# Mathematical methods for economic theory

Martin J. Osborne

## 6.3 The envelope theorem

In economic theory we are often interested in how the maximal value of a function depends on some parameters.

Consider, for example, a firm that can produce output with a single input using the production function f. The standard theory is that the firm chooses the amount x of the input to maximize its profit pf(x) − wx, where p is the price of output and w is the price of the input. Denote by x*(wp) the optimal amount of the input when the prices are w and p. An economically interesting question is: how does the firm's maximal profit pf(x*(wp)) − wx*(wp) depend upon p?

We have already answered this question in an earlier example. To do so, we used the chain rule to differentiate pf(x*(wp)) − wx*(wp) with respect to p, yielding

f(x*(wp)) + x*'p(wp)[pf'(x*(wp)) − w],
and then used the fact that x*(wp) satisfies the first-order condition pf'(x*(wp)) − w = 0 to conclude that the derivative is simply f(x*(wp)).

That is, the fact that the value of the variable satisfies the first-order condition allows us to dramatically simplify the expression for the derivative of the firm's maximal profit. On this page I describe results that generalize this observation to an arbitrary maximization problem.

### Unconstrained problems

Consider the unconstrained maximization problem
maxx f(xr),
where x is a n-vector and r is a k-vector of parameters. Assume that for any vector r the problem has a unique solution; denote this solution x*(r). Denote the maximum value of f, for any given value of r, by f*(r):
f*(r) = f(x*(r), r).
We call f* the value function.
Example
Let n = 1, k = 2, and f(xr) = xr1 − r2x, where 0 < r1 < 1. This function is concave (look at its second derivative), and any solution of maxx f(xr) is positive, so a solution satisfies the first-order condition r1xr1−1 − r2 = 0. Thus x*(r) = (r1/r2)1/(1 − r1) is the solution of the problem, so that the value function of f is
 f*(r) = (x*(r))r1 − r2x*(r) = (r1/r2)r1/(1 − r1) − r2(r1/r2)1/(1 − r1).
We wish to find the derivatives of f* with respect to each parameter rh for h = 1, ..., k. First, because x*(r) is a solution of the problem when the parameter vector is r, if f is differentiable then x*(r) satisfies the first-order conditions
f'i(x*(r), r) = 0 for i = 1, ..., n.
Now, if the function x* is differentiable then differentiating f* with respect to rh using the chain rule we have
 f*h'(r) = ∑ni=1 f'i(x*(r), r)·(∂xi*/∂rh)(r) + f'n+h(x*(r), r).
The first term corresponds to the change in f* caused by the change in the solution of the problem that occurs when rh changes; the second term corresponds to the direct effect of a change in rh on the value of f.

Given the first-order conditions, this expression simplifies to

 f*h'(r) = f'n+h(x*(r), r) for h = 1, ..., k.
Note that the derivative on the right-hand side is the partial derivative of f with respect to rh (the n + hth variable in the vector (xr)), holding x fixed at x*(r). The argument I have given for this result assumes that the maximization problem has a unique solution x* and this solution is differentiable (in r). These assumptions are not required for the conclusion; only the differentiability of f* is required. A precise statement (with proof) follows.
Proposition (Envelope theorem for an unconstrained maximization problem)
Let f be a function of n + k variables, let r be a k-vector, and let the n-vector x* be a maximizer of f(xr). Assume that the partial derivative f'n+h(x*, r) (i.e. the partial derivative of f with respect to rh at (x*, r)) exists. Define the function f* of k variables by
f*(r) = maxx f(xr) for all r.
If the partial derivative f*h'(r) exists then
 f*h'(r) = f'n+h(x*, r).
Proof
We have f(x*, r) = f*(r) and f(x*, s) ≤ f*(s) for all s. Put differently, f(x*, s) − f*(s) ≤ 0 for all s and f(x*, r) − f*(r) = 0. Thus r maximizes f(x*, s) − f*(s). Because the partial derivatives f'n+h(x*, r) and f*h'(r) exist, the partial derivative of f(x*, s) − f*(s) with respect to sh at r exists, so that by a previous result we have f'n+h(x*, r) − f*h'(r) = 0.
This result says that the change in the maximal value of the function as a parameter changes is the change caused by the direct impact of the parameter on the function, holding the value of x fixed at its optimal value; the indirect effect, resulting from the change in the optimal value of x caused by a change in the parameter, is zero.

The next two examples illustrate how the result simplifies the calculation of the derivatives of the value function.

Example
Consider the problem studied at the start of this section, in which a firm can produce output, with price p, using a single input, with price w, according to the production function f. The firm's profit when it uses the amount x of the input is π(x, (wp)) = pf(x) − wx, and its maximal profit is
 π*(w, p) = pf(x*(w, p)) − wx*(w, p),
where x*(wp) is the optimal amount of the input at the prices (wp). The function π* is known as the firm's profit function. By the envelope theorem, the derivative of this function with respect to p is the partial derivative of π with respect to p evaluated at x = x*(wp), namely
f(x*(wp)).
In particular, the derivative is positive: if the price of output increases, then the firm's maximal profit increases.

Also by the envelope theorem the derivative of the firm's maximal profit with respect to w is

x*(pw).
(This result is known as Hotelling's Lemma, after Harold Hotelling, 1895–1973.) In particular, this derivative is negative: if the price of the input increases, then the firm's maximal profit decreases.

A consequence of Hotelling's Lemma is that we can easily find the firm's input demand function x* if we know the firm's profit function, even if we do not know the firm's production function: we have x*(pw) = −π*w'(pw) for all (pw), so we may obtain the input demand function by simply differentiating the profit function.

Example
Consider the earlier example, in which f(xr) = xr1 − r2x, where 0 < r1 < 1. We found that the solution of the problem
maxx f(xr)
is given by
x*(r) = (r1/r2)1/(1 − r1).
Thus by the envelope theorem, the derivative of the maximal value of f with respect to r1 is the derivative of f with respect to r1 evaluated at x*(r), namely
(x*(r))r1ln x*(r),
or
(r1/r2)r1/(1 − r1)ln (r1/r2)1/(1 − r1).
(If you have forgotten how to differentiate xr1 with respect to r1 (not with respect to x!), remind yourself of the rules.)

If you approach this problem directly, by calculating the value function explicitly and then differentiating it, rather than using the envelope theorem, you are faced with the task of differentiating

 (x*(r))r1 − r2x*(r) = (r1/r2)r1/(1 − r1) − r2(r1/r2)1/(1 − r1)
with respect to r1, which is much more difficult than the task of differentiating the function f partially with respect to r1.
A trivial example in which f does not have a unique maximizer is f(xr) = r. In this case, the value of f does not depend on x, so that every value of x maximizes f. Thus a straightforward application of the chain rule to demonstrate the envelope theorem is not possible for this example. But the assumptions in my statement of the result are satisfied, and indeed the conclusion holds: f*(r) = r, so that f*'(r) = 1, and fr(x*, r) = 1.

Here is an example in which f has a unique maximizer, but this maximizer is not a differentiable function of r.

Example
Define the function f of two variables by
f(xr) = −(x − r)2 if r ≥ 0 −(x + r)2 if r < 0 .
For each value of r, this function has a unique maximizer, given by x*(r) = |r|. Thus the maximizer is not differentiable at r = 0. However, the partial derivative f'r(x*(0), 0) exists (f(0, r) = −r2 for all r) and f*(r) = 0 for all r, so that the assumptions of the result are satisfied. The conclusion holds for r = 0, the point at which x* is not differentiable: f*'(0) = 0 and f'r(0, 0) = 0.
Why is the result called the envelope theorem? The American Heritage Dictionary (3ed) gives one meaning of “envelope” to be “A curve or surface that is tangent to every one of a family of curves or surfaces”. In the following figure, each black curve is the graph of f as a function of r for a fixed value of x. (Only a few values of x are considered; one can construct as many as one wishes.) Each of these graphs shows how f changes as r changes, for a given value of x. To find the solution of the maximization problem for any given value of r, we find the highest function for that value of r. For example, for r = r', the highest function is the one colored blue. The graph of the value function f* is the locus of these highest points; it is the envelope of the graphs for each given value of x. From the figure, the envelope theorem is apparent: the slope of the envelope at any given value of r is the slope of the graph of f(x*(r), r). (For example, the slope of the envelope at r' is the slope of the blue curve at r'.)

### Constrained problems

We may apply the same arguments to maximization problems with constraints. Consider the problem
maxx f(xr) subject to gj(xr) = 0 for j = 1, ..., m,
where x is an n-vector, r is a k-vector of parameters, and f and gj for j = 1, ..., m are continuously differentiable functions. Assume that for every value of r the problem has a single solution, and denote this solution x*(r). As before, denote the maximum value of f, for any given value of r, by f*(r):
f*(r) = f(x*(r), r).
Call f* the value function.

We want to calculate the derivatives f*h'(r) for h = 1, ..., k of the function f*. If x* is differentiable, then using the chain rule we have

 f*h'(r) = ∑ni=1 f'i(x*(r),r)·(∂x*i/∂rh)(r) + f'n+h(x*(r), r).
Now, if the Jacobian matrix of the constraints has rank m then by an earlier result for any value of r there are unique numbers λ1(r), ..., λm(r) such that the solution x*(r) satisfies the first-order conditions
f'i(x*(r), r) − m
j=1
λj(r)(∂gj/∂xi)(x*(r), r)
= 0 for i = 1, ..., n.
Thus we have
 f*h'(r) = ∑ni=1[∑mj=1λj(r)(∂gj/∂xi)(x*(r), r)] · (∂x*i/∂rh)(r) + f'n+h(x*(r), r).
Reversing the sums, we obtain
 f*h'(r) = ∑mj=1λj(r)[∑ni=1 (∂gj/∂xi)(x*(r),r) · (∂x*i/∂rh)(r)] + f'n+h(x*(r),r).
Now, for all j = 1, ..., m we have gj(x*(r), r) = 0 for all r. Differentiating this equation with respect to rh we get
n
i=1
(∂gj/∂xi)(x*(r), r)
· (∂xi*/∂rh)(r) + (∂gj/∂rh)(x*(r), r) = 0.
Hence
 f*h'(r) = −∑mj=1 λj(r)(∂gj/∂rh)(x*(r), r) + f'n+h(x*(r), r).
Now define the Lagrangean function L by
 L(x, λ, r) = f(x, r) − ∑mj=1λjgj(x, r),
where λ = (λ1, ..., λm). Then we have
f*h'(r) = L'n+m+h(x*(r), λ(r), r).
We have proved the following result.
Proposition (Envelope theorem for constrained maximization problems)
Let f and g1, ..., gm be continuously differentiable functions of n + k variables, with m ≤ n. Suppose that for all values of the k-vector r the problem
maxx f(xr) subject to gj(xr) = 0 for j = 1, ..., m,
where x is an n-vector, has a unique solution, which is differentiable in r. Denote this solution x*(r) and define the function f* of k variables by
 f*(r) = maxx f(x, r) subject to gj(x, r) = 0 for j = 1, ..., m.
Suppose that the rank of the m × n matrix in which the (ij)th component is (∂gj/∂xi)(x*(r), r) is m. Define the function L by
 L(x, λ, r) = f(x, r) − ∑mj=1λjgj(x, r) for every (x, λ, r),
where x is an n-vector, λ is an m-vector, and r is a k-vector. Then
 f*h'(r) = L'n+m+h(x*(r), λ(r), r) for h = 1, ..., k,
where λ(r) = (λ1(r), ..., λm(r)) and λj(r) is the value of Lagrange multiplier associated with the jth constraint at the solution of the problem.
Example
Consider a utility maximization problem:
maxx u(x) subject to p·x = w.
where x is a vector (a bundle of goods), p is the price vector, and w is the consumer's wealth (a number). Denote the solution of the problem by x*(pw), and denote the value function by v, so that
v(pw) = u(x*(pw)) for every (pw).
The function v is known as the indirect utility function.
(∂v/∂pi)(pw) = −λ*(pw)xi*(pw)
(since u does not depend independently on p or w) and
(∂v/∂w)(pw) = λ*(pw).
Thus
 (∂v/∂pi)(p, w) (∂v/∂w)(p, w)
= −xi*(pw).
That is, if you know the indirect utility function then you can recover the demand functions. This result is known as Roy's identity (after René Roy, 1894–1977).