28
Oct 17

The Lagrangian multiplier interpretation

Motivation. Consider the problem of utility maximization $\max u(x,y)$ under the budget constraint $p_xx+p_yy=M.$ Suppose the budget $M$ changes a little bit and let us see how the solution of the problem depends on changes in the budget. The maximized value of the utility will depend on $M$, so let us reflect this dependence in the notation $U(M)=u(x^\ast(M),y^\ast(M))$ where, for each $M$, $(x^\ast(M),y^\ast(M))$ is the maximizing bundle. The value of the Lagrangian multiplier $\lambda$ in the FOC's $\frac{\partial L}{\partial x}=0$, $\frac{\partial L}{\partial y}=0$, $\frac{\partial L}{\partial \lambda }=0$ will also depend on $M$: $\lambda =\lambda(M).$ The property that we will prove is

$\frac{dU}{dM}=\lambda (M),$

that is, the Lagrangian multiplier measures the sensitivity of the maximized utility function to changes in the budget.

General formulation

The main problem is to maximize $f(x,y)$ subject to $g(x,y)=0$. However, instead of a fixed constraint we consider a set of constraints perturbed by a constant $c$: $g(x,y)+c=0.$ Here $c$ varies in a small neighborhood $(-\varepsilon ,\varepsilon)$ of zero (which corresponds to varying $M$ in a neighborhood of some $M_{0}$ in the motivating example). Now everything depends on $c$, and differentiation with respect to $c$ will give the desired result.

Employing the constraint

As before, we assume the implicit function existence condition:

$\left( \frac{\partial g}{\partial x},\frac{\partial g}{\partial y}\right)\neq 0.$

Changing the notation, if necessary, we can think that it is the last component of the gradient that is not zero: $\frac{\partial g}{\partial y}\neq 0.$ This condition applies to the perturbed constraint too and guarantees existence of the implicit function, which now depends also on $c$: $y=y(x,c)$. Plugging it into the constraint we get $g(x,y(x,c))+c=0$, and differentiation yields

(1) $\frac{\partial g}{\partial y}\frac{\partial y}{\partial c}+1=0.$

Employing the FOC's

Critical assumption. For each $c\in (-\varepsilon ,\varepsilon )$ the perturbed maximization problem has a solution. At least in our motivating example, this assumption is satisfied.

The maximized objective function is denoted $F(c)=f(x,y(x,c)).$ At each $c,$ we have the right to use the FOC's for the Lagrangian. One of the FOC's is

$\frac{\partial f}{\partial y}+\lambda (c)\frac{\partial g}{\partial y}=0.$

To make use of (1), multiply this by $\frac{\partial y}{\partial c}$:

$0=\frac{\partial f}{\partial y}\frac{\partial y}{\partial c}+\lambda(c)\frac{\partial g}{\partial y}\frac{\partial y}{\partial c}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial c}-\lambda (c).$

It follows that

$\frac{dF}{dc}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial c}=\lambda (c).$

With $c=0$ we have $\lambda(c)=\lambda$, the Lagrange multiplier for the unperturbed problem, and as a result

$\frac{dF}{dc}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial c}=\lambda.$

20
Oct 17

Optimization with constraints: economic and financial examples

The economic and financial examples below provide ample motivation for using the optimization theory with constraints. For geometry, check out the post on level sets, isoquants and indifference curves.

Examples in Economics

Consumption theory. Let $u(x,y)$ be a utility function of consuming two goods $x,y.$ Denote $p_x,\ p_y$ their prices and $M$ the amount available for consumption. Then the budget constraint is $p_xx+p_yy=M.$ The utility maximization problem is to

(1) maximize $u(x,y)$ subject to $p_xx+p_yy=M.$

The dual problem to the utility maximization problem is the expenditure minimization problem:

(2) minimize expenditure $p_xx+p_yy$ subject to the desired level of consumption $u(x,y)=const$

where the $const$ is chosen by the consumer.

Production theory. Let $f(x,y)$ be a production function, that is, $x,y$ are two inputs (for example, capital and labor) and the value $f(x,y)$ gives the output produced using these inputs. $M$ in this context means the budget available for employing the inputs. The output maximization problem is to

(3) maximize $f(x,y)$ subject to the cost constraint $p_xx+p_yy=M.$

This is absolutely similar to (1). Moreover, often both $u(x,y)$ and $f(x,y)$ are modeled as the Cobb-Douglas function $Ax^\alpha y^\beta.$

It's easy to formulate the dual problem of cost minimization:

(4) minimize the cost of production $p_xx+p_yy$ subject to the desired level of output $f(x,y)=const$

where the $const$ is chosen by the producer.

Examples in Finance

Now we consider investor optimization problems. There are $m$ stocks and the stock returns are denoted $R_1,...,R_m.$ They are random because they are not predictable. The investor has initially $M$ dollars to invest in these stocks. Suppose he chooses to invest $M_1$ in stock 1,..., $M_m$ in stock $m,$ so that the initial investment is split as $M_1+...+M_m=M.$ Dividing this equation through by $M$ we get $M_1/M+...+M_m/M=1.$ Here $s_i=M_i/M$ are called shares of the total investment. They obviously satisfy

(5) $0 and $s_1+...+s_m=1.$

Working with shares is better because they don't depend on $M$ and the results are equally applicable to small and large investors. Shares of investment are determined at the time investment is made and are deterministic. The total return on the portfolio is defined by $R=s_1R_1+...+s_mR_m.$

Riskiness of the portfolio is measured by standard deviation $\sigma (R).$ The basic fact is that a higher expected return $ER$ is associated with a higher risk $\sigma (R).$ Now we can state two optimal portfolio choice problems:

(6) maximize $ER$ subject to the chosen level of risk $\sigma (R)=const$

and

(7) minimize risk $\sigma (R)$ subject to the chosen level of expected return $ER=const.$

In both cases the investor varies the shares $s_{i}$ so as to obtain the best allocation of his money. The constants are chosen subjectively. A more objective problem obtains if we use what is called an expected return per unit of risk: $\frac{ER}{\sigma (R)}.$ Instead of the above two problems we can solve the maximization problem

(8) $\max \frac{ER}{\sigma (R)}$

In addition to explicitly written constraints in (6)-(7) we remember that the shares satisfy (5), which contains inequality constraints.