Multivariate Calculus Solution 1

Transcription

1 Math Camp Multivariate Calculus Solution Hessian Matrices Math Camp In st semester micro, you will solve general equilibrium models. Sometimes when solving these models it is useful to see if utility functions are concave. One way of testing for concavity involves calculating the function s Hessian. Find the Hessian matrices of the following utility functions (these functions were used in previous homeworks and tests:. (Core Eam U(, u u u u Therefore H ] (Final Eam U(, + δ α α δ α u u (α δ α u u Therefore H ] (α δ α 3. (Homework problem U(,, 3 + δ 3 δ 3 3 u u 3 u 3 u δ 3 3 u 3 u u 3 u u 3 3 Therefore 3 H δ 3 3 If you find any typo please me: Maria_Jose_Boccardi@Brown.edu

2 Solutions Multivariate Calculus 4. (Midterm Eam U(, 3 ln( + 3 ln( 3 3 u u 3 3 u Therefore H ] 3 3 Slutsky equation The Slutsky Equation breaks changes in demand into income effects and substitution effects. Last semester for one of the homework problems, we were asked to calculate the Slutsky equation to find the substitution effect and the income effect. In the problem, the utility function was the same as in question 3 above, and it can be shown that the direct demand functions are (p, p, w w 3p (p, p, w w 3p Let the function (p, w : R 3 R be the combined demand function, where p (p, p. The Slutsky equation is as follows: D p + D w D p v, where the subscripts denote which derivative it is respect to (Note: these are total derivatives, not directional derivatives and v is the indirect utility function. Find D p (the total effect, D w (the wealth effect, and use the above Slutsky equation to compute D p v (the substitution effect. Hint: D p is a matri, D w is, and is. This involves matri multiplication, something we haven t covered yet, but that you should already know. (p, w ( w 3p w 3p ( (p, w (p, w Therefore Then we have that D p D w ( p p p p ( D w w w ( 3p 3p D p + D w ( w, w 3p 3p ( w, w 3p 3p ( w 9p w 9p p ( w 3p w 3p ( 3p 3p ( w 9p w 9p p w 9p p w 9p w 9p p 4w 9w D p v

3 Math Camp 3 Gradient If the graph of a function F : R 3 R lives in R 4, in which space does the gradient F live? The gradient lives in R 3 Integration Evaluate the following integrals:. 4 ydyd ˆ ˆ 4 ydyd ˆ ˆ ˆ ( y 4 d ( (4 d ( d ] ydyd ˆ ydyd ] y d ( ( ] d 3 4] d ] 3. dyd for the region bounded by y and y 3 In figure 3 we can see the integration region, where the two curves intersect at 3 and 3, therefore we can set the integral as ˆ 3 3 ˆ 3 dyd Therefore we have that

4 4 Solutions Multivariate Calculus Figure : Region ˆ 3 3 ˆ 3 dyd ˆ 3 3 ˆ 3 3 ˆ 3 3 y] 3 d (3 ] d 3 3 ] d ( 3 3 ] ( 3 4 ( ( ( y sin(yddy Let s first echange the order of integration, see graph 4, therefore we have that y sin(yddy ˆ Let s focus first in the inner integral, then we have that ˆ ˆ sin(ydy sin(ydy sin(ydyd cos(y ] cos( ] + cos(] cos( ] cos(

5 Math Camp 5 Figure : Region Now going back to the other integral and using substitution we have that ˆ sin(ydyd cos( ] d d ] f( cos u g( du d ˆ g( cos( d cos( d cos( d + cos udu g( sin u] + sin Sketch the following regions:., y 7 See figure., y See figure 3., y + See figure 3

6 6 Solutions Multivariate Calculus Figure 3: Region Figure 4: Region

7 Math Camp 7 Figure 5: Region 4., y See figure 4 Integrate the following: See figure for the integration region. f(, y sin( over the area, y with respect to first Therefore we can rewrite the integral as ˆ y sin( ddy There is not an easy way to solve it, see f(, y sin( over the area, y Therefore we can rewrite the integral as with respect to y first ˆ ˆ sin( dyd

8 8 Solutions Multivariate Calculus Figure 6: Region Figure 7: Region

9 Math Camp 9 and we have that ˆ ˆ sin( dyd ˆ ˆ sin( y ] d sin( ] d f( 4 sin( u g( g ( du d ˆ 4 ] 4 sin(u du ] 4 4 cos(u cos(4] 4 Differentiate the following:. λe λy dy with respect to. We need to use Leibnitz s Integral rule. Remind that d dy ˆ b(y a(y f(, yd ˆ b(y a(y f db d + fb(y, y] fa(y, y]da y dy dy which can be rewritten as d d ˆ b( a( f(, ydy ˆ b( a( f db da dy + f, b(] f, a(] d d

10 Solutions Multivariate Calculus Therefore we have that d d λe λy dy. e ey dy with respect to. λe λy dy f, ] λyλe λy dy λe λ λ ye λy dy λe λ λ ye λy dy λe λ f(y y g (y e λy g(y e λy λ ] ˆ ] λ y e λy e λy λ λ dy λe λ ye ] λy ˆ ] λ e λy dy λe λ ] e λ ˆ ] e λ λ e λy dy λe λ ] e λ e λ ] e λy dy λe λ ] ] e λ e λ e λy ] λe λ λ ( ] ] e λ + λ e e d d ˆ e e y dy ˆ e ˆ e f dy + f, e ] de f, ]d d d e y + ye y ] dy + e e e e Working with pdf In st semester econometrics, you will be asked to integrate probability density functions (or p.d.f.s in order to find the probability that certain events will occur. Sometimes it is necessary to transform the random variables. For eample, if we are given the density functions for X and X, and Y and Y as functions of X and X, we can then transform this system to solve for the p.d.f. s of Y and Y. The following problems are taken from p.d.f. s in the first semester econometrics tetbook, section 3.7. For each of the following, Find X and X as functions of Y and Y Find the determinant of the Jacobian Sketch the region of integration in terms of X and X Sketch the region of integration in terms of Y and Y Evaluate the new integral in terms of Y and Y

11 Math Camp. f(x, X e X X over the region X >, X >, where Y X X +X and Y X + X. Since we have that Y X X +X and Y X + X, we can solve for X and X and obtain that X Y Y and X Y ( Y. Computing the Jacobian we have that therefore X Y Y X Y Y X Y Y X Y ( Y J X X X X Y ( Y + Y Y Y Y Y Y Y The original integration region was given by X > and X >, therefore we have that Y Y > and Y ( Y > which is possible if and only if Y > and < Y < Y. Therefore we can rewrite the integration as ˆ ˆ e X X dx dx ˆ ˆ ˆ ˆ e YY ( YY Y dy dy e Y Y dy dy e Y Y Y ] dy e Y Y dy e Y Y + ˆ e Y dy e Y e + e. f(x, X 8X X over the region X X, where Y X X and Y X. Since we have that Y X X and Y X, we can solve for X and X and obtain that X Y Y and X Y. Computing the Jacobian we have that therefore X Y Y X Y Y X Y X Y J X X X X Y Y Y Y Y The original integration region was given by X X, therefore we have that Y Y Y which is possible if and only if Y and Y. Therefore we can rewrite the integration as ˆ X 8X X dx dx 8Y Y 3 dy dy 4Y Y 3 ] dy 4Y 3 ]dy Y 4 8Y Y Y Y dy dy

12 Solutions Multivariate Calculus Hint: As a check, realize that all p.d.f. s integrate to one. So the original integrals with respect to X and X, as well as the transformed integral with respect to Y and Y, should integrate to one. Try it!