclude. E) (10 points). Prove formally that if f is a concave function, then f satisfies M.K.9. Hint: Use the fact that f is differentiable.
If f is concave, then for all x, x , f (x)(x  x) f (x )  f (x). Therefore, if f (x )  f (x) > 0, then f (x)(x  x) f (x )  f (x) > 0. Hence M.K.9 is satisfied. 7 Problem 5 (50 points). Fix > 0 and define h(; x, y) = e(x+y) + x2 + y4 . Now consider the unconstrained maximization problem minimize h(; , ) on R2 For your convenience, Fig. 6 on the last page of the exam plots z = e . Also recall that
de d (5) = e . A) (6 points). Compute h(; 0, 0). e0 + 02 + 04 e0 h(; 0, 0) = = = 1 B) (6 points). Use your answer to part A) to prove that the solution to (5) is the same as the following constrained minimization problem: minimize h(; , ) such that x [1, 1] and y [1, 1].
If either x > 1 or y > 1, then either x2 or y4 exceeds 1, while the other is positive. Moreover, h(; , ) is nonnegative function. Hence, h(; x, y) > e(x+y) + 1 > 1 = h(; 0, 0) . Conclude that a necessary condition for h(; , ) to be minimized at (x, y) is that max(x, y) 1. C) (6 points). Use your answer to part B) to prove that (5) has a solution. Here and later in the question, you may take it as given that h is a continuously differentiable function on R R R++ .
Since h(; , ) is continuous, and the set [1, 1]2 is a compact set, the Weierstrass theorem implies that h(; , ) attains a minimum on this set. Part B) then implies that this minimum is in fact an unconstrained minimum for h(; , ). This minimum is the solution to (5). D) (6 points). For arbitrary x, y R2 , compute the Hessian of h(; , ) at (x, y). Hh(; x, y) = 2 e(x+y) + 2 2 e(x+y) 2 e(x+y) 2 e(x+y) + 12y2 E) (6 points). Use your answers to C) and D) to prove that (5) has a unique solution. 8 The first principal minor of Hh(; x, y) is 2 e(x+y) + 2 which is positive. The second principal minor is det(Hh(x,y) (; x, y)) = 24y2 + (2 + 12y2 )2 e(x+y) > 0 This establishes that h(, , ) is strictly convex. A strictly convex function has at most one global minimum. Combining this with C), we have established that (5) has a unique solution. F) (6 points). a) Write down the first order conditions for (5).
The first order conditions are h(; x, y) x h(; x, y) y = = e(x+y) + 2x e(x+y) + 4y3 = = 0 0 b) Prove that a solution to these first order conditions exist and is unique.
For a strictly convex function, the first order conditions are necessary and sufficient for a unique solution. Since we've proved in C) that (5) has a solution, it follows that there exists a unique pair (x, y) which solves the equation system in F). c) Show that the unique solution to the FOC can be written as the level set corresponding to 0 of a function from R3 to R2 (to be denoted by f ).
Define the function f : R3 R2 , defined by f (, x, y) = + 2x (x+y) + 4y3 e e(x+y) For each > 0, the FOC specify that f (, x, y) must equal zero. Hence these conditions can be written as the level set corresponding to 0 of f . G) (6 points). Show that for all > 0, you can apply the implicit function theorem to the function f given in Fc).
Consider the function f (, , ). Clearly, J f(x,y) (, , ) is the Hessian Hh(x,y) specified above. We've already checked that this matrix has a positive determinant. Hence the condition of the implicit function is satisfied. H) (8 points). Given > 0, let (x (), y ()) denote the solution to (5). Use Cramer's Rule to find an expression for dx () d . 9 Plugging through the implicit function theorem, we obtain dx () d = 12y ()2 e(x ()+y ()) 1 + (x () + y ()) 24y ()2 + (2 + 12y ()2 )2 e(x ()+y ()) 10 1 0  0 F IGURE 6. The graph of z = e ...
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This note was uploaded on 08/01/2008 for the course ARE 211 taught by Professor Simon during the Fall '07 term at Berkeley.
 Fall '07
 Simon

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