Mathematical methods for economic theory

Martin J. Osborne
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2.4 Exercises on differentials and comparative statics

  1. Find the differentials of the following.
    1. z = xy2 + x3.
    2. z = a1x12 + ... + anxn2 (where a1, ..., an are constants).
    3. z = A(α1x1−ρ + ... + αnxn−ρ)−1/ρ (where A, ρ, and α1, ..., αn are constants). [This is a constant elasticity of substitution function.]

    Solution

    1. dz = (y2 + 3x2)dx + 2xy dy.
    2. dz = 2a1x1dx1 + ... + 2anxndxn.
    3. dz = A(α1x1−ρ + ... + αnxn−ρ)−1−1/ρ(α1x1−ρ−1dx1 + ... + αnxn−ρ−1dxn).

  2. Consider the system of equations
    xu3 + v  =  y2
    3uv − x  =  4
    1. Take the differentials of both equations and solve for du and dv in terms of dx and dy.
    2. Find ∂u/∂x and ∂v/∂x using your result in part (a).

    Solution

    1. Taking differentials of the equations we obtain
      u3dx + 3xu2du + dv  =  2ydy
      3vdu + 3udv − dx  =  0
      or, rearranging these equations,
      3xu2du + dv  =  −u3dx + 2ydy
      3vdu + 3udv  =  dx
      or
      left parenthesis 3xu2 1 right parenthesis
      3v 3u
      left parenthesis du right parenthesis
      dv
      		 = 
      left parenthesis −u3dx + 2ydy right parenthesis
      dx
      		.
      Inverting the matrix on the left, we obtain du = −((3u4 + 1)/D)dx + (6yu/D)dy and dv = ((3xu2 + 3u3v)/D)dx − (6yv/D)dy, where D = 9xu3 − 3v.
    2. ∂u/∂x = −(3u4 + 1)/D and ∂v/∂x = (3xu2 + 3u3v)/D.

  3. The equilibrium values of the variables x, y, and λ are determined by the following set of three equations:
    U'1(x, y)  =  λp
    U'2(x, y)  =  λq
    px + qy  =  I
    where p, q, and I are parameters and U is a twice differentiable function. Find ∂x/∂p. [Use Cramer's rule to solve the system of equations you obtain.]

    Solution

    Taking differentials we obtain
    U"11(x,y)dx + U"12(x,y)dy = pdλ + λdp
    U"21(x,y)dx + U"22(x,y)dy = qdλ + λdq
    xdp + pdx + ydq + qdy = dI
    or, setting dq = dI = 0,
    left parenthesis U"11 U"12 −p right parenthesis
    U"21 U"22 −q
    p q 0
    left parenthesis dx right parenthesis
    dy
    dλ
    		 =
    left parenthesis λdp right parenthesis
    0
    −xdp
    		.
    Using Cramer's rule we find that
    dx = (λq2 + qxU"12 − pxU"22)dp/Δ,
    where Δ is the determinant of the matrix on the left-hand side of the equation, or
    ∂x
    ∂p
    		 =
    λq2 + x(qU"12 − pU"22)
    Δ
    		.

  4. The equilibrium values of the variables Y, C, and I are given by the solution of the three equations
    Y  =  C + I + G
    C  =  f(Y, T, r)
    I  =  h(Y, r)
    where T, G, and r are parameters and f and h are differentiable functions. How does Y change when T and G increase by equal amounts?

    Solution

    Taking differentials we obtain
    dY  =  dC + dI + dG
    dC  =  f'1(Y,T,r)dY + f'2(Y,T,r)dT + f'3(Y,T,r)dr
    dI  =  h'1(Y,r)dY + h'2(Y,r)dr,
    or
    dY(1−f'1(Y,T,r) − h'1(Y,r)) = f'2(Y,T,r)dT + (f'3(Y,T,r) + h'2(Y,r))dr + dG,
    or
    dY = 
    f'2(Y,T,r)dT + (f'3(Y,T,r) + h'2(Y,r))dr + dG
    1−f'1(Y,T,r) − h'1(Y,r)
    Thus if dT = dG and dr = 0 we have
    dY = 
    (1 + f'2(Y,T,r))dT
    1−f'1(Y,T,r) − h'1(Y,r)
    (Given that the question asks only about changes in T and G, we could ignore the change dr in r from the beginning.)

  5. An industry consists of two firms. The optimal output of firm 1 depends on the output q2 of firm 2 and a parameter α of firm 1's cost function: q1 = f(q2,α), where f'1(q2,α) < 0 and f'2(q2,α) > 0 for all q2 and all α. The optimal output of firm 2 depends on the output q1 of firm 1: q2 = g(q1), where g'(q1) < 0. The equilibrium values of q1 and q2 are thus determined as the solution of the simultaneous equations
    q1  =  f(q2, α)
    q2  =  g(q1).
    Is this information sufficient to determine whether an increase in α increases or decreases the equilibrium value of q1?

    Solution

    We have
    dq1  = f'1(q2,α)dq2 + f'2(q2,α)dα
    dq2  = g'(q1)dq1
    so that
    ∂q1/∂α = f'2(q2,α) / [1−f'1(q2,α)g'(q1)].
    Since we don't know if f'1g' is less than or greater than 1, we don't know if this is positive or negative.

  6. The variables x and y are determined by the following pair of equations:
    f(x)  = g(y)
    Ay + h(x)  = β
    where f, g, and h are given differentiable functions, β is a constant, and A is a parameter. By taking differentials, find ∂x/∂A and ∂y/∂A.

    Solution

    We have
    f'(x)dx  = g'(y)dy
    Ady + ydA + h'(x)dx  = 0
    so that, solving these two equations for dx and dy we deduce that
    ∂x/∂A = −g'(y)y/[Af'(x) + g'(y)h'(x)]
    and
    ∂y/∂A = −f'(x)y/[Af'(x) + g'(y)h'(x)].

  7. The optimal advertising expenditure of politician 1 depends on the spending s2 of politician 2 and a parameter α: s1 = f(s2, α), where 0 < f'1(s2, α) < 1 and f'2(s2, α) < 0 for all s2 and all α. The optimal expenditure of politician 2 depends on the spending s1 of politician 1 and a parameter β: s2 = g(s1, β), where 0 < g'1(s1, β) < 1 and g'2(s1, β) < 0. The equilibrium values of s1 and s2 are given by the solution of the simultaneous equations
    s1  =  f(s2, α)
    s2  =  g(s1, β).
    Does an increase in α (holding β constant) necessarily increase or necessarily decrease the equilibrium value of s1?

    Solution

    We have
    ds1  =  f'1(s2, α)ds2 + f'2(s2, α)dα
    ds2  =  g'1(s1, β)ds1 + g'2(s1, β)dβ
    so that
    ∂s1/∂α = f'2(s2, α)/[1 − f'1(s2, α)g'1(s1, β)].
    Since 0 < f'1 < 1 and 0 < g'1 < 1 and f'2 < 0, an increase in α reduces the equilibrium value of s1.

  8. The equilibrium outputs q1 and q2 of two firms satisfy
    q1  = b(q2, c1)
    q2  = b(q1, c2),
    where b is a differentiable function that is decreasing in each of its arguments and satisfies b'1(q, c) > −1 for all q and c, and c1 and c2 are parameters.
    1. Find the differentials of the pair of equations.

    2. Find the effect on the values of q1 and q2 of equal increases in c1 and c2 starting from a situation in which c1 = c2 and an equilibrium in which q1 = q2.

    Solution

    1. The differentials of the pair of equations are
      dq1  = b'1(q2, c1)dq2 + b'2(q2, c1)dc1
      dq2  = b'1(q1, c2)dq1 + b'2(q1, c2)dc2.

    2. We may write these equations as
      left parenthesis 1 −b'1(q2, c1) right parenthesis
      −b'1(q1, c2) 1
      left parenthesis dq1 right parenthesis
      dq2
      		=
      left parenthesis b'2(q2, c1)dc1 right parenthesis
      b'2(q1, c2)dc2
      		,
      so that
      left parenthesis dq1 right parenthesis
      dq2
      		= (1/Δ)
      left parenthesis 1 b'1(q2, c1) right parenthesis
      b'1(q1, c2) 1
      left parenthesis b'2(q2, c1)dc1 right parenthesis
      b'2(q1, c2)dc2
      		,
      where Δ = 1 − b'1(q2, c1)b'1(q1, c2). Hence
      dq1  = (1/Δ)[b'2(q2, c1)dc1 + b'1(q2, c1)b'2(q1, c2)dc2]
      dq2  = (1/Δ)[b'1(q1, c2)b'2(q2, c1)dc1 + b'2(q1, c2)dc2].

      To find the effect on q1 and q2 of small and equal increases in c1 and c2, set dc1 = dc2 = dc. Given q1 = q2 = q and c1 = c2 = c we have b'2(q1, c2) = b'2(q2, c1), so that

      dq1  = (1/Δ)b'2(q, c)[1 + b'1(q, c)]dc
      dq2  = (1/Δ)b'2(q, c)[b'1(q, c) + 1]dc,
      so that dq1 = dq2 < 0. That is, both outputs decrease, and do so by the same amount.

  9. The equilibrium values of the variables Y and r are given by the solution of the two equations
    I(r)  = S(Y)
    aY + L(r)  = M
    where a > 0 and M are parameters, I is an increasing differentiable function, and S and L are decreasing differentiable functions. How do Y and r change when M increases (holding a constant)?

    Solution

    Taking differentials we obtain
    I'(r)dr  = S'(Y)dY
    adY + L'(r)dr  = dM
    or
    left parenthesis I'(r) −S'(Y) right parenthesis
    L'(r) a
    left parenthesis dr right parenthesis
    dY
    		 = 
    left parenthesis 0 right parenthesis
    dM
    		.
    Solving for dr we get
    ∂r/∂M = S'(Y)/[aI'(r) + L'(r)S'(Y)] < 0.
    Similarly, solving for dY we get
    ∂Y/∂M = I'(r)/[aI'(r) + L'(r)S'(Y)] > 0.

  10. Consider a market containing two goods. Denote the prices of these goods by p and q. Suppose that the demand for each good depends on p, q, and the amount of advertising expenditure a on good 1, and that the supply of each good depends only on the price of that good. Denoting the demand functions by x and y and the supply functions by s and t, for any given value of a a market equilibrium is a pair (p, q) or prices such that
    x(p, q, a)  = s(p)
    y(p, q, a)  = t(q).

    How does the equilibrium price p of good 1 change as a changes?

    Assume that x'p < 0, x'q > 0, x'a > 0, s' > 0, y'q < 0, y'p > 0, y'a < 0, and t' > 0. (What are the economic interpretations of these assumptions?) Assume also that (x'p − s')(y'q − t') − x'qy'p > 0 for all (p, q, a). Does the equilibrium price of good 1 necessarily increase if a increases?

    Solution

    Take differentials of the system of equations (noting that p and q are variables, and a is a parameter):
    x'p(p, q, a)dp + x'q(p, q, a)dq + x'a(p, q, a)da  = s'(p)dp
    y'p(p, q, a)dp + y'q(p, q, a)dq + y'a(p, q, a)da  = t'(q)dq.

    Now solve for dp and dq. Writing the system in matrix form (omitting the arguments of the functions), we have

    left parenthesis x'p − s' x'q right parenthesis
    y'p y'q − t'
    left parenthesis dp right parenthesis
    dq
    		 = 
    left parenthesis −x'ada right parenthesis
    −y'ada
    		,
    so that
    left parenthesis dp right parenthesis
    dq
    		 = (1/Δ)
    left parenthesis y'q − t' −x'q right parenthesis
    −y'p x'p − s'
    left parenthesis −x'ada right parenthesis
    −y'ada
    		,
    where Δ = (x'p − s')(y'q − t') − x'qy'p. Hence
    dp = (1/Δ)[−(y'q − t')x'a + x'qy'a]da.

    Under the stated assumptions, Δ > 0, but the coefficient of da is not necessarily positive—if x'q and/or y'a are large enough then it could be negative. Thus the price of good 1 may decrease when a increases: if the (positive) effect of an increase in a on the demand for good 1 is small while the (negative) effect on the demand for good 2 is large, then the equilibrium price of good 1 may fall when a increases.

    If, however, x'qy'a is small then the equilibrium price of good 1 definitely increases.