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Lecture 0 Mathematical Essentials
1
Lecture 0. Mathematical Essentials
Throughout this course we will be dealing with models involving constrained optimization. In this
section we outline a rather straightforward method for finding solutions to these problems and
analyzing the comparative statics of these solutions. The formal solution to these problems is most
easily understood in terms of the method of lagrange multipliers. In general we will consider a
problem of the form
(0.1) Max F(X1 ...Xk,B) s.t G(X1...Xk,B) £ M
x
where X represents the individual's decision variables (how much of each good to consume for the
generic consumer problem or how much of each type of product to produce) and B represent the
parameters influencing the individual's decision such as the prices of individual products or the market
rates of return on alternative investments. The function F represents the individual's objective while his
choice of the X's is restricted to combinations that satisfy G(X,B) £ M. M gives the level of this
individual's constraint.
The solution to this problem can be found by setting up the Lagrangian function
(0.2) L = F(X1 ...Xk,B) + l ( M - G(X1...Xk,B))
The solution for the optimal levels of X will be a critical point of L (i.e. a point where the derivatives of
the Lagrangian with respect to X and l are simultaneously zero. Hence these conditions are
(0.3) dL/dX1 = dF/d X1 - l dG/ d X1 = 0 => dF/d X1 = l dG/ d X1
dL/dX2 = dF/d X2 - l dG/ d X2 = 0 => dF/dX2 = l dG/dX2
.
.
.
dL/dXk = dF/ dXk - l dG/ dXk = 0 => dF/dXk = l dG/dXk
dL/dl = M - G(X,B) = 0 => M = G(X,B)
The first k equations tell us that the optimal solution requires the impact of each of the X variables on
the objective function, dF/dXj, to be proportionate to its effect on the constraint function, dG/dXj (with
l being the factor of proportionality). The necessity of this condition follows from the following
argument: if we increase Xi by some small amount dXi then the value of our constraint function would
fall be dG/dXi * dXi which would then allow us to change Xj by an amount - dG/dXi / dG/dXj dXi (to
keep the value of G(X,B) = M). The net effect of this on the objective function would be
Lecture 0 Mathematical Essentials
2
(0.4) dF = dF/dXi dXi - dF/dXj dG/dXi / dG/dXj dXi.
At the optimum dF must be zero which implies that
(0.5) dF/dXi / dFdXj = dG/dXi / dG/dXj
or that the effects of the x variables on the objective must be proportional to their effect on the
constraint. In fact it should be noted that the argument used here to derive equation (0.5), where we
evaluated the change in the objective function by changing the X variables in such a way as to continue
to satisfy the constraint, G(X,B) = M will often be used to derive necessary conditions for
maximization when it is impractical or unnecessary to derive the full set of conditions for maximization.
The k+1 first order conditions in (0.3) give us k+1 equations to solve for the k+1 variables, X1 ...Xk
and L. Since the first order conditions depend on the parameters of the model, B and M, the solutions
for X1 ...Xk and L will also be functions of B and M. We will typically denote these solutions by
(0.6) X1 = X1 *(B,M)
X2 = X2*(B,M)
.
.
X k = Xk*(B,M)
l = l*(B,M)
and the optimized value of the Lagrangian by
(0.7) L*(B,M) = F(X1(B,M)...Xk(B,M),B) + l(B,M)*( M - G(X1(B,M)...Xk(B,M),B) ).
In practice we are often interested in how the X variables respond to changes in the underlying
parameters B and M. These can be found by totally differentiating the equations in 0.3 to yield
(0.8)
Fxx Gx Gx
Gx
dX
d
Fxb Gxb
Gb
dB
dM
- -
-
æ
è ç
ö
ø ÷
æ
è ç
ö
ø ÷
=
- +
-
æ
è çö ø ÷
æ
è ç
ö
ø ÷
l
l
l
0
0
1
which implies
(0.9)
dX
d
Fxx Gxx Gx
Gx
Fxb Gxb
Gb
dB
l dM
æ l l
è ç
ö
ø ÷
=
- -
-
æ
è ç
ö
ø ÷
- +
-
æ
è ç
ö
ø ÷
æ
è ç
ö
ø ÷
-
0
0
1
1
Often we are also interested in how the value of the maximized the optimized Lagrangian, L* (equal in
value to the optimized objective function since M - G(X1(B,M)...Xk(B,M),B) = 0) responds to
Lecture 0 Mathematical Essentials
3
changes in the underlying parameters, B. By the envelope theorem (or by total differentiation and
wholesale cancellation using the equations in 0.3) we then have
(0.8) dL*/dB = dF/dB - ldG/dB and
dL*/dM = l.
Hence, the derivatives of the maximized Lagrangian with respect to the parameters B and M are equal
to the derivatives of the original Lagrangian. Intuitively since we have maximized with respect to the X
variables (and hence the derivatives with respect to these variables are zero) any changes in the X
variables (or L) have only second order effects on the objective. Understanding and being able to
implement the techniques illustrated in this section should allow you to analyze most of the problem we
cover in this course.
Lecture 0 Mathematical Essentials
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