Problem Set Solutions for the video course A Mathematics Course for Political and Social Research

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1 Problem Set Solutions for the video course A Mathematics Course for Political and Social Research David A Siegel July 0, 05 Associate Professor, Department of Political Science, Duke University, Durham, NC 7708; davidsiegel@dukeedu; web:

2 Lecture Identify whether each of the following is a constant or a variable: (a) Variable (b) Constant (c) Variable Identify whether each of the following is a variable or a value of a variable: (a) Value (of the variable Age) (b) Variable (taking the values of candidates or parties) (c) Variable (d) Value (of the variable Education) Identify whether each of the following indicators is measured at a nominal, ordinal, interval, or ratio level Note also whether each is a discrete or a continuous measure: (a) Ratio, discrete (b) Nominal, discrete (c) Ratio, continuous (d) Ordinal, discrete 4 Let A = (6, 8) and B = [7, 9] (a) Neither (b) (6, 9] (c) [7, 8) (d) Any (x, y) in which x (6, 8), y [7, 9] 5 Simplify or evaluate each of the following: (a) x (b) a(b a) (c) 4 + 4(5) = 5 (d) 0 (e) ± (f) = (g) = (h) x( + 4(y ) + z)

3 6 Ratio: :9 Proportion: = Percentage: 00% 547% % ( ) 00% 0094 = 94% 8 (x 4)(x+) 7(x+) = x y (φ )(φ + 5) y = x = 4 or x = x = 5 4 ± θ > Lecture For each pair of ordered sets, state whether it represents a function or a correspondence: (a) Function, since no element of the first set maps to more than one element of the second (b) Correspondence, since the element in the first set maps to three values (0, 5, ) in the second f(x) = x, g(x) = ln x, and h(x) = x : Simplify or evaluate the following expressions, using f(x) = x, g(x) = ln(x), and h(x) = x : (a) (ln(x)) (b) ln( x ) = ln(x) (c) x x = x x (the latter as long as x 0) (d) ln(x) x (e) (x ) = x 6 = x 6 (f) x + (g) x for x > 0 Either find the following limits or show that they do not exist (a) 9

4 (b) 0 (c) 4 7 (d) Does not exist (would be 0 ) (e) 4 4 For each of the following sets, state whether they are (a) open, closed, both, or neither; (b) bounded; (c) compact; (d) convex: (a) (a) neither, (b) bounded, (c) not compact, (d) convex (b) (a) closed, (b) bounded, (c) compact, (d) convex (c) (a) neither, (b) bounded, (c) not compact, (d) not convex (d) (a) open (it s (, 5)), (b) bounded, (c) not compact, (d) convex 5 For each of the following functions, state whether or not it is continuous on the domain provided (a) Yes (b) This function is not continuous because it has the value 0 at x =, but approaches as it approaches x = from the right (One could make it continuous by either adding from the equation for all x (to get x + ), or by subtracting to the equation for all x > (to get x ) Lecture Use the definition of the derivative to find the derivative of y with respect to x for the following: (a) dy/dx = 0 (b) dy/dx = x + (c) dy/dx = 6x + 0x (d) dy/dx = 8x 6x (e) dy/dx = 0x + 6x + For each of the following, find the partial derivative with respect to both x and z (a) x (b) x = x, z = 4z = 0, z = z (c) x = z, z = xz (d) x = x z, z = x z 4

5 4 Lecture 4 Differentiate the following: f (x) = x 4 f (x) = 4ax f (x) = x 4 f (x) = ( x)(x + 5x 7) + (x x )(6x + 0x) 5 f (x) = 5(x + ) 4 6 f (x) = (( 4x + x )(x x) ( x + x )(x ))/(x x) 7 f (x) = (x 5) (x ) 8 f (x) = (4x x) ln(x 5) + (x4 x +)(x) x 5 9 f (x) = (5x 4 x)e 5x5 x 0 f (x) = xe x ln(x) + x e x ln(x) + xe x + 5x 4 f (x) = ln(c)c x (x ) x ln(c) ) (log c (x)) f (x) = 6x (log c (x))+x ( 5 Lecture 5 Integrate the following: (a) F (x) = 5 x x + C (b) F (x) = x + 5x x4 + 6 x6 + C (c) F (x) = ln( x ) + x + C (d) F (x) = e x + C (e) F (x) = e x5 + + C (f) F (x) = (x ln(x) x) + C Evaluate the following integrals: (a) 0 (b) = Note the integrals evaluated at x = cancel This is = 0 (c) = 0, the same as for the previous question 5

6 (d) (x ln(x) x x ) e This is e e e = (e ) (e) 4 ( x + 6x + ) + C (f) x e x (xe x e x ) + C = e x (x x + ) + C 6 Lecture 6 Find all extrema (local and global) of the following functions on the specified domains, and state whether each extremum is a minimum or maximum and whether each is only local or global on that domain Global minimum at x = 5, global maximum at the boundary x = 0 Global minimum at x = 0, global maximum at x = Global minimum at x = 0, global maximum at x = 4 Global maximum at x =, global minimum at x = 0 5 Global maximum at x =, global minimum at x = 6 Three stationary points at x =,, Global minimum at x =, global maximum at x =, local maximum at x =, local minimum at x = 7 Lecture 7 Identify each of the following as a classical (objective), empirical (objective), or subjective probability claim: (a) Empirical (objective) (b) Classical (objective) (c) Subjective (presumably not based on a forecasting model) Identify the following as simple or compound events: (a) Simple (b) Compound (c) Compound (d) Simple Characterize the following as independent, mutually exclusive, and/or collectively exhaustive: (a) Mutually exclusive and collectively exhaustive values of a variable (b) Independent 6

7 (c) Mutually exclusive values of a variable 4 Let P r(a) = 045, P r(b) = 0, and P r(c) = 0, (a) 0 (b) 067 (c) 0 (d) 0068 (e) 07 (f) = 07 (g) = Compute each of the following: (a) 60 (b) 90 (c) 0 (d) 5 (e) 8 (f) 0 (g) A committee contains ten legislators with six men and four women Find the number of ways that a delegation of five: (a) This is the number of ways 5 elements can be chosen from 0, or 0 5 = 5 (b) We have the joint probability of two independent events: choosing women from 4 and men from 6 This is: ( 6 4 )( ) = 0 6 = 0 (c) There are two ways to do this: have women or have 4 women on the committee ( We add the number of ways that each can occur For three women: 6 )( 4 ) = 5 4 = 60 For four women: ( 6 4 )( 4) = 6 = 6 So the total is , 9 9 In one legislature, 0% of the legislators are conservatives, 0% are liberals, and 60% are independents In a recent vote to expand counterinsurgency operations, 80% of conservatives, 0% of liberals, and 50% of independents voted in favor (a) P r(v ) = P r(v C)P r(c) + P r(v L)P r(l) + P r(v I)P r(i) = = 05 7

8 (b) P r(c V ) = (c) 008 (d) /06 : or 667: (/)/(/4) = 8 P R(V C)P r(c) P R(V ) = (8)() 5 = 0 An unemployed person is 8% more likely than an employed person, and someone with a college education is 55% less likely than a person with only a high school education, to vote for the National Front 8 Lecture 8 We use a Poisson distribution with these data: 0 5 e 0 5! = 005 We use a binomial distribution: ( 5 5) () 5 (8) 0 = 0000 We use a negative binomial distribution: + P (Y =, 0) = ( ) 4 = = 4! (!)(!) 005 = = 00 4 Bach: EU(B) = p(b)eu(bt ) + ( p(b))eu(ba), where B is Bach, T is Together and A is alone EU(B) = () + 8() = = 4 EU(S) = p(b)eu(sa) + ( p(b))eu(st ), where S is Stravinsky EU(S) = ( ) + 8(6) = = 44 So she should choose Stravinsky because 44 > 4, even though she personally likes Bach much better 5 First we use Bayes rule to figure out the posterior probability of B s being strong after P r(m S)P r(s) A observes maneuvers This is P r(s M) = = (8)() = P r(m S)P r(s)+p r(m W )P r(w ) (8)()+()(8) Next we ask what this posterior probability must be for A to want to start a war This is an expected utility comparison A gets 0 for doing nothing, and expects to get P r(s M) ( ) + ( P r(s M)) () for starting a war after observing maneuvers This becomes + = < 0, so A would prefer not to start a rebellion after observing maneuvers, despite having an initially low prior belief about the strength of the state This is because the maneuvers are very informative 8

9 9 Lecture 9 This is: (x µ) f(x)dx = 0 (x µx + µ x µ )dx = ( 4 x4 µx + µ x µ x ) 0 = ( µ + 4 µ µ ) = ( + ) = 0 () Assume a policy outcome, x, depends on a choice of policy, p, and an exogenous economic shock, ɛ, so that x = p + ɛ Further assume that ɛ is distributed uniformly on [, ] (a) (b) No, still 0 (c) It decreases At p = 0, for [, ]: ( ɛ )dɛ = 6 ɛ = ( ) = 6 For [, ]: 4 ( ɛ )dɛ = ɛ = 6 ( 8 8) = = 4 She is effectively risk averse around her ideal point of zero, so prefers the less uncertain case xdx = = It would not change It would still be equal to the midpoint of the distribution, which is still 50 A risk neutral person cares only about the expected value of the lottery xdx = 600 / 00 / = xdx = / 00 / 00 = 86 This is greater then before; she is risk averse, and prefers the less risky lottery to the more risky lottery, given equal expected values of the lotteries 0 Lecture 0 4 Let: a = 5, b = 5 5, c = (,, ), d = (6, 4, ), and e = (0, 0, 0, 40)T Calculate each of the following, indicating that it s not possible if there is a calculation you cannot perform 6 (a)

10 8 (b) (c) Not possible to add because they have different dimensions 4 6 (d) 0 8 (e) = (f) ()(6) + ()(4) + ()() = 6 (g) 80 Identify the following matrices as diagonal, identity, square, symmetric, triangular, or none of the above (note all that apply) (a) This is a square, (lower) triangular matrix (b) This matrix has none of these properties (c) This is a symmetric, square matrix Write down the transpose of matrices A through C from the previous problem 4 0 (a) A T = (b) B T = (c) C = Given the following matrices, perform the calculations below A = [ ] B = C = [ (a) Not possible to add because they have different dimensions [ ] 5 (b) 4 8 ] 0

11 [ (c) (d) [ ] ] (e) Cannot multiply due to incompatible dimensions [ ] 4 (f) (g) [ ] 0 (h) 4 44 [ ] 0 (i) 4 44 (j) 5 Find the determinants and, if they exist, the inverses of the following matrices: [ ] (a) Det(A) = 6 A = (b) Det(B) = 0 No inverse exists because the matrix is singular by virtue of its determinant of zero (c) Det(C) = ()( 0) ()(4 4) + ()(0 6) = = 0 M = 8, M = 0, M = 4, M =, M = 5, M =, M =, M = 0, M = 4 (d) C = (e) We cannot take the determinant of a matrix that is not square, and so we cannot find its inverse Lecture 4 4 Let: a = 5, b = 5 5, c = 4 0, and d = 0 0 Answer each of the 4 following questions, saying that it s not possible if there is a calculation you cannot perform

12 (a) sa + tb = 4s + t s + 5t 5s + 5t s + t (b) No, only two at most are For example, one can write d = b and c = a b (c) These span a two-dimensional space, as there are two, and only two, linearly independent vectors (If you wrote c and d in terms of a and b in the previous problem, then the latter two vectors are independent and span a two-dimensional space) Solve the following systems of equations using substitution or elimination, or both: (a) x =, y = 6, z = 7 (b) x =, y =, z = (c) x = x, y = x, z = x The third equation is a multiple of the first equation, so we only have two equations in three unknowns To solve, we leave x as a parameter and solve for y and z in terms of x to get this answer (d) The first and second equations are contradictory, as the first implies 4x 4y z = 9 rather than = 0 as the second one states Thus these equations have no solutions 5 0 Let A =, B =, and c = Calculate each of the following, 6 4 indicating that it s not possible if there is a calculation you cannot perform (a) A is full rank () as its determinant is non-zero and so it is non-singular (b) B has rank It is singular, so its columns (or rows) are linearly dependent Removing one of them leaves one linearly independent column (or row), making its rank ( ) (c) A =, and A 0 5 c = = (d) Since B has determinant 0 it is singular, and does not have an inverse Thus we can t use matrix inversion As one of the rows is a multiple of the other we only have one unique equation for two variables, indicating an infinite number of solutions to the system of equations represented by the matrix 4 det(a) =, det(b ) = 48, det(b ) = 4, det(b ) = x = 4, y =, z = Lecture λ =,, associated eigenvectors, Let Q =, so Q = 0 Now A = QDQ so D = Q AQ =, as required 0

13 λ 0 First we find the eigenvalues A λi = 0 λ 0 A λi = ( λ)( 0 0 λ λ)( λ) Setting this equal to zero gives the three eigenvalues λ = 0,, Next we find the eigenvectors For each eigenvalue we solve the equation Ax = λx where we let x = a, moving around the location of the if we get a contradiction This produces b a system of three equations for each eigenvalue that look like: +a = λ, a = λa, a = λb For λ =, the first equation yields a = 0, the second a = a, and the third a = b The solution to these equations set a = b = 0, so the corresponding eigenvector is x = 0 For λ =, the first equation yields a =, the second a = a, and the 0 third a = b The solution to these equations set a = b =, so the corresponding eigenvector is x = For λ = 0, the first two equations are contradictory, so a we try a different location for the in the eigenvector We ll try x = b Now the system of three equations is a + b = 0, b = 0, b = 0 The solution to this set of 0 equations is a = b = 0, so the corresponding eigenvector to 0 is x = 0 Yes, it is ergodic because it is aperiodic and all states communicate To find the steady state π, we solve Mπ = π, where π = This matrix equation yields the system a of equations a ( = ), 5 + a = a which has ( the unique solution a = 5 Thus 6 6 ) our steady state is 5 or, after normalizing it, Yes, it is ergodic because it is aperiodic and the first state is a distinguished state, reachable with positive probability from all other states To find the steady state π, we solve Mπ = π, where π = a This matrix equation yields the system of equations b + 75a + 5b =, 5a = a, and5b = b which has the unique solution a = 0, b = 0 Thus our steady state is 0, as suggested by the form of the transition matrix, 0 which basically transfers all probability to the first, distinguished state, which is an absorbing state

14 Lecture The partial derivatives are, respectively, β + β x and β + β x x They correspond to the marginal effects of, respectively, x and x, holding all other variables constant Note that the model is non-linear in both variables The marginal effect of x depends on the value of x in a substantial way, increasing rapidly as x moves away from zero in either direction The marginal effect of x depends on the values of both x and x, implying an even more complex effect x = x e x x x x + x e x ln x x = xe x x x ex x = ln x + x ex ln x x (ln x ) x = 4x 6 ln(x )x x x = x = x x x + x x 4 First do the integral over x, because the bounds of the inner integral contains an y So y x dy = x y = [y ] Now do the outer integral over y This is 0 y [y ]dy = 0 (y5 y )dy This equals ( 6 (6 0) ( 0)) = 8 4 Lecture 4 = xz df = + dz = xz + x dx x z dx x ( + x ey ) 6xe yz f = 8x ze yz 8x ye yz 6e yz 6xze yz 6xye yz H = 6xze yz 8x z e yz 8x e yz ( + yz) At the point (,, ) this is H = 6xye yz 8x e yz ( + yz) 8x y e yz 6e 6e 6e 6 8e 8e or 8e The eigenvalues of this matrix are 0, (+ ), ( 6e 8e 8e ) Since these signs change, the matrix is indeterminate and we have a saddle point z 0 x 4 J = 0 z y y xy 0 5 Lecture 5 (x, y ) = (5, ) is the only critical point, and it is a minumum 4

15 x =, y = 4, λ = 4 x =, y =, λ = 4 x =, y = 6 Lecture 6 This is dx (b) b = 4b + b Write f = (ln(bx (b) + )) x (b) = 0, so the implicit function theorem says that x (b) = b b = (ln(bx (b) + )) x (b), = b bx (b)+ x (ln(bx (b) + )) x (b), so bx (b)+ x = (ln(bx x (b)+)) (b) bx (b)+ So x (b) b (ln(bx x (b)+)) (b) bx (b)+ f = u x = x c x = 0 f = u x = x c x = 0 Now we can apply the implicit function theorem in more than one dimension We need to compute two Jacobians The first is with respect to b: J x This is ( c c ) We need the inverse of this matrix, so we first compute the determinant: 4c c Now we can find the inverse: J b = c 4c c c x 0 Finally, we need to find the second Jacobian, J c = 0 x We can put these together to get: J x (c) = 4c x x 4c c If the costs are sufficiently high then each person s x 4c x choice decreases in both people s costs However, if they are sufficiently low (4c c < ), then each person s choice increases in both costs This kind of result indicates the power of considering both choices at the same time 5