San José State University
Department of Economics
Thayer Watkins
Silicon Valley

The Slutsky Equation

First consider the following optimization problem and its comparative statics:

Maximize U(x1, x2)
with respect to x1 and x2
subject to the constraint that
p1x1 + p2x2 = y

The first order conditions for the maximizing values of x1 and x2 are:

∂u/∂x1 - λp1 = 0
∂u/∂x2 - λp2 = 0

where λ is the Lagrangian multiplier.

Comparative Statics

The differentiation of the first order equations with respect to p1 yields

u1,1∂x1/∂p1 + u1,2∂x2/∂p1 − p1∂λ/∂p1 = λ
u2,1∂x1/∂p1 + u2,2∂x2/∂p1 − p2∂λ/∂p1 = 0

where ui,j is ∂2u/∂xj∂xi.

The differentiation of the budget constraint with respect to p1 yields an equation that can be put into the form:

-p1(∂x1/&partt;p1) − p2(∂x2/∂p1) = x1.

These equations in matrix form are:

| u1,1u1,2 -p1 || ∂x1/∂p1 |  | λ  |
| u2,1u2,2 -p1 | | ∂x2/∂p1 | = | 0  |
| −p1-p2    0 || ∂λ/∂p1    | | x1 |


Let the 3×3 matrix on the left, which happens to be called a bordered Hessian matrix, be denoted as H. The column vector of the x1, x2 and λ values can be represented as partial derivatives with respect to p1 and are denoted as ∂X/∂p1. The matrix equation then become

H(∂X/∂p1) = (λ, 0 , x1)T,

where the column vector on the right is represented in the form of a transpose of a row vector.

Thus the solution is ∂X/∂p1 = H-1(λ, 0 , x1)T
which can be decomposed into two terms; i.e.,

∂X/∂p1 = H-1(λ, 0 , 0)T + H-1(0, 0 , x1)T

These two terms represent the substitution effect and the income effect, respectively. This assertion is to be proven.

The Income Effect

When the first order conditions are differentiated with respect to income y the result is:

| u1,1u1,2 -p1 || ∂x1/∂y | | 0 |
| u2,1u2,2 -p1 | | ∂x2/∂y | = | 0  |
| -p1-p2    0 || ∂λ/∂y   | | -1 |



∂X/∂y = H-1(0, 0 , -1)T

The second term in the equation for the effect of a change in p1,

H-1(0, 0 , x1)T
can be represented as
-x1H-1(0, 0 , -1)<T
and hence as

Thus this term is the income effect. To establish the substitution effect another optimization problem has to be considered.

Minimizing the Cost of
Achieving a Given Utility Level

The optimization problem is:

Minimize C = p1x1 + p2x2
with respect to x1 and x2
subject to the constraint
u(x1, x2) = u0.

The first order conditions for this optimization problem are:

p1 - (1/μ)∂u/∂x1 = 0
p2 - (1/μ)∂u/∂x2 = 0

where the Lagrangian multiplier has been expressed as (1/μ) in order that the first conditions can be written as

∂u/∂x1 - μp1 = 0
∂u/∂x2 - μp2 = 0

The differentiation of these equations with respect to p1 yields essentially the same first two equation as for the first optimization problem. In the previous case the constraint was differentiated with respect to p1. In this case the result is

∂u/∂x1(∂x1/∂p1) + ∂u/∂x2(∂x2/∂p1) = 0
but because of the first order conditions that
∂u/∂x1 = μp1 and ∂u/∂x2 = μp2
the condition can be expressed as
-p1∂x1/∂;ppppp1 - p2∂x2/∂p1 = 0

The matrix form of the equation is thus

H(∂X/∂p1) = (μ, 0, 0)T
and their solution is
∂X/∂p1 = H-1(μ, 0, 0)T

This is the same as the other term in the comparative statics analysis for the first optimization problem with λ=μ.

The same results apply for changes in p2.

Thus we have Slutsky's equation:

(∂X/∂pi)y = (∂X/∂pi)u - xi(∂X/∂y)P

where ∂X/∂pi and ∂X/∂y represents the impact of a change in the price pi and money income y on the vector of quantities demanded and the Lagrangian multiplier. The notation ( )y, ( )u and ( )P indicates that the derivatives inside of the parentheses are with, respectively, money income y held constant, utility (real income) u held constant, and all prices P held constant.


In the preceding only a change in p1 was considered. There are analogous equations for the impact of a change in p2. Rather than present those equations separately it is more interesting to present the comparative statics analysis of price and money income changes.

The full set of equations which derive from the first order conditions is:

| u1,1u1,2   -p1 |    | λ    0       0 |
| u2,1u2,2   -p1 | | ∂X/∂p1 ∂X/∂p2 ∂X/∂y | = | 0    λ       0 |
| -p1-p2     0 |      | x1   x2    -1 |


where X is the column vector (x1, x2, λ )T.

Thus the solution is

    | λ    0       0 |
| ∂X/∂p1 ∂X/∂p2 ∂X/∂y | = H-1 | 0    λ       0 |
     | x1   x2    -1 |


where H is the bordered Hessian matrix.

HOME PAGE OF applet-magic
HOME PAGE OF Thayer Watkins