Lecture 11: Implicit and Inverse Function Theorems

Learning objectives

State local implicit and inverse function results; derive comparative statics; analyze equilibrium systems using Jacobians; distinguish local from global uniqueness.

1. Implicit function theorem

Let $F(x,y)=0$, where $F$ is continuously differentiable and $F_y\ne0$ at a solution $(x_0,y_0)$. Locally, the equation defines $y=h(x)$, with

\[ h'(x)=-\frac{F_x}{F_y}. \]

The theorem is local: it does not establish a globally defined function or global uniqueness.

2. Comparative statics

If an equilibrium condition is $F(y,\theta)=0$, then

\[ \frac{dy^*}{d\theta}=-\frac{F_\theta}{F_y} \]

at a regular equilibrium. The sign depends on both partial derivatives. For a system $F(x,\theta)=0$, the derivative is $dx^*/d\theta=-F_x^{-1}F_\theta$ when the Jacobian is nonsingular.

3. Inverse function theorem

If $f:\mathbb R^n\to\mathbb R^n$ is continuously differentiable and its Jacobian is nonsingular at $x_0$, then it has a local differentiable inverse near that point. The derivative of the inverse is the inverse Jacobian.

4. Equilibrium systems

For multiple markets or endogenous variables, write equilibrium conditions as a vector system. The Jacobian describes local responses and regularity. Singular Jacobians signal failure of this local theorem; they may correspond to bifurcation, redundant equations, or other structural features requiring separate analysis.

Worked example

Let equilibrium satisfy $F(y,\theta)=y^2+\theta y-1=0$. Then $dy/d\theta=-y/(2y+\theta)$ wherever $2y+\theta\ne0$. Evaluate the derivative at a specified equilibrium branch.

Problem set

Derive the scalar implicit derivative.
Compute comparative statics for a supply-demand equilibrium.
State the nonsingularity condition for a vector system.
Explain why a local inverse need not be globally unique.
Interpret a singular equilibrium Jacobian.

Computational lab

Solve nonlinear equilibrium systems numerically. Compare finite-difference comparative statics with derivatives from the implicit function theorem.

Coding exercises

These exercises apply the calculus of this lecture to real economic data in Python, mixing symbolic (sympy), numerical (scipy) and statistical (statsmodels) tools. Install once with pip install sympy scipy numpy pandas statsmodels wooldridge. Try the starter code first, then open the solution. Every printed result below comes from running the solution code exactly as shown.

Exercise 11.1 — Comparative statics of the Solow steady state with the implicit function theorem

Data: macrodata — U.S. investment share of GDP and population growth, used to calibrate the model.

In the Solow model, steady-state capital per worker $k^*$ solves $F(k,s)=s\,k^{\alpha}-(n+\delta)k=0$.

Calibrate $s$ as the average ratio of real investment to real GDP and $n$ as average annual population growth; set $\alpha=1/3$, $\delta=0.05$.
Solve for $k^*$ numerically (brentq), without using the closed form.
Use the implicit function theorem, $dk^*/ds=-F_s/F_k$, to compute the effect of a higher saving rate, and the elasticity $\frac{dk^*}{ds}\frac{s}{k^*}$. Compare with a finite difference and with the closed-form elasticity $1/(1-\alpha)$.

import numpy as np, statsmodels.api as sm
from scipy import optimize
m = sm.datasets.macrodata.load_pandas().data
s = (m["realinv"] / m["realgdp"]).mean(); n = np.log(m["pop"]).diff().mean() * 4
a, dl = 1/3, 0.05
F = lambda k, s: s * k**a - (n + dl) * k
# TODO: brentq for k*, IFT derivative: -F_s/F_k with F_s = k^a, F_k = s a k^(a-1) - (n+dl)

Solution, output and discussion

import numpy as np, statsmodels.api as sm
from scipy import optimize
m = sm.datasets.macrodata.load_pandas().data
s = (m["realinv"] / m["realgdp"]).mean(); n = np.log(m["pop"]).diff().mean() * 4
a, dl = 1/3, 0.05
F = lambda k, s: s * k**a - (n + dl) * k
kstar = lambda s: optimize.brentq(F, 1e-6, 1e6, args=(s,), xtol=1e-14)
k = kstar(s)
print(f"calibration: s = {s:.4f}, n = {n:.4f}, alpha = {a:.3f}, delta = {dl}")
print(f"steady state k* = {k:.4f}  (closed form {(s/(n+dl))**(1/(1-a)):.4f})")
Fs, Fk = k**a, s * a * k**(a - 1) - (n + dl)
dk = -Fs / Fk
h = 1e-6; fd = (kstar(s + h) - kstar(s - h)) / (2 * h)
print(f"IFT: dk*/ds = {dk:.4f}  vs finite difference {fd:.4f};  F_k = {Fk:.4f} < 0 (stable)")
print(f"elasticity dk*/ds * s/k* = {dk * s / k:.4f}  vs closed form 1/(1-alpha) = {1/(1-a):.4f}")

Output:

calibration: s = 0.1328, n = 0.0110, alpha = 0.333, delta = 0.05
steady state k* = 3.2142  (closed form 3.2142)
IFT: dk*/ds = 36.3174  vs finite difference 36.3174;  F_k = -0.0406 < 0 (stable)
elasticity dk*/ds * s/k* = 1.5000  vs closed form 1/(1-alpha) = 1.5000

Calibrated to the U.S. (investment share 13%, population growth 1.1%), the numerical root matches the closed-form steady state. The implicit function theorem gives $dk^*/ds=-F_s/F_k$ without solving for $k^*$ explicitly, and it agrees with the finite difference. $F_k<0$ at the steady state, which is both the condition for the IFT to apply and the condition for stability. The elasticity of steady-state capital with respect to the saving rate is exactly $1/(1-\alpha)=1.5$.

Exercise 11.2 — Yield to maturity as an inverse function, and duration

Data: intqrt — U.S. 12-month Treasury yield (average, used as a reference yield).

The price of a bond is a decreasing function of its yield, $P(y)=\sum_t c_t/(1+y)^t$. Its inverse $y(P)$ is the yield to maturity.

For a 10-year bond with a 6% annual coupon and face value 100, compute $P$ at the average 12-month yield.
Invert numerically: find $y$ from $P$, and check that the round trip recovers the yield.
By the inverse function theorem, $dy/dP=1/P'(y)$. Compute it, and the modified duration $-P'(y)/P$. Use duration to approximate the price change from a 1-point yield rise and compare with the exact change.

import numpy as np, wooldridge as wd
from scipy import optimize
y0 = wd.data("intqrt")["r12"].mean() / 100
cf = np.r_[np.full(9, 6.0), 106.0]; t = np.arange(1, 11)
P = lambda y: np.sum(cf / (1 + y)**t); dP = lambda y: -np.sum(t * cf / (1 + y)**(t + 1))
# TODO: price, inverse via brentq, dy/dP = 1/dP, modified duration

Solution, output and discussion

import numpy as np, wooldridge as wd
from scipy import optimize
y0 = wd.data("intqrt")["r12"].mean() / 100
cf = np.r_[np.full(9, 6.0), 106.0]; t = np.arange(1, 11)
P = lambda y: np.sum(cf / (1 + y)**t); dP = lambda y: -np.sum(t * cf / (1 + y)**(t + 1))
p0 = P(y0)
yinv = optimize.brentq(lambda y: P(y) - p0, -0.5, 1.0, xtol=1e-15)
print(f"yield {100*y0:.4f}% -> price {p0:.4f};  inverted yield {100*yinv:.10f}%")
print(f"dy/dP = 1/P'(y) = {1/dP(y0):.6f}  (yield change per $1 of price)")
D = -dP(y0) / p0
print(f"modified duration {D:.4f} years")
print(f"1-point yield rise: duration approximation {-D*0.01*100:.3f}% of price, exact {100*(P(y0+0.01)/p0 - 1):.3f}%")

Output:

yield 6.9269% -> price 93.4675;  inverted yield 6.9269354824%
dy/dP = 1/P'(y) = -0.001483  (yield change per $1 of price)
modified duration 7.2163 years
1-point yield rise: duration approximation -7.216% of price, exact -6.890%

Because the price is strictly decreasing in the yield, the inverse function exists, and the numerical inversion recovers the yield to ten decimals. The inverse function theorem gives $dy/dP=1/P'(y)$: a $1 price fall raises the yield by about 15 basis points. Modified duration (7.2 years) predicts a 7.2% price fall for a 1-point yield rise, against an exact fall of 6.9%. The gap is convexity: the price–yield curve is convex, so duration overstates losses and understates gains (Finance, Lecture 7).

Reading

Simon and Blume, Mathematics for Economists; Chiang and Wainwright, Fundamental Methods of Mathematical Economics; Apostol, Calculus, Vol. 2.