Basic Mathematical Tools Outline Summation Operation

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Basic Mathematical Tools

Read Wooldridge, Appendix A

Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat

Outline

I. Summation Operation and Descriptive Statistics

II. Properties of Linear Functions III. Proportions and Percentages IV. Special Functions

V. Differential Calculus

I. Summation II. Linear III. Prop&Perc IV. SpecFunc V. Calculus

2 Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat

I. Summation Operation and Descriptive Statistics

• Summation operator () involves the sum of many numbers.

• Given a sequence of n numbers {x i ; i=1, …, n}

• The sum of these numbers x i = x 1 + x 2 + …. + x n

3

1 n

 i

Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat I. Summation Operation and Descriptive Statistics

Summation Operation

• Property s.1: For any constant c,

• The sum of n constants (c) equals the product of n and c

1 n

i

c nc



 

4

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Summation Operation

• Property s.2:

The sum of c times x i equals c time the sum of x i .

1 1

n n

i i

cx c x

 

  

5

Summation Operation

• Property s.3: If {(x i ,y i ): i=1, …,n} is a set of n pairs of numbers and a and b are constants, then

1 1 1

( )

n n n

i i i i

i i i

ax by a x b y

  

  

  

I. Summation II. Linear III. Prop&Perc IV. SpecFunc V. Calculus

6

Summation Operation

• Notes that the sum of ratios is not the ratio of the sums.

• Example: n = 2

1 1

1 n n i

i i

n

i i

x x

y y



 

 

7

Summation Operation

• Note that the sum of the squares is not the square of the sum.

• Example: n = 2

x 1 2 + x 2 2  (x 1 + x 2 ) 2

2 2

1 1

( )

n n

i i

x x

 

  

8

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Summation Operation and Descriptive Statistics

• Given a sequence of n numbers {x i ; i=1, …, n}, the average or mean can be written as

• Average is computed by adding them up and dividing by n

1

1 n

i i

x x

n 

 

9 Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat I. Summation Operation and Descriptive Statistics

Descriptive Statistics: Sample average

• When the x i are a sample of data on a particular variable, we call this the sample average or sample mean.

• Sample average is an example of a descriptive statistic.

• Sample average is a statistic that describes the central tendency of the set of n points.

Descriptive Statistics: Sample median

• Other measure of central tendency is sample median.

• Example: Given numbers, {‐4, 8, 2, 0, 21, ‐10, 18}

– Sample mean = 35/7 = 5 – Sample median = 2

• Ordered sequence {‐10, ‐4, 0, 2, 10, 18, 21}

Descriptive Statistics: Sample median

• Steps in finding sample median

Step 1: order the values of the x i from smallest to largest.

Step 2: if n is odd, the sample median is the middle number of the ordered observations.

Step 3: if n is even, the median is defined to be the average of the two middle values.

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Descriptive Statistics: Sample median

• If 21 is changed to 42

• Numbers {‐4, 8, 2, 0, 42, ‐10, 18}

– Sample mean = 56/7 = 8 – Sample median = 2

• Ordered sequence {‐10, ‐4, 0, 2, 10, 18, 42}

• Sample median:

Good point: it is less sensitive than sample average to changes in the extreme values in a list of numbers. Examples are median housing values or median income.

Summation Operation and Descriptive Statistics

• Deviations

• Deviations can be found by taking each observation and subtracting off the sample average

i i

d   x x

Summation Operation and Descriptive Statistics

• Properties d1: Given {x i ; i=1, …, n}, The sum of the deviations equal zero.

1 1

0

n n

i i

d x x

 

  

 

Deviations and Demean Sample

Example: n =5

x 1 = 6, x 2 = 1, x 3 = ‐2, x 4 = 0, x 5 = 5

Demean sample is {4, ‐1, ‐4, ‐2, 3} x  ?

1

0

n i i

x x



  

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Descriptive Statistics: Algebraic Fact

• Properties d2: Given {x i ; i=1, …, n},

the sum of squared deviations is the sum of squared x i minus n times the squared of sample mean.

• Show!

2 2 2

1 1

( ) ( )

n n

i i

x x x n x

 

  

 

I. Summation II. Linear III. Prop&Perc IV. SpecFunc V. Calculus

Descriptive Statistics: Algebraic Fact

• Properties d3: Given {(x i ,y i ): i=1, …,n}, It can be shown that

1 1

( )( ) ( )

n n

i i i i

i i

x x y y x y y

 

   

 

1 1

( ) ( )

n n

i i i i

i i

x x y x y n x y

 

      

Summation Deviation

s.1 d.1

s.2 d.2 2

2

s.3 d.3

Summary: Summation and Deviation

19

Problem A.1

A.1 The following table contains monthly housing expenditures for 10 families.

(i) Find the average monthly housing expenditure. [ans.]

(ii) Find the median monthly housing expenditure. [ans.]

I. Summation II. Linear III. Prop&Perc IV. SpecFunc V. Calculus

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Problem A.1 continue

• (iii) If monthly housing expenditures were measured in

hundreds of dollars , rather than in dollars, what would be the average and median expenditures?

[ans.]

• (iv) Suppose that family number 8 increases its monthly housing expenditure to $900 dollars , but the

expenditures of all other families remain the same.

Compute the average and median housing expenditures. [ans.]

Solution A.1 (i)

(i) $566.

1

1 n

i i

x x

n 

 

Family

Housing Expenditures

1 300

2 440

3 350

4 1100

5 640

6 480

7 450

8 700

9 670

10 530

Sum 5,660

Mean 566

Solution A.1 (ii)

(ii) 505

Steps in finding sample mean

• Step 1: order the values of the x i from smallest to largest.

{300, 350, 440, 450, 480, 530 , 640, 670, 700, 1100,}

• Step 3: if n is even, the median is defined to be the average of the two middle values.

The two middle numbers are 480 and 530; when these are averaged, we obtain 505, or $505.

Solution A.1 (iii)

(iii)

• $566 and $505 (in dollars), respectively

• 5.66 and 5.05 (in hundreds of dollars), respectively.

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Solution A.1 (iv)

(iv)

• The average increases to

$586 from $566.

• while the median is unchanged ($505).

{300, 350, 440, 450,

480, 530 , 640, 900, 670, 1100,}

Family

Housing Expenditures

1 300 300

2 440 440

3 350 350

4 1100 1100

5 640 640

6 480 480

7 450 450

8 700 900

9 670 670

10 530 530

Sum 5,660 5,860

Mean 566 586

II. Properties of Linear Functions

A linear function can be written as y =  0 +  1 x

• y and x are variables;

•  0 and  1 are parameters;

– 

₀

is called the intercept;

– 

₁

is called the slope.

• We say that y is a linear function of x.

26 Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat II. Properties of Linear Functions

Linear Functions

• y =  0 +  1 x

y =  1 x

 denotes “change”.

• The change in y is always  1 times the change in x, x.

• In other words, the marginal effect of x on y is constant and equals to  1 .

Example A.2 Linear Housing Expenditure Functions

• Relationship between monthly housing (dollar) expenditure and monthly income (dollar)

– housing = 

₀

+ 

₁

income – housing = 164 + 0.27income

• Interpret:  1 = 0.27 or slope

– When family income increases by 1 dollar, housing expenditure will go up by 0.27 dollar or 27 cents

• What if family income increases by $300??

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Example A.2 Linear Housing Expenditure Functions

– housing = 

₀

+ 

₁

income – housing = 164 + 0.27income

• Interpret:  0 = 164 or intercept

– When income=0, housing expenditures equal $164.

– A family with no income spends $164 on housing.

Example A.2 Linear Housing Expenditure Functions

• housing = 164 + 0.27income

• MPC and APC

– 

₁

is the marginal propensity to consume (MPC).



₁

= 0.27

– The average propensity to consume can be written as

• Note that (1) APC is not constant (2) APC>MPC

(3) APC gets closer to MPC as income increases.

sin 164

hou g 0.27

income  income 

Linear functions

• A linear function can have more than two variables.

• y =  0 +  1 x 1 +  2 x 2

–  0 is called the intercept (the value of y when x 1 =0 and x 2 =0)

–  1 and  2 are slopes.

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Linear functions

• y =  0 +  1 x 1 +  2 x 2

• The change in y, for given changes in x 1 and x 2 is

y =  1 x 1 +  2 x 2

• If x 2 does not change (x 2 =0), then

y =  1 x 1 if x 2 =0.

• If x 1 does not change (x 1 =0), then

y =  2 x 2 if x 1 =0.

Linear functions

•  1 is the slope of the relationship in the direction of x 1

•  1 is how y changes with x 1 , holding x 2 fixed. We called the

partial effect of x 1 on y .

• The notion of ceteris paribus .

Note that partial effect involves holding some factors fixed.

1 2

1

.... 0

y if x

   x  



Example A.2: Demand for Compact Discs

• y =  0 +  1 x 1 +  2 x 2

• quantity = 120 – 9.8price + 0.03income price: dollars per disc income: measured in dollars

• Interpretation

 – 9.8 is the partial effect of price on quantity. Holding income fixed, when the price of compact discs increases by one dollar, the quantity demanded falls by 9.8.

 Interpret 

₂

Example A.2: Demand for Compact Discs

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Problem A.2

• A.2 Suppose the following equation describes the relationship between the average number of classes missed during a semester (missed) and the distance from school (distance, measured in miles):

missed = 3 + 0.2distance.

(i) Sketch this line, being sure to label the axes. How do you interpret the intercept in this equation? [ans.]

Problem A.2 continue

• (ii) What is the average number of classes missed for someone who lives five miles away? [ans.]

• (iii) What is the difference in the average number of classes missed for someone who lives 10 miles away and someone who lives 20 miles away? [ans.]

Solution A.2 (i)

(i)

missed = 3 + 0.2distance

• This is just a standard linear equation with intercept equal to 3 and slope equal to .2.

• The intercept is the number of missed classes for a student who lives on campus.

39

Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat II. Properties of Linear Functions

Solution A.2 (ii)

(ii)

distance = 5

missed = 3 + 0.2distance

• 3 + .2(5) = 4 classes.

40

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Solution A.2 (iii)

(iii)

missed = 3 + 0.2distance Difference

distance = 20: missed = 3 + 0.2(20) distance = 10: missed = 3 + 0.2(10)

• 10(.2) = 2 classes.

41

Problem A.3

A.3 In Example A.2, quantity of compact disks was related to price and income by

quantity = 120 – 9.8price +.03 income.

What is the demand for CDs if price = 15 and income = 200? What does this suggest about using linear functions to describe demand curves? [ans.]

Solution A.3

A.3 quantity = 120 – 9.8price + 0.3income

• If price = 15 and income = 200,

quantity = 120 – 9.8(15) + .03(200)

= –21,

• This is nonsense.

• This shows that linear demand functions generally cannot describe demand over a wide range of prices and income.

I. Summation II. Linear III. Prop&Perc IV. SpecFunc V. Calculus

III. Proportions and Percentages

• Proportions and percentages play an important role in applied economics.

• Examples in the form of percentages

– inflation rates,

– unemployment rates, and – entrance acceptance rates.

44 Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat III. Proportions and Percentages

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Proportions and Percentages

• A proportion is the decimal form of percent.

A percentage is simply obtained by multiplying a proportion by 100.

• When using percentages, we often need to convert them to decimal form.

For example: find interest income

– if the annual return on time deposit is 7.6% and we save 30,000 baht at the beginning of the year,

– 7.6% = 0.76 (percentage = proportion) – our interest income is 30,000*0.076 = 2280

Proportions and Percentages

Example: Admission Rate

• Proportion :

– the proportion of applicants who are admitted to MABE is 0.50.

• Percentage:

– 50 percent of the applications are admitted to MABE program.

• If there are 200 applicants, how many students are admitted?

Proportions and Percentages

• A report by popular media can be incorrect.

Which one is correct?

• In Thailand, the percentage of high school dropout is .20.

• Her percentage in econometric exam is .75.

Proportions

• Find changes in various quantities.

• Let x be annual income

x

₀

is the initial value x

₁

is the subsequent value.

The proportionate change in x in moving from x 0 to x 1 is

This is sometimes called the relative change.

• Note that to get the proportionate change we simply divide the change in x, or x, by its initial value.

1 0

0 0

x x x

x x

  

I. Summation II. Linear III. Prop&Perc IV. SpecFunc V. Calculus

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Proportions

• For example, if income goes from $30,000 per year to $36,000 per year, what is the proportionate change?

x

₀

= 30,000 (initial value: income in 1994) x

₁

= 36,000 (subsequent value: income in 1995 proportionate change = 6,000/30,000 = 0.2

1 0

0 0

x x x

x x

  

Percentages

• The percentage change in x in going from x 0 to x 1 is

• It is simply 100 times the proportionate change.

• %x is read as the percentage change in x.

0

% 100 * x

x x

  

Percentages and Proportions

• Suppose the real GDP in 2010 and 2011 are 300 and 315 trillion baht, respectively. What are the proportionate change and the percentage change in real GDP in 2011?

– proportionate change = 15/300 = 0.05 – percentage change = 100*(0.05) = 5%

• True or False: The percentage change in GDP is 0.05.

0

% 100 * x

x x

  

1 0

0 0

x x x

x x

  

Percentage Point Change vs. Percentage Change

• Percentage Point Change vs. P ercentage Change

• Variable in interest is itself a percentage!

• How to find percentage change?

Let x be unemployment rate. Suppose x

₂₀₀₂

is 4% and

x

₂₀₀₃

is 5%,

what is the percentage point change? What is the percentage change?

1) percentage point change = x = 5%‐4% = 1%

2) percentage change = 100*x/x

₀

= 1%/4% = 25%

0

% 100 * x

x x

  

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Percentages

• Unemployment rate has increased by one percentage point.

• But unemployment rate has increased by 25 percent!

• Summary

– The percentage point change is the change in the percentages.

– The percentage change is the change relative to the initial value (x

₂₀₀₂

)

1 0

0 0

100* x x 100* x

x x

  

Proportions and Percentages

• Example: During General Chavalit administration (1997), the value added tax was increased from 7% to 10%.

• Who is correct?

– Supporters

This is simply a three percentage point increase! or an increase of three satang on the baht.

– Opponents

This is a 43% increase in value added tax!

0

% 100 * x

x x

  

Problem A.4

• A.4 Suppose the unemployment rate in the United States goes from 6.4% in one year to 5.6% in the next.

(i) What is the percentage point decrease in the unemployment rate? [ans.]

(ii) By what percent has the unemployment rate fallen? [ans.]

55

Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat III. Proportions and Percentages

Solution A.4 (i)

(i)

• The percentage point change is 5.6 – 6.4 = –.8,

• or an eight‐tenths of a percentage point decrease in the unemployment rate.

56

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Solution A.4 (ii)

(ii)

• The percentage change in the unemployment rate is

100[(5.6 – 6.4)/6.4] = –12.5%.

57

Problem A.5

• A.5 Suppose that the return from holding a particular firm’s stock goes from 15% in one year to 18% in the following year.

– The majority shareholder claims that “the stock return only increased by 3%,”

– while the chief executive officer claims that “the return on the firm’s stock has increased by 20%.”

– Reconcile their disagreement. [ans.]

I. Summation II. Linear III. Prop&Perc IV. SpecFunc V. Calculus

Solution A.5

A.5

• The CEO is referring to the change relative to the initial return of 15%.

• The majority shareholder is referring to the percentage point increase in the stock return,

– To be precise, the shareholder should specifically refer to a 3 percentage point increase.

I. Summation II. Linear III. Prop&Perc IV. SpecFunc V. Calculus

59

IV. Some Special Functions and Their Properties

• Linear function

– y = 

₀

+ 

₁

x

₁

– Interpret:

• One unit change in x results in a same change in y, regardless of the starting value of x.

• Marginal effect of x on y is constant

• Example: utility function – the notion of diminishing marginal returns is not consistent with a linear relationship.

A. quadratic B. Logarithm C. Exponential

60 Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat IV. Some Special Functions and Their Properties

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Some Special Functions and Their Properties

• Many economic phenomena requires the use of nonlinear functions

• Nonlinear function is characterized by the fact that the change in y for a given change in x depends on the starting value of x.

• Quadratic functions

• Natural Logarithm

• Exponential Function

Quadratic functions

• One way to capture diminishing returns is to include a quadratic term to a linear function.

y =  0 +  1 x +  2 x 2

• Given  1 >0 and  2 <0

The relationship between y and x has the parabolic shape. Let

 0 =6  1 =8 and  2 = ‐2 y = 6 + 8x ‐ 2x 2

Graph: Quadratic functions

Quadratic functions

• When  1 >0 and  2 <0, the maximum of the function occurs at

• This is called the turning point.

If y = 6 + 8x ‐ 2x 2 (so  1 =8 and  2 = ‐2), the largest value of y occur at

x* = 8/4 = 2

y* = 6+8(2)‐2(2 2 ) = 14

1 2

* ( 2 )

x 

 



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Quadratic functions

•  2 <0. This implies a diminishing marginal effect of x on y – Suppose we start at a low value of x and then increase x by

some amount, say c

• This has a larger effect on y than if we start at a higher value of x and increase x by the same amount. (See graph)

– Once x>x*, an increase in x actually decreases y.

Quadratic functions

• A diminishing marginal effect of x on y is the same as saying that – the slope of the function decreases as x increases.

• Calculus: the derivative of the quadratic function:

y = 

₀

+ 

₁

x + 

₂

x

²

• For “small” changes in x, the approximate slope of the quadratic function is

1

2

slope y x

x  

   



Example A.4: A quadratic wage function

• wage hourly wage

• exper years in the workforce wage = 5.25 + .48exper – 0.008exper

²

Interpretation:

1) Since 

₂

<0, this implies the diminishing marginal effect of exper on wage.

2) exper* = .48/[2(.008)] = 30. This is turning point.

– exper has a positive effect on wage up to the turning point.

3) The marginal effect of exper on wage depends on years of experience.

1

^st

year (x

₀

=0, x

₁

=1, x=1) wage = .48 – 2*0.008(0) = .48 5

^th

year (x

₄

=4, x

₅

=5 x=1) wage = .48 – 2*0.008(4) = .416 4) At 30 years, an additional year of experience would lower the wage.

32

^th

year (x

31

=31, x

32

=32 x=1) wage = .48 – 2*0.008(31) = ‐.016 A. quadratic B. Logarithm C. Exponential

Quadratic functions

• y =  0 +  1 x +  2 x 2

• When  1 <0 and  2 >0, the graph of the quadratic function has U‐shape.

1) there is an increasing marginal return.

2) the minimum of the function is at the point

1 2

* ( 2 )

x 

 



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Natural Logarithm

• The natural logarithm, an important nonlinear function, plays an important role in econometric analysis.

• We denotes the natural logarithm as the log function y = log(x)

• Other common symbols include log

_e

(x) and ln(x).

– Most calculators use ln(x).

– Different symbols are useful when we use logarithm with different bases.

A. quadratic B. Logarithm C. Exponential

Natural Logarithm

• Properties

1) The log function is defined only for positive values of x (x>0).

See graph of a log function for 0<x<1 log(x) < 0 for x=1 log(x) = 0 for x>1 log(x) >0

2) When y= log(x), the effect of x on y never becomes negative.

y = log(x)

• The relationship between x and y displays diminishing returns.

• The slope of the function gets closer and closer to zero as x gets large

1 y x x

 



Graph: log function

Natural Logarithm

• What is the difference in slopes between the quadratic and the log function?

Ans. The marginal effect of x on y of the log function never becomes negative.

• Some useful algebraic facts:

l.1) log(x

₁

x

₂

) = log(x

₁

) + log(x

₂

) l.2) log(x

₁

/x

₂

) = log(x

₁

) – log(x

₂

) l.3) log(x

^c

) = clog(x) for any constant c.

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Natural Logarithm

• Let x

₀

and x

₁

be positive values.

• The difference in logs can be used to approximate proportionate changes.

for small changes in x.

• Note that log(x) = log(x

₁

)‐log(x

₀

)

• Approximate percent change is

100 log(x) ≈ %x for small changes in x.

1 0

0 0

log( ) log( ) x x x

x x

 

  

Natural Logarithm

• How good is the approximation?

• Example, for a small change: x 0 =40 and x 1 =41

– Exact percentage change

= 100*(41‐40)/40 = 2.5%

– Approximate percentage change

= 100*[log(41)‐log(40)] = 2.47%

• Example, for a large change: x 0 =40 and x 1 =60

– Exact percentage change

= 100*(60‐40)/40 = 50%

– Approximate percentage change

= 100*[log(60)‐log(40)] = 40.55%

Natural Logarithm

• Elasticity of y with respect to x can be written as

– It is the percentage change in y when x increases by 1%

• If y is a linear function, y =  0 +  1 x, then the elasticity is

– It depends on the value of x.

1 1

0 1

y x x x

x y  y  x

 

  

 

%

y x y

x y x

  

 

Natural Logarithm

• Constant Elasticity Model log(y) =  0 +  1 log(x)

• The slope or elasticity is approximately equal to

• It is the elasticity of y with respect to x

1

log( ) log( ) y

   x



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Example A.5: Constant Elasticity Demand Function

• Let q quantity demanded (unit) p price (dollar)

log(q) = 4.7 – 1.25log(p)

– The price elasticity of demand = ‐1.25 – Interpret

• A 1% percent increase in price leads to a 1.25% fall in the quantity demanded.

I. Summation II. Linear III. Prop&Perc IV. SpecFunc V. Calculus

Natural Logarithm

• Log‐level model

log(y) = 

₀

+ 

₁

x

log(y) = 

₁

x 100log(y) = 100

₁

x

%y = (100

₁

)x

• The slope or the semi‐elasticity of y with respect to x is

• The semi‐elasticity is the percentage change in y when x increases by one unit. It is equal to 100

₁

.

1

% y 100

x 

  



Example A.6: Logarithmic Wage Equation

• Hourly wage and years of education are related by log(wage) = 2.78 + .094 educ

%(wage) = 100(.094)educ

%(wage) = 9.4educ

• Interpret:

– One more year education increases hourly wage by about 9.4%

Natural Logarithm

• Level‐log function y =  0 +  1 log(x)

y = 

₁

log(x)

100y =  1 100 log(x)

• Using approximation, 100y =  1 %x

y = ( 1 /100)(%x)

• Interpret

– 

₁

/100 is the unit change in y when x increases by 1%

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Example A.7: Labor Supply Function

• Let hours hours worked per week wage hourly wage

hours = 33 + 45.1log(wage)

hours = (45.1/100)(%wage)

hours = 0.451(%wage)

• Interpret

1% increase in wage increases the weekly hours worked by about 0.45, or slightly less than one‐half hour.

Summary A. quadratic B. Logarithm C. Exponential

Exponential Function

• The exponential function is related to the log function.

• For example, given log(y) that is a linear function of x,

– How to find y itself as a function of x?

• We write the exponential function as y = exp(x)

– Other notation can be written as

y = e x

• Facts

– exp(0) =1;

– exp(1) = 2.7183

– exp(x) is defined for any value of x.

– exp(x) is always greater than 0

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Graph: Exponential Function

• The exponential function is the inverse of the log function in the following sense:

log[exp(x)] = x, for all x exp[log(x)] = x, for x>0



In other words, the log

“undoes” the exponential, and vice versa.



The exponential function is sometimes called the anti‐log function.

Exponential Function

• Given a function,

log (y)= 

₀

+ 

₁

x

• It is equivalent to the function y = exp(

₀

+ 

₁

x)

• If 

₁

>0, then x has an increasing marginal effect on y.

• Some algebraic facts:

e.1) exp(x

₁

+ x

₂

) = exp(x

₁

)∙exp(x

₂

) or e

^x1+ x2

= e

^x1

∙e

^x2

e.2) exp(x

₁

‐ x

₂

) = exp(x

₁

)/exp(x

₂

) or e

^x1‐^x2

= e

^x1

/e

^x2

e.3) exp[clog(x)] = x

^c

or e

^clog(x)

= x

^c

Logarithm Exponential

l.1 log(x 1 x 2 ) = log(x 1 ) + log(x 2 ) e.1 exp(x 1 + x 2 ) = exp(x 1 )∙exp(x 2 ) l.2 log(x 1 /x 2 ) = log(x 1 ) – log(x 2 ) e.2 exp(x 1 ‐ x 2 ) = exp(x 1 )/exp(x 2 ) l.3 log(x c ) = clog(x) e.3 exp[clog(x)] = x c

Summary: Logarithm and Exponential

88

Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat IV. Some Special Functions and Their Properties

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Problem A.6

A.6 Suppose that Person A earns $35,000 per year and Person B earns $42,000.

(i) Find the exact percentage by which Person B’s salary exceeds Person A’s. [ans.]

(ii) Now use the difference in natural logs to find the approximate percentage difference. [ans.]

I. Summation II. Linear III. Prop&Perc IV. SpecFunc V. Calculus

Solution A.6 (i)

(i)

100[42,000 – 35,000)/35,000]

= 20%.

90

Solution A.6 (ii)

(ii)

• The approximate proportionate change is log(42,000) – log(35,000) =.182,

so the approximate percentage change is %18.2.

• [Note: log() denotes the natural log.]

91

Problem A.7

• A.7 Suppose the following model describes the relationship between annual salary (salary) and the number of previous years of labor market experience (exper):

log(salary) = 10.6 + .027exper.

(i) What is salary when exper = 0? when exper = 5? (Hint: You will need to exponentiate.) [ans.]

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Problem A.7 continue …

(ii) Use equation (A.28) to approximate the percentage increase in salary when exper increases by five years. [ans.]

(iii) Use the results of part (i) to compute the exact percentage difference in salary when exper = 5 and exper = 0. Comment on how this compares with the approximation in part (ii).

[ans.]

Solution A.7 (i)

(i) log(salary) = 10.6 + .027exper

• When exper = 0,

log(salary) = 10.6; therefore, salary = exp(10.6) = $40,134.84.

• When exper = 5,

salary = exp[10.6 +.027(5)]

= $45,935.80.

I. Summation II. Linear III. Prop&Perc IV. SpecFunc V. Calculus

94

Solution A.7 (ii)

(ii)

100log(y)= 100 1 x (A.28)

• The approximate proportionate increase is .027(5) = .135,

so the approximate percentage change is 13.5%.

95

Solution A.7 (iii)

(iii) log(salary) = 10.6 + .027exper exper=0 log(salary) = 10.6 + .027exper

salary = exp(10.6) = $40,134.84.

exper=5 log(salary) = 10.6 + .027exper

salary = exp[10.6 +.027(5)] = $45,935.80.

• Exact percentage increase

100[(45,935.80‐40,134.84)/40,134.84)

= 14.5%,

so the exact percentage increase is about one percentage point higher.

96

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Problem A.8

A.8 Let

grthemp denote the proportionate growth in

employment, at the county level, from 1990 to 1995, and let

salestax denote the county sales tax rate, stated as a proportion.

Interpret the intercept and slope in the equation grthemp = .043 –.78salestax. [ans.]

Solution A.8

A.8 grthemp = –.78(salestax).

• Since both variables are in proportion form, we can multiply the equation through by 100 to turn each variable into percentage form.

• Slope = –.78.

– So, a one percentage point increase in the sales tax rate (say, from 4%

to 5%) reduces employment growth by –.78 percentage points.

• Intercept = .043

– When salestax = 0, the proportionate growth in employment is .043.

98

V. Differential Calculus

 Let y = f(x) for some function f



For small changes in x

 df/dx is the derivative of the function f, evaluated at the initial point x 0 .

99 Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat V. Differential Calculus

0

y 1 x

  x 

• Example y = log(x)

• For a small change, evaluated at the initial point x 0 ,

,

which is the approximation of the proportionate change in x.

0

log( ) x

x x

   1

dy dx  x

Differential Calculus

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• Let y = f(x)

DC.1 Let f(x) = c = 0 or = 0

– The derivative of a constant c is zero.

DC.2 Let f(x) = log(x) = or =

– The derivative of the log function of x is one over x.

DC.3 Let f(x) = exp(x) = exp(x) or = exp(x)

– The derivative of the exponential function of x is the exponential function of x

DC.4 Let f(x) = x

^c

= cx

^c‐1

or = cx

^c‐1

Differential Calculus

• Other Important Rules DC.5 Let y =cf(x) = c

– The derivative of a constant times any function is that same constant times the derivative of the function,

– Example: y =cf(x) = x

^c

= c = cx

^c‐1

DC.6 Let y = f(x)+g(x) = +

– The derivative of the sum of two functions is the sum of the derivatives DC.7 Let y = z(f(x)) =

– Chain Rule

Chain Rule

DC.7 Let y = f(z(x)) = = Example:

y = exp( 0 +  1 x) z =  0 +  1 x

= = x

= exp

= exp x

Find the derivatives of the following functions:

– y =  0 +  1 x+  2 x 2 – y =  0 +  1 /x – y =  0 +  1 – y =  0 +  1 log(x) – y = exp( 0 +  1 x)

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DC.1 Let f(x) = c = 0 DC.5 Let y =cf(x) = c

DC.2 Let f(x) = log(x) = DC.6 Let y = f(x)+g(x) = +

DC.3 Let f(x) = exp(x) = exp(x) DC.7 Let y = z(f(x)) =

DC.4 Let f(x) = x

^c

= cx

^c‐1

Example DC.7: y = exp(

₀

+ 

₁

x)

= = x

= exp = exp x

Summary: Calculus

105

Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat V. Differential Calculus

Differential Calculus

• Notion of a partial derivative

• Suppose that y = f(x

₁

, x

₂

)

• Two partial derivatives

1) The partial derivative of y with respect to x

₁

= (where x

₂

is treated as a constant) 2) The partial derivative of y with respect to x

₂

=

(where x

₁

is treated as a constant)

²

y x



106 1



 y x

Differential Calculus

• We can approximate the change in y as

• Example

y = 

₀

+ 

₁

x

₁

+ 

₂

x

₂

1) What is the partial derivative of y with respect to x

₁

? 2) What is the partial derivative of y with respect to x

₂

?

Differential Calculus

• y = f(x 1 ,…, x n )

Example:

log( ) =  0 +  1 educ +  2 exper +  3 tenure +  4 age

log( ) = .514 + .078educ +.002exper +.0088tenure +.0033age 1) What is the partial derivative of y with respect to x 1 ? 2) What is the partial derivative of y with respect to x 4 ?

108

x

₂

,…, x

_n

fixed

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• Example

1) What is the partial derivative of y with respect to x 1 ? 2) What is the partial derivative of y with respect to x 2 ?

Example A.8 wage function with interaction

wage = 3.1 + .41educ + .19exper ‐ .004exper

²

+ .007educ  exper

• Find the partial effect of exper on wage:

wage/exper = .19 ‐ .008exper +.007educ

• At the initial values (exper

₀

=5, educ

₀

=12), the approximate change in wage is

wage/exper = 23.4 cents per hour

• Given exper

₀

=5, educ

₀

=12; and exper

₁

=6, educ

₁

=12, the exact change in wage is

wage = 23 cents per hour

• Minimizing and maximizing functions. Let f(x

₁

, x

₂

, …, x

_k

) is the differentiable function of k variables.

• A necessary condition for x

₁^*

, x

₂^*

, …, x

_k^*

to optimize f over all possible values of x

_j

is

• Notes

1) All of the partial derivatives of f must be zero when they are evaluated at the x

_h^*

.

2) These are called the first order conditions.

3) We hope to solve above equations for the x

_h^*

.

I. Summation II. Linear III. Prop&Perc IV. SpecFunc V. Calculus

Problem A.9

A.9 Suppose the yield of a certain crop (in bushels per acre) is related to fertilizer amount (in pounds per acre) as

yield = 120 +.19

(i) Graph this relationship by plugging in several values for fertilizer. [ans.]

(ii) Describe how the shape of this relationship compares with a linear function between yield and fertilizer. [ans.]

fertilizer

Basic Mathematical Tools

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113

(i)

• The relationship between yield and fertilizer is graphed below.

fertilizer

0 50 100

120 121

yield ₁₂₂

Solution A.9 (i)

Solution A.9 (ii)

(ii)

• Compared with a linear function, the function yield = .120 + .19

has a diminishing effect.

• The slope approaches zero as fertilizer gets large.

– The initial pound of fertilizer has the largest effect, and each additional pound has an effect smaller than the previous pound.

114

fertilizer