Mathematical methods for economic theory

Martin J. Osborne
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2.5 Exercises on homogeneous functions

  1. Determine whether or not each of the following functions is homogeneous, and if so of what degree.
    1. 3x + 4y.
    2. 3x + 4y − 2.
    3. (√x + √y + √z)/(x + y + z).

    Solution

    1. Degree 1: 3(tx) + 4(ty) = t(3x + 4y).
    2. Not homogeneous: Suppose, to the contrary, that there exists some value of k such that 3(tx) + 4(ty) − 2 = tk(3x + 4y − 2) for all t and all (x, y). Then, in particular, 6x + 8y − 2 = 2k(3x + 4y − 2) for all (x, y) (taking t = 2), and hence 6 − 2 = 2k(3 − 2), or 2k = 4 (taking (x, y) = (1, 0)), and 8 − 2 = 2k(4 − 2), or 2k = 3 (taking (x, y) = (0, 1)). These two conditions are inconsistent, so the function is not homogeneous of any degree.
    3. Degree −1/2: (√(tx) + √(ty) + √(tz))/(tx + ty + tz) = t−1/2(√x + √y + √z)/(x + y + z).

  2. Determine whether or not each of the following functions is homogeneous, and if so of what degree.
    1. 2x2 + xy.
    2. x2 + x3.
    3. [g(x1, ..., xn)]p where g is homogeneous of degree n.

    Solution

    1. Homogeneous of degree 2: 2(tx)2 + (tx)(ty) = t2(2x2 + xy).
    2. Not homogeneous: Suppose, to the contrary, that there exists some value of k such that (tx)2 + (tx)3 = tk(x2 + x3) for all t and all x. Then, in particular, 4x2 + 8x3 = 2k(x2 + x3) for all x (taking t = 2), and hence 6 = 2k (taking x = 1), and 20/3 = 2k (taking x = 2). These two conditions are inconsistent, so the function is not homogeneous of any degree.
    3. Homogeneous of degree np: (g(tx1, ..., txn))p = (tng(x1, ..., xn)))p = tnp(g(x1, ..., xn))p.

  3. Is the function 30x1/2y3/2 − 2x3/y homogeneous of any degree? If so, which degree? (If not, give a complete argument why not.)

    Solution

    We have 30(tx)1/2(ty)3/2 − 2(tx)3/(ty) = t2[30x1/2y3/2 − 2x3/y], so the function is homogeneous of degree 2.

  4. The function f(x1, ..., xn), with domain equal to the set of n-tuples of positive numbers, is homogeneous of degree k and f(x1, ..., xn) > 0 for all (x1, ..., xn) in the domain. Is there any value of k for which the function g defined by g(x1, ..., xn) = ln f(x1, ..., xn) is homogeneous of some degree?

    Solution

    In order for g to be homogeneous of degree m we need
    g(tx1, ..., txn) = tmg(x1, ..., xn) for all (x1, ..., xn) in the domain and all t > 0.
    Translating this condition into a condition involving f, we get
    ln f(tx1, ..., txn) = tmln f(x1, ..., xn) for all (x1, ..., xn) in the domain and all t > 0.
    Using the fact that f is homogeneous of degree k, this condition is equivalent to
    ln tkf(x1, ..., xn) = tmln f(x1, ..., xn) for all (x1, ..., xn) in the domain and all t > 0
    or
    ln tk + ln f(x1, ..., xn) = tmln f(x1, ..., xn) for all (x1, ..., xn) in the domain and all t > 0
    or
    ln tk = (tm − 1)ln f(x1, ..., xn) for all (x1, ..., xn) in the domain and all t > 0.
    Thus if k = 0 then g is homogeneous of degree 0 (m = 0). If k ≠ 0, the left-hand side depends on t, so that we need m ≠ 0 (because the right-hand side is independent of t if m = 0). But if m ≠ 0 then the right-hand side depends on (x1, ..., xn) (because k ≠ 0 means that f is not constant) whereas the left-hand side does not, so that no value of m satisfies the equation for all (x1, ..., xn) and all t > 0.

  5. Suppose that f(x1, ..., xn) is homogeneous of degree r. Show that each of the following functions h(x1, ..., xn) is homogeneous, and find the degree of homogeneity.
    1. h(x1, ..., xn) = f(xm
      1      , ..., xm
      n      ) for some number m.
    2. h(x1, ..., xn) = [f(x1, ..., xn)]p for some number p.

    Solution

    1. We have f((tx1)m, ..., (txn)m) = f(tmxm
      1      , ..., tmxm
      n      ) = (tm)rf(xm
      1      , ..., xm
      n      ) = tmrh(x1, ..., xn), so h is homogeneous of degree mr.
    2. We have h(tx1, ..., txn) = [f(tx1, ..., txn)]p = [trf(x1, ..., xn)]p = trp[f(x1, ..., xn)]p = trph(x1, ..., xn), so that h is homogeneous of degree rp.

  6. Is the sum of two homogeneous functions necessarily homogeneous?

    Solution

    No. Consider, for example, h(x) = 1 + x. The constant function f(x) = 1 is homogeneous of degree 0 and the function g(x) = x is homogeneous of degree 1, but h is not homogeneous of any degree. (If h were homogeneous of degree k, then we would have 1 + tx = tk(1 + x) for all t and all x, which implies in particular that 1 + 2x = 2k(1 + x) (taking t = 2), which in turn implies that 1 = 2k (taking x = 0) and 3 = 2(2k), which are inconsistent.)

  7. Is the product of two homogeneous functions, with possibly different degrees of homogeneity, necessarily homogeneous? Necessarily not homogeneous?

    Solution

    Let f be homogeneous of degree k and g be homogeneous of degree m. Then f(tx) = tkf(x) for all x and all t > 0, and g(tx) = tmg(x) for all x and all t > 0. Now let h(x) = f(x)g(x). Then
    h(tx) = f(tx)g(tx) = tk+mf(x)g(x) = tk+mh(x),
    so that h is homogeneous of degree k + m.

    1. Is the function x1/2y + x3/2 homogeneous of any degree? (If so, which degree?)
    2. A firm's (differentiable) production function f(x1, ..., xn) is homogeneous of degree 1 (where xi is the amount of input i). Assume that the price (in terms of output) of each input i is its marginal product f'i(x1, ..., xn). What is the relation between the total cost of the firm's inputs and its output?

    Solution

    1. Yes: (tx)1/2(ty) + (tx)3/2 = t3/2(x1/2y + x3/2), so that the function is homogeneous of degree 3/2.
    2. The total cost of the firm's inputs is
      ∑n
      i=1       xi
      ∂f
      ∂xi
      and the firm's output is f(x1, ..., xn). The function f is homogeneous of degree 1, so the two amounts are equal.

    1. Is the function (x3 − y3)/(x1/2 + y1/2) homogeneous of any degree? (If so, which degree?)
    2. Is the function x3y3 + x1/2 homogeneous of any degree? (If so, which degree?)
    3. A consumer's (differentiable) demand function for some good is f(p1, ..., pn, w), where pi is the price of the ith good, and w is the consumer's wealth. This function f is homogeneous of degree 0. Is there any necessary relationship between ∑n
      i=1      (pif'i(p1, ..., pn, w)) and wf'n+1(p1, ..., pn, w)?

    Solution

    1. Yes: ((tx)3 − (ty)3)/((tx)1/2 + (ty)1/2) = t5/2(x3 − y3)/(x1/2 + y1/2), so that the function is homogeneous of degree 5/2.
    2. No, it is not homogeneous of any degree. Suppose, to the contrary, it is homogeneous of degree k. Then for some value of k we have (tx)3(ty)3 + (tx)1/2 = tkx3y3 + x1/2 for all t and all (x, y). In particular, taking t = 4 we have 4096x3y3 + 2x1/2 = 4k(x3y3 + x1/2) for all (x, y), and hence 2 = 4k (taking (x, y) = (1, 0)) and 4098 = 2(4k) (taking (x, y) = (1, 1)), which are inconsistent.
    3. By Euler's theorem we have ∑n
      i=1      pif'i(p1, ..., pn, w) + wf'n+1(p1, ..., pn, w) = 0. (Note that f has n + 1 arguments.)

  10. The function g(x,y) is homogeneous of degree 1. Is the function f(x,y) = aln(g(x,y)/x) homogeneous of any degree?

    Solution

    We have f(tx,ty) = aln(g(tx,ty)/tx) = aln(tg(x,y)/tx) = f(x,y), given that g is homogeneous of degree 1. Thus f is homogeneous of degree 0.

  11. A consumer's utility function is homogeneous of some degree. She purchases the bundle of goods that maximizes her utility subject to her budget constraint. The bundle of goods she purchases when the prices are (p1, ..., pn) and her income is y is (x1, ..., xn). What do you know about the bundle she purchases when the same prices are (p1, ..., pn) and her income takes other values?

    Solution

    Because the consumer's utility function is homogeneous, the slope of her indifference curves is constant along any given ray through the origin. Thus if the bundle of goods she purchases when the prices are (p1, ..., pn) and her income is y0 is (x1, ..., xn), then the bundle she purchases when the prices are (p1, ..., pn) and her income is y1 is ((y1/y0)x1, ..., (y1/y0)xn).

  12. The function g is defined by
    g(x, y) = f(x, y) − aln(x + y),
    where a is a constant and f satisfies the condition
    xf'x(x,y) + yf'y(x,y) = a for all (x, y).
    Show that g is homogeneous of degree 0.

    Solution

    We have
    xg'x(x,y) + yg'y(x,y) = x(f'x(x,y) − a/[x + y]) + y(f'y(x,y) − a/[x + y]) = a − a = 0
    for all (x,y). Thus by Euler's theorem g is homogeneous of degree 0.

  13. The function g(x,y) is homogeneous of degree r. Is the function f defined by f(x,y) = g(x,y)/(xy) homogeneous of any degree?

    Solution

    We have
    f(kx,ky) = g(kx,ky)/(kxky) = krg(x,y)/(k2xy) = kr−2g(x,y)/(xy) = kr−2f(x,y).
    Thus f is homogeneous of degree r−2.

  14. The twice-differentiable function f(x, y) is homogeneous of degree k, and its second derivatives are continuous. Show that
    x2f"11(x, y) + 2xyf"12(x, y) + y2f"22(x, y) = k(k − 1)f(x, y) for all (x, y).

    Solution

    The fact that f is homogeneous of degree k means that f'1 and f'2 are homogeneous of degree k − 1 (by a proved result). Thus by Euler's theorem applied to f'1 and to f'2 we have
    xf"11(x, y) + yf"12(x, y)  = (k − 1)f'1(x, y) for all (x, y)
    xf"21(x, y) + yf"22(x, y)  = (k − 1)f'2(x, y) for all (x, y).
    Now the sum of x times the first equation and y times the second equation is
    x2f"11(x, y) + 2xyf"12(x, y) + y2f"22(x, y) = (k − 1)[xf'1(x, y) + yf'2(x, y)] for all (x, y)
    (given that f"12(x, y) = f"21(x, y), by Young's theorem). Finally, the term in brackets on the right-hand side of this equation is equal to kf(x, y) by Euler's theorem, yielding the required result.

  15. A firm uses two inputs to produce a single output. Its production function f is homogeneous of degree 1. An implication of the homogeneity of f, which you are not asked to prove, is that the partial derivatives f'x and f'y with respect to the two inputs are homogeneous of degree zero. Use Euler's theorem to find an expression for the cross partial derivative f"xy(x, y) in terms of x, y, and f"xx(x, y).

    Solution

    Because f'x is homogeneous of degree 0, we have
    xf"xx(x, y) + yf"xy(x, y) = 0,
    so that f"xy(x, y) = −(x/y)f"xx(x, y).