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4. Models with Multiple Explanatory Variables

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Chapter 2 assumed that the dependent variable (Y) is affected by only ONE explanatory variable (X).

Sometimes this is the case. Example:Age = Days Alive/365.25

Usually, this is not the case. Example: midterm mark depends on:

- how much you study
- how well you study
- intelligence, etc

Demand = f( price of good, price of substitutes, income, price of compliments)

Consumption = f( income, tastes, wages)

Graduation rates = f( tuition, school quality, student quality)

Christmas present satisfaction = f (cost, timing, knowledge of person, presence of card, age, etc.)

It is often impossible analyze ONE variable’s impact if ALL variables are changing.

Instead, we analyze one variable’s impact, assuming ALL OTHER VARIABLES REMAIN CONSTANT

We do this through the partial derivative.

This chapter uses the partial derivative to expand the topics introduced in chapter 2.

4.1 Derivatives of Functions of More Than One Variable

4.2 Applications Using Partial Derivatives

4.3 Partial and Total Derivatives

4.4 Unconstrained Optimization

4.5 Constrained Optimization

Consider the function z=f(x,y). As this function takes into account 3 variables, it must be graphed on a 3-dimensional graph.

A partial derivative calculates the slope of a 2-dimensional “slice” of this 3-dimensional graph.

The partial derivative ∂z/∂x asks how x affects z while y is held constant (ceteris paribus).

In taking the partial derivative, all other variables are kept constant and hence treated as constants (the derivative of a constant is 0).

There are a variety of ways to indicate the partial derivative:

- ∂y/∂x
- ∂f(x,z)/∂x
- fx(x,z)
Note: dy=dx is equivalent to ∂y/∂x if y=f(x); ie: if y only has x as an explanatory variable.

(Therefore often these are used interchangeably in economic shorthand)

Let y = 2x2+3xz+8z2

∂y/ ∂x = 4x+3z+0

∂y/ ∂z = 0+3x+16z

(0’s are dropped)

Let y = xln(zx)

∂ y/ ∂ x = ln(zx) + zx/zx

= ln(zx) + 1

∂ y/ ∂ z = x(1/zx)x

=x/z

Let y = 3x2z+xz3-3z/x2

∂ y/ ∂ z=3x2+3xz2-3/x2

∂ y/ ∂ x=6xz+z3+6z/x3

Try these:

z=ln(2y+x3)

Expenses=sin(a2-ab)+cos(b2-ab)

Higher order partial derivates are evaluated exactly like normal higher order derivatives.

It is important, however, to note what variable to differentiate with respect to:

From before:

Let y = 3x2z+xz3-3z/x2

∂ y/ ∂ z=3x2+3xz2-3/x2

∂2y/ ∂ z2=6xz

∂2y/ ∂ z ∂ x=6x+3z2+6/x3

From before:

Let y = 3x2z+xz3-3z/x2

∂ y/ ∂ x=6xz+z3+6z/x3

∂2y/ ∂ x2=6z-18z/x4

∂2y/ ∂ x ∂ z=6x+3z2+6/x3

Notice that d2y/dxdz=d2y/dzdx

This is reflected by YOUNG’S THEOREM: order of differentiation doesn’t matter for higher order partial derivatives

As many real-world situations involve many variables, Partial Derivatives can be used to analyze our world, using tools including:

- Interpreting coefficients
- Partial Elasticities
- Marginal Products

Given a function a=f(b,c,d), the dependent variable a is determined by a variety of explanatory variables b, c, and d.

If all dependent variables change at once, it is hard to determine if one dependent variables has a positive or negative effect on a.

A partial derivative, such as ∂ a/ ∂ c, asks how one explanatory variable (c), affects the dependent variable, a, HOLDING ALL OTHER DEPENDENT VARIABLES CONSTANT (ceteris paribus)

A second derivative with respect to the same variable discusses curvature.

A second cross partial derivative asks how the impact of one explanatory variable changes as another explanatory variable changes.

Ie: If Happiness = f(food, tv),

∂2h/ ∂ f ∂tv asks how watching more tv affects food’s effect on happiness (or how food affects tv’s effect on happiness). For example, watching TV may not increase happiness if someone is hungry.

Consider the following formula for corn production:

Corn = 500+100Rain-Rain2+50Scare*Fertilizer

Corn = bushels of corn

Rain = centimeters of rain

Scare=number of scarecrows

Fertilizer = tonnes of fertilizer

Explain this formula

1) Intercept = 500

-if it doesn’t rain, there are no scarecrows and no fertilizer, the farmer will harvest 500 bushels

2) ∂Corn/∂Rain=100-2Rain

-positive until Rain=50, then negative

-more rain increases the harvest at a decreasing rate until rain hits 50cm, then additional rain decreases the harvest at an increasing rate

3) ∂2Corn/∂Rain2=-2<0, (concave)

-More rain has a DECREASING impact on the corn harvest OR

-More rain DECREASES rain’s impact on the corn harvest by 2

4) ∂Corn/∂Scare=50Fertilizer

-More scarecrows will increase the harvest 50 for every tonne of fertilizer

-if no fertilizer is used, scarecrows are useless

5) ∂2Corn/∂Scare2=0 (straight line, no curvature)

-Additional scarecrows have a CONSTANT impact on corn’s harvest

6) ∂2Corn/∂Scare∂Fertilizer=50

-Additional fertilizer increases scarecrow’s impact on the corn harvest by 50

7) ∂Corn/∂Fertilizer=50Scare

-More fertilizer will increase the harvest 50 for every scarecrow

-if no scarecrows are used, fertilizer is useless

8) ∂2Corn/∂Fertilizer2=0, (straight line)

-Additional fertilizer has a CONSTANT impact on corn’s harvest

9) ∂2Corn/∂Fertilizer ∂Scare =50

-Additional scarecrows increase fertilizer’s impact on the corn harvest by 50

Consider the demand formula:

Q = β1 + β2 Pown + β3 Psub + β4 INC

(Quantity demanded depends on a product’s own price, price of substitutes, and income.)

Here ∂ Q/ ∂ Pown= β2 = the impact on quantity when the product’s price changes

Here ∂ Q/ ∂ Psub= β3 = the impact on quantity when the substitute’s price changes

Here ∂ Q/ ∂ INC= β4 = the impact on quantity when income changes

Furthermore, partial elasticities can also be calculated using partial derivatives:

Own-Price Elasticity = ∂ Q/ ∂ Pown(Pown/Q)

= β2(Pown/Q)

Cross-Price Elasticity = ∂ Q/ ∂ Psub(Psub/Q)

= β3(Psub/Q)

Income Elasticity = ∂ Q/ ∂ INC(INC/Q)

= β4(INC/Q)

A favorite function of economists is the Cobb-Douglas Production Function of the form

Q=aLbKcOf

Where L=labour, K=Capital, and O=Other (education, technology, government, etc.)

This is an attractive function because if b+c+f=1, the demand function is homogeneous of degree 1. (Doubling all inputs doubles outputs…a happy concept)

Consider a production function for university degrees:

Q=aLbKcCf

Where

L=Labour (ie: professors), K=Capital (ie: classrooms)C=Computers

Finding partial derivatives:

∂ Q/ ∂ L=abLb-1KcCf

=b(aLbKcCf)/L

=b(Q/L)

=b* average product of labour

-in other words, adding an additional professor will contribute a fraction of the average product of each current professor

-this partial derivative gives us the MARGINAL PRODUCT of labour

For example, if 20 professors are employed by the department, and 500 students graduate yearly, and b=0.5:

∂ Q/ ∂ L=0.5(500/20)

=12.5

Ie: Hiring another professor will graduate 12.5 more students. The marginal productof professors is 12.5

Consider the function Q=f(L,K,O)

The partial derivative reveals the MARGINAL PRODUCT of a factor, or incremental effect on output that a factor can have when all other factors are held constant.

∂ Q/ ∂ L=Marginal Product of Labour (MPL)

∂ Q/ ∂ K=Marginal Product of Capital (MPK)

∂ Q/ ∂ O=Marginal Product of Other (MPO)

Since the “Professor Elasticity” (or PE) is defined as:

PE= ∂ Q/ ∂ L(L/Q)

We can find that

PE=b(Q/L)(L/Q)

=b

The partial elasticity with respect to labour is b.

The partial elasticity with respect to capital is c

The partial elasticity with respect to other is f

We can highlight elasticities by using logs:

Q=aLbKcCf

Converts to

Ln(Q)=ln(a)+bln(L)+cln(k)+fln(C)

We now find that:

PE= ∂ ln(Q)/ ∂ ln(L)=b

Using logs, elasticities more apparent.

Consider a log-log demand example:

Ln(Qdx)=ln(β1)+β2 ln(Px)+ β3 ln(Py)+ β4 ln(I)

We now find that:

Own Price Elasticity = β2

Cross-Price Elasticity = β3

Income Elasticity = β4

Considering the demand for the ipad, assume:

Ln(Qdipad)=2.7-1ln(Pipad)+4ln(Pnetbook)+0.1ln(I)

We now find that:

Own Price Elasticity = -1, demand is unit elastic

Cross-Price Elasticity = 4, a 1% increase in the price of netbooks causes a 4% increase in quantity demanded of ipads

Income Elasticity = 0.1, a 1% increase in income causes a 0.1% increase in quantity demanded for ipads

Often in econometrics, one variable is influenced by a variety of other variables.

Ie: Happiness =f(sun, driving)

Ie: Productivity = f(labor, effectiveness)

Using TOTAL DERIVATIVES, we can examine how growth of one variable is caused by growth in all other variables

The following formulae will combine x’s impact on y (dy/dx) with x’s impact on y, with other variables held constant (δy/δx)

Assume you are increasing the square footage of a house where

AREA = LENGTH X WIDTH

A=LW

If you increase the length,

the change in area is equal

to the increase in length

times the current width:

dL

Length

Width

Area

Notice that:

δA/δL=W, (partial derivative, since width is constant)

Therefore the increase in area is equal to:

dA=(δA/δL)dL

Length

A=LW

If you increase the width,

the change in area is equal

to the increase in width

times the current length:

Width

Area

dW

Notice that:

δA/δW=L, (partial derivative, since length is constant)

Therefore the increase in area is equal to:

dA=(δA/δW)dW

Next we combine the two effects:

Length

A=LW

An increase in both length

and width has the following

impact on area:

Width

Area

dW

dL

Now we have:

dA=(δA/δL)dL+(δA/δW)dW+(dW)dL

But since derivatives always deal with instantaneous slope and small changes, (dW)dL is small and ignored, resulting in:

dA=(δA/δL)dL+(δA/δW)dW

Length

dA=(δA/δL)dL+(δA/δW)dW

Effectively, we see that change in the dependent variable (A), comes from changes in the independent variables (W and L). In general, given the function z=f(x,y) we have:

Width

Area

dW

dL

In a joke factory,

QJokes=workers(funniness)

You employ 500 workers, each of which can create 100 funny jokes an hour.

How many more jokes could you create if you increase workers by 2 and their average funniness by 1 (perhaps by discovering any joke with an elephant in it is slightly more funny)?

The key advantage of the total derivative is it takes variable interaction into account.

The partial derivative (δz/δx) examines the effect of x on z if y doesn’t change. This is the DIRECT EFFECT.

However, if x affects y which then affects z, we might want to measure this INDIRECT EFFECT.

We can modify the total derivative to do this:

- Here we see that x’s total impact on z is broken up into two parts:
- x’s DIRECT impact on z (through the partial derivative)
- x’s INDIRECT impact on z (through y)
- Obviously, if x and y are unrelated, (δy/δx)=0, then the total derivative collapses to the partial derivative

Assume Happiness=Candy+3(Candy)Money+Money2

h=c+3cm+m2

Furthermore, Candy=3+Money/4 (c=3+m/4)

The total derivative of happiness with regards to money:

Total derivatives can also give us the relationship between elasticity and revenue that we found in Chapter 2.2.3:

Unconstrained optimization falls into two categories:

- Optimization using one variable (ie: changing wage to increase productivity, working conditions are constant)
- Optimization using two (or more) variables (ie: changing wage and working conditions to maximize productivity)

For a multivariable case where only one variable is controlled, optimization steps are easy:

Consider the function z=f(x)

1) FOC:

Determine where δz/δx=0 (necessary condition)

2) SOC:

δ2z/δx2<0 is necessary for a maximum

δ2z/δx2>0 is necessary for a minimum

3) Determine max/min point

Substitute the point in (2) back into the original equation.

Let productivity = -wage2+10wage(working conditions)2

P(w,c)=-w2+10wc2

If working conditions=2, find the wage that maximizes productivity

P(w,c)=-w2+40w

1) FOC:

δp/δw =-2w+40=0

w=20

2) SOC:

δ2p/δw2= -2 < 0, a maximum exists

P(w,c)=-w2+10wc2

w=20 (maximum confirmed)

3) Find Maximum

P(20,4)=-202+10(20)(2)2

P(20,4)=-400+800

P(20,4)=400

Productivity is maximized at 400 when wage is 20.

For a multivariable case where only two variable are controlled, optimization steps are more in-depth:

Consider the function z=f(x,y)

1) FOC:

Determine where δz/δx=0 (necessary condition)

And

Determine where δz/δy=0 (necessary condition)

For a multivariable case where only two variable are controlled, optimization steps harder:

Consider the function z=f(x,y)

2) SOC:

δ2z/δx2<0 and δ2z/δy2<0 are necessary for a maximum

δ2z/δx2>0 and δ2z/δy2>0 are necessary for a minimum

Plus, the cross derivatives can’t be too large compared to the own second partial derivatives:

If this third SOC requirement is not fulfilled, a SADDLE POINT occurs, where z is a maximum with regards to one variable but a minimum with regards to the other. (ie: wage maximizes productivity while working conditions minimizes it)

Vaguely, even though both variables work to increase z, their interaction with each other outweighs this maximizing effect

Let P(w,c)=-w2+wc-c2 +9c , maximize productivity

1) FOC:

δp/δw =-2w+c=0

2w=c

δp/δc=w-2c+9=0

w=2c-9

w=2(2w)-9

-3w=-9

w=3

2w=c

6=c

P(w,c)=-w2+wc-c2 +9c

δp/δw =-2w+c=0

δp/δc=w-2c+9=0

w=3, c=6 (possible max/min)

2) SOC:

δ2p/δw2= -2 < 0

δ2p/δc2= -2 < 0, possible max

Maximum confirmed

P(w,c)=-w2+wc-c2 +9c

w=3, c=6 (confirmed max)

3) Find productivity:

Productivity is maximized at 27 when wage=3 and working conditions=6.

Typically constrained optimization consists of maximizing or minimizing an objective function with regards to a constraint, or

Max/min z=f(x,y)

Subject to (s.t.): g(x,y)=k

Where k is a constant

Often economic agents are not free to make any decision they would like. They are CONSTRAINED by factors such as income, time, intelligence, etc.

When optimizing with constraints, we have two general methods:

- Internalizing the constraint
- Creating a Lagrangeian function

If the constraint can be substituted into the equation to be optimized, we are left with an unconstrained optimization problem:

Example:

Bob works a full week, but every Saturday he has seven hours left free, either to watch TV or read. He faces the constrained optimization problem:

Max. Utility=7TV-TV2+Read (U=7TV-TV2+R)

s.t. 7=TV+Read (7=TV+R)

Max. U=7TV-TV2+R

s.t. 7=TV+R

We can solve the constraint:

R=7-TV

And substitute into the objective function:

U=-TV2+7TV+(7-TV)

U=-TV2+6TV+7

Max. U=7TV-TV2+R

s.t. 7=TV+R

U=-TV2+6TV+7

We can then perform unconstrained optimization:

FOC:

δU/ δTV=-2TV+6=0

TV=3

R=7-TV

R=7-3

R=4

Max. U=7TV-TV2+R

s.t. 7=TV+R

U=-TV2+6TV+7, TV=3, R= 4

δU/ δTV=-2TV+6

SOC:

δ2U/ δTV2=-2<0, concave max.

Evaluate:

U=7TV-TV2+R

U=7(3)-32+4

U=21-9+4=16

Max. U=7TV-TV2+R

s.t. 7=TV+R

U=-TV2+6TV+7, TV=3, R= 4

δU/ δTV=-2TV+6

δ2U/δTV2=-2<0, concave max.

U=21-9+4=16

Utility is maximized at 16 when Bob watches 3 hours of TV and reads for 4 hours.

Substituting the constraint into the objective function may not be applicable for a variety of reasons:

- The substitution makes the objective function unduly complicated, or substitution is impossible
- You want to evaluate the impact of the constraint
- The constraint is an inequality
- Your exam paper asks you to do so
In this case, you must construct a Lagrangian function.

Given the optimization problem:

Max/min z=f(x,y)

s.t. g(x,y)=k (Where k is a constant)

The Lagranean (Lagrangian) function becomes:

L=z*=z(x,y)+λ(k-g(x,y))

Where λ is known as the Lagrange Multiplier.

We then continue with FOC’s and SOC’s.

L=z*=z(x,y)+λ(k-g(x,y))

FOC’s:

Note that the third FOC simply returns the constraint, g(x,y)=k

Typically, one will solve for λ in the first two conditions to find a relationship between x and y, then use this relationship with the third condition to solve for x and y.

L=z*=z(x,y)+λ(k-g(x,y))

After finding FOC’s, to confirm a maximum or minimum, the SOC is employed.

This SOC must be negative for a maximum and positive for a minimum

Note that for more terms, this function becomes exponentially complicated.

SOC’s:

Max. U=7TV-TV2+R

s.t. 7=TV+R

L=z*=z(x,y)+λ(k-g(x,y))

L=7TV-TV2+R+λ(7-TV-R)

FOC:

Since the second order condition is negative, the points found are a maximum.

Notice that we found the same answers as internalizing the constraint.

The Lagrange Multiplier, λ, provides a measure of how much of an impact relaxing the constraint would make, or how the objective function changes if k of g(x,y)=k is marginally increased.

The Lagrange multiplier answers how much the maximum or minimum changes when the constraint g(x,y)=k increases slightly to g(x,y)=k+δ

- This means that if Bob gets an extra hour, his maximum utility will increase by approximately 1.
- (Alternately, if Bob loses an hour of leisure, his maximum utility will decrease by approximately 1.)
- Check:
- If 8=TV+R, TV=3.5, R=4.5, U=16.75 (utility increases by approximately 1)

Often constraints are INEQUALITIES

ie: You can spend UP TO $200 on Christmas presents.

When the constraint is an inequality, one must ask, IS THE CONSTRAINT BINDING?

Example One: Constraint is binding

Maximize utility subject to a budget ≤ $100.

Assume that λ=2.5; an extra $1 in the Christmas budget will increase utility by 2.5

The constraint IS BINDING – your answer is correct.

Example Two: Constraint is not binding

Maximize utility subject to time ≤ 10 hours.

Assume that λ=-5; an extra hour would DECREASE utility by 5. OR

One fewer hour would INCREASE utility by 5. You don’t WANT to use all the available hours

The constraint IS NOT BINDING – therefore:

Redo the problem WITHOUT the constraint.

Inequality procedures can be summarized as follows (Table 4.1 in text):