Outline
I. Summation Operation and Descriptive Statistics
II. Properties of Linear Functions
III. Proportions and Percentages
IV. Special Functions
V. Differential Calculus
Basic Mathematical Tools
Read Wooldridge, Appendix A
I. Summation
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
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 Operation
• Summation operator () involves the sum of many numbers.
• Property s.1: For any constant c,
n
 c  nc
• Given a sequence of n numbers
{xi; i=1, …, n}
i 1
• The sum of these numbers
• The sum of n constants (c) equals the product of n and c
n
 x = x
i
1
+ x2 + …. + xn
i1
I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
3
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
4
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Summation Operation
Summation Operation
• Property s.2:
• Property s.3: If {(xi,yi): i=1, …,n} is a set of n pairs of numbers and a and b are constants, then
n
n
 cxi  c xi
i 1
n
n
n
 (ax  by )  a x  b y
i 1
i
i 1
i
i 1
i
i 1
i
The sum of c times xi equals c time the sum of xi.
I. Summation
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation Operation and Descriptive Statistics
I. Summation
II. Linear
Summation Operation
xi
i 1
i

n
x
x
i 1
n
i
i 1
II. Linear
i
n
2
 ( xi ) 2
i 1
i
• Example: n = 2
x12 + x22  (x1 + x2 ) 2
• Example: n = 2
I. Summation
6
y
i 1
I. Summation Operation and Descriptive Statistics
V. Calculus
• Note that the sum of the squares is not the square of the sum.
n
y
IV. SpecFunc
Summation Operation
• Notes that the sum of ratios is not the ratio of the sums.
n
III. Prop&Perc
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
7
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Descriptive Statistics: Sample average
Summation Operation and Descriptive Statistics
• When the xi are a sample of data on a particular variable, we call this the sample average or sample mean.
• Given a sequence of n numbers {xi; i=1, …, n}, the average or mean can be written as
1
x 
n
• Sample average is an example of a descriptive statistic.
n

i 1
xi
• Sample average is a statistic that describes the central tendency of the set of n points.
• Average is computed by adding them up and dividing by n
I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Descriptive Statistics: Sample median
I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
Descriptive Statistics: Sample median
• Other measure of central tendency is sample median.
• Steps in finding sample median
• Example: Given numbers, {‐4, 8, 2, 0, 21, ‐10, 18}
Step 2: if n is odd, the sample median is the middle number of the ordered observations.
Step 1: order the values of the xi from smallest to largest.
– Sample mean = 35/7 = 5
– Sample median = 2
Step 3: if n is even, the median is defined to be the average of the two middle values.
• Ordered sequence {‐10, ‐4, 0, 2, 10, 18, 21}
I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
10
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
V. Calculus
11
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Summation Operation and Descriptive Statistics
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}
•
Deviations
•
Deviations can be found by taking each observation and subtracting off the sample average
di  xi  x
• 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.
I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
13
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Example: n =5
x1 = 6, x2 = 1, x3 = ‐2, x4 = 0, x5 = 5
x ?
n
 di   xi  x  0
i 1
II. Linear
Deviations and Demean Sample
• Properties d1: Given {xi; i=1, …, n},
The sum of the deviations equal zero.
n
I. Summation
I. Summation Operation and Descriptive Statistics
Demean sample is {4, ‐1, ‐4, ‐2, 3}
i 1
n
x x 0
i 1
I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
15
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
i
II. Linear
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Descriptive Statistics: Algebraic Fact
Descriptive Statistics: Algebraic Fact
• Properties d2: Given {xi; i=1, …, n},
the sum of squared deviations is the sum of squared xi minus n times the squared of sample mean.
• Properties d3: Given {(xi,yi): i=1, …,n},
It can be shown that
n
n
 (x  x )   x
2
i 1
i
i 1
i
2
 n( x )
n
n
i 1
i 1
 ( xi  x )( yi  y )   xi ( yi  y )
2
n
n
i 1
i 1
  ( xi  x ) yi   xi yi  n( x  y )
• Show!
I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
17
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
s.1 s.2 d.1 s.3
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
A.1 The following table contains monthly housing expenditures for 10 families.
Deviation
d.2 II. Linear
Problem A.1
Summary: Summation and Deviation
Summation
I. Summation
I. Summation Operation and Descriptive Statistics
2
2
(i) Find the average monthly housing expenditure. [ans.]
d.3 (ii) Find the median monthly housing expenditure. [ans.]
I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
19
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
20
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Problem A.1 continue
Solution A.1 (i)
• (iii) If monthly housing expenditures were measured in hundreds of dollars, rather than in dollars, what would be the average and median expenditures? [ans.]
(i) $566.
1 n
x   xi
n i 1
• (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.]
I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
Family
21
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Solution A.1 (ii)
I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
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
III. Prop&Perc
IV. SpecFunc
V. Calculus
22
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Solution A.1 (iii)
(ii) 505
(iii)
• $566 and $505 (in dollars), respectively
• 5.66 and 5.05 (in hundreds of dollars), respectively.
Steps in finding sample mean
• Step 1: order the values of the xi 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.
I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
23
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
III. Prop&Perc
IV. SpecFunc
V. Calculus
24
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
II. Properties of Linear Functions
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,} I. Summation
II. Linear
I. Summation Operation and Descriptive Statistics
Family
Housing
Expenditures
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
III. Prop&Perc
IV. SpecFunc
V. Calculus
A linear function can be written as
y = 0 + 1x
• 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.
25
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
IV. SpecFunc
V. Calculus
26
– housing = 0 + 1income
– housing = 164 + 0.27income
• Interpret: 1 = 0.27 or slope • The change in y is always 1 times the change in x, x.
– When family income increases by 1 dollar, housing expenditure will go up by 0.27 dollar or 27 cents
• In other words, the marginal effect of x on y is constant and equals to 1.
II. Linear
III. Prop&Perc
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
• Relationship between monthly housing (dollar) expenditure and monthly income (dollar)
• y = 0 + 1x
y = 1x
 denotes “change”.
I. Summation
II. Linear
Example A.2 Linear Housing Expenditure Functions
Linear Functions
II. Properties of Linear Functions
I. Summation
II. Properties of Linear Functions
III. Prop&Perc
IV. SpecFunc
V. Calculus
• What if family income increases by $300?? 27
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Properties of Linear Functions
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
28
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Example A.2 Linear Housing Expenditure Functions
Example A.2 Linear Housing Expenditure Functions
– housing = 0 + 1income
– 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.
I. Summation
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
29
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
II. Properties of Linear Functions
I. Summation
II. Linear
Example A.2 Linear Housing Expenditure Functions III. Prop&Perc
IV. SpecFunc
V. Calculus
30
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
II. Properties of Linear Functions
Linear functions
• housing = 164 + 0.27income
• MPC and APC
• A linear function can have more than two variables.
• y = 0 + 1x1 + 2x2
– 1 is the marginal propensity to consume (MPC).
1 = 0.27
– The average propensity to consume can be written as
– 0 is called the intercept (the value of y when x1=0 and x2=0)
– 1 and 2 are slopes.
hou sin g
164

 0.27
income
income
• Note that (1) APC is not constant (2) APC>MPC (3) APC gets closer to MPC as income increases.
I. Summation
II. Properties of Linear Functions
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
31
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Properties of Linear Functions
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
32
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Linear functions
Linear functions
• 1 is the slope of the relationship in the direction of x1
• y = 0 + 1x1 + 2x2
1 
• The change in y, for given changes in x1 and x2 is
y = 1 x1 + 2 x2
• 1 is how y changes with x1, holding x2 fixed. We called the partial effect of x1 on y.
• If x2 does not change (x2 =0), then
y = 1 x1 if x2 =0.
• The notion of ceteris paribus.
Note that partial effect involves holding some factors fixed.
• If x1 does not change (x1 =0), then
y = 2 x2 if x1 =0.
I. Summation
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
33
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
II. Properties of Linear Functions
y
.... if x2  0
x1
I. Summation
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
34
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
II. Properties of Linear Functions
Example A.2: Demand for Compact Discs
Example A.2: Demand for Compact Discs
• y = 0 + 1x1 + 2x2
• 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
I. Summation
II. Properties of Linear Functions
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
35
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Properties of Linear Functions
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
36
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Problem A.2
Problem A.2 continue
• 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):
• (ii) What is the average number of classes missed for someone who lives five miles away? [ans.]
missed = 3 + 0.2distance.
• (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.]
(i) Sketch this line, being sure to label the axes. How do you interpret the intercept in this equation? [ans.]
I. Summation
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
37
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
II. Properties of Linear Functions
I. Summation
II. Linear
III. Prop&Perc
Solution A.2 (i)
Solution A.2 (ii)
(i) missed = 3 + 0.2distance
(ii)
distance = 5
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.
I. Summation
II. Properties of Linear Functions
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
39
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
IV. SpecFunc
V. Calculus
38
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
II. Properties of Linear Functions
• 3 + .2(5) = 4 classes.
I. Summation
II. Properties of Linear Functions
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
40
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Solution A.2 (iii)
Problem A.3
(iii) A.3 In Example A.2, quantity of compact disks was related to price and income by
missed = 3 + 0.2distance
Difference
distance = 20: missed = 3 + 0.2(20)
distance = 10: missed = 3 + 0.2(10)
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.]
• 10(.2) = 2 classes.
I. Summation
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
41
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
II. Properties of Linear Functions
A.3 quantity = 120 – 9.8price + 0.3income
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
42
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
• Proportions and percentages play an important role in applied economics. • 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. Proportions and Percentages
Solution A.3
II. Properties of Linear Functions
I. Summation
II. Properties of Linear Functions
III. Prop&Perc
IV. SpecFunc
V. Calculus
43
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
• Examples in the form of percentages
– inflation rates, – unemployment rates, and – entrance acceptance rates.
I. Summation
III. Proportions and Percentages
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
44
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Proportions and Percentages
Proportions and Percentages
• A proportion is the decimal form of percent. Example: Admission Rate
A percentage is simply obtained by multiplying a proportion by 100. • Proportion: – the proportion of applicants who are admitted to MABE is 0.50.
• When using percentages, we often need to convert them to decimal form. • Percentage:
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
I. Summation
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
– 50 percent of the applications are admitted to MABE program.
• If there are 200 applicants, how many students are admitted?
45
I. Summation
Proportions and Percentages
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
46
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
III. Proportions and Percentages
Proportions
• Find changes in various quantities.
• A report by popular media can be incorrect.
• Let x be annual income x0 is the initial value
x1 is the subsequent value.
Which one is correct?
• In Thailand, the percentage of high school dropout is .20.
• Her percentage in econometric exam is .75. The proportionate change in x in moving from x0 to x1 is
x1  x0 x

x0
x0
This is sometimes called the relative change.
•
I. Summation
III. Proportions and Percentages
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
47
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Note that to get the proportionate change we simply divide the change in x, or x, by its initial value.
I. Summation
III. Proportions and Percentages
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
48
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Proportions
Percentages • The percentage change in x in going from x0 to x1 is
• For example, if income goes from $30,000 per year to $36,000 per year, what is the proportionate change?
%x  100*
x0 = 30,000 (initial value: income in 1994) x1 = 36,000 (subsequent value: income in 1995
x
x0
• It is simply 100 times the proportionate change.
proportionate change = 6,000/30,000 = 0.2
x1  x0 x

x0
x0
I. Summation
II. Linear
• %x is read as the percentage change in x.
III. Prop&Perc
IV. SpecFunc
V. Calculus
49
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
– percentage change = 100*(0.05) = 5%
%x  100*
III. Prop&Perc
II. Linear
III. Prop&Perc
IV. SpecFunc
50
• How to find percentage change?
Let x be unemployment rate. Suppose x2002 is 4% and x2003 is 5%, what is the percentage point change? What is the percentage change?
x
x0
1) percentage point change = x = 5%‐4% = 1%
2) percentage change = 100*x/x0 = 1%/4% = 25%
%x  100*
V. Calculus
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
V. Calculus
• Variable in interest is itself a percentage!
• True or False: The percentage change in GDP is 0.05.
I. Summation
IV. SpecFunc
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
• Percentage Point Change vs. Percentage Change
• 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?
x1  x0 x

x0
x0
II. Linear
Percentage Point Change vs. Percentage Change
Percentages and Proportions
– proportionate change = 15/300 = 0.05
I. Summation
III. Proportions and Percentages
51
I. Summation
III. Proportions and Percentages
II. Linear
x
x0
III. Prop&Perc
IV. SpecFunc
V. Calculus
52
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Percentages
Proportions and Percentages
• Unemployment rate has increased by one percentage point.
• But unemployment rate has increased by 25 percent!
• Example: During General Chavalit administration (1997), the value added tax was increased from 7% to 10%.
• Summary
• Who is correct?
– The percentage point change is the change in the percentages.
– The percentage change is the change relative to the initial value (x2002)
100*
– 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! x1  x0
x
 100*
x0
x0
%x  100*
I. Summation
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
53
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
III. Proportions and Percentages
Problem A.4
x
x0
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Solution A.4 (i)
• A.4 Suppose the unemployment rate in the United States goes from 6.4% in one year to 5.6% in the next.
(i) • The percentage point change is 5.6 – 6.4 = –.8, (i) What is the percentage point decrease in the unemployment rate? [ans.]
• or an eight‐tenths of a percentage point decrease in the unemployment rate.
(ii) By what percent has the unemployment rate fallen? [ans.]
I. Summation
III. Proportions and Percentages
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
56
Solution A.4 (ii)
Problem A.5
(ii) • The percentage change in the unemployment rate is 100[(5.6 – 6.4)/6.4] = –12.5%.
• 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
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
III. Proportions and Percentages
A.5
• The CEO is referring to the change relative to the initial return of 15%. IV. SpecFunc
III. Prop&Perc
IV. SpecFunc
V. Calculus
V. Calculus
58
A. quadratic
B. Logarithm
C. Exponential
– y = 0 + 1x1
– 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
– To be precise, the shareholder should specifically refer to a 3 percentage point increase.
II. Linear
III. Prop&Perc
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
• Linear function
• The majority shareholder is referring to the percentage point increase in the stock return, I. Summation
II. Linear
IV. Some Special Functions and Their Properties
Solution A.5
III. Proportions and Percentages
I. Summation
III. Proportions and Percentages
59
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
• Example: utility function – the notion of diminishing marginal returns is not consistent with a linear relationship.
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Some Special Functions and Their Properties
A. quadratic
B. Logarithm
C. Exponential
Quadratic functions
• Many economic phenomena requires the use of nonlinear functions • One way to capture diminishing returns is to include a quadratic term to a linear function.
y = 0 + 1x + 2x2
• 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.
• Given 1>0 and 2 <0
The relationship between y and x has the parabolic shape. Let 0=6 1=8 and 2 = ‐2
• Quadratic functions
• Natural Logarithm
• Exponential Function
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
A. quadratic
B. Logarithm
C. Exponential
y = 6 + 8x ‐ 2x2 IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Graph: Quadratic functions
A. quadratic
B. Logarithm
C. Exponential
I. Summation
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
IV. Some Special Functions and Their Properties
Quadratic functions
A. quadratic
B. Logarithm
C. Exponential
• When 1>0 and 2 <0, the maximum of the function occurs at
x* 
1
(2  2 )
• This is called the turning point.
If y = 6 + 8x ‐ 2x2 (so 1=8 and 2 = ‐2), the largest value of y occur at
x* = 8/4 = 2 y* = 6+8(2)‐2(22) = 14
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Quadratic functions
A. quadratic
B. Logarithm
C. Exponential
Quadratic functions
• 2 <0. This implies a diminishing marginal effect of x on y
•
A. quadratic
B. Logarithm
C. Exponential
A diminishing marginal effect of x on y is the same as saying that – the slope of the function decreases as x increases.
– Suppose we start at a low value of x and then increase x by some amount, say c
•
Calculus: the derivative of the quadratic function: y = 0 + 1x + 2x2
•
For “small” changes in x, the approximate slope of the quadratic function is
y
 1  2  2 x
slope 
x
• 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.
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
Example A.4: A quadratic wage function
•
•
wage
exper
V. Calculus
65
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
A. quadratic
B. Logarithm
C. Exponential
I. Summation
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.
x* 
wage = .48 – 2*0.008(0) = .48
wage = .48 – 2*0.008(4) = .416
4) At 30 years, an additional year of experience would lower the wage.
I. Summation
II. Linear
V. Calculus
66
A. quadratic
B. Logarithm
C. Exponential
1) there is an increasing marginal return.
2) the minimum of the function is at the point
3) The marginal effect of exper on wage depends on years of experience.
IV. Some Special Functions and Their Properties
IV. SpecFunc
• y = 0 + 1x + 2x2
• When 1<0 and 2>0, the graph of the quadratic function has U‐shape.
wage = 5.25 + .48exper – 0.008exper2
32th year (x31=31, x32=32 x=1) III. Prop&Perc
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Quadratic functions
hourly wage
years in the workforce
1st year (x0=0, x1=1, x=1) 5th year (x4=4, x5=5 x=1) II. Linear
IV. Some Special Functions and Their Properties
1
(2  2 )
wage = .48 – 2*0.008(31) = ‐.016
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
A. quadratic
B. Logarithm
C. Exponential
Natural Logarithm
Natural Logarithm
A. quadratic
B. Logarithm
C. Exponential
• Properties
•
•
1) The log function is defined only for positive values of x (x>0).
See graph of a log function
The natural logarithm, an important nonlinear function, plays an important role in econometric analysis.
for 0<x<1 for x=1
for x>1 We denotes the natural logarithm as the log function
y = log(x)
•
2) When y= log(x), the effect of x on y never becomes negative.
Other common symbols include loge(x) and ln(x).
y = log(x)
– Most calculators use ln(x).
– Different symbols are useful when we use logarithm with different bases.
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
log(x) < 0
log(x) = 0
log(x) >0
III. Prop&Perc
IV. SpecFunc
V. Calculus
•
•
69
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Graph: log function
A. quadratic
B. Logarithm
C. Exponential
y 1

x 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
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Natural Logarithm
A. quadratic
B. Logarithm
C. Exponential
• 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(x1x2) = log(x1) + log(x2)
l.2) log(x1/x2) = log(x1) – log(x2)
l.3) log(xc) = clog(x) for any constant c.
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
A. quadratic
B. Logarithm
C. Exponential
Natural Logarithm
A. quadratic
B. Logarithm
C. Exponential
Natural Logarithm
•
Let x0 and x1 be positive values.
• How good is the approximation? •
The difference in logs can be used to approximate proportionate changes. • Example, for a small change: x0=40 and x1=41 log( x1 )  log( x0 ) 
x1  x0 x

x0
x0
•
Note that log(x) = log(x1)‐log(x0)
•
Approximate percent change is
100 log(x) ≈ %x
I. Summation
II. Linear
– Exact percentage change = 100*(41‐40)/40 = 2.5%
– Approximate percentage change = 100*[log(41)‐log(40)] = 2.47%
for small changes in x.
• Example, for a large change: x0=40 and x1=60 – Exact percentage change = 100*(60‐40)/40 = 50%
– Approximate percentage change = 100*[log(60)‐log(40)] = 40.55%
for small changes in x.
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
IV. Some Special Functions and Their Properties
Natural Logarithm
A. quadratic
B. Logarithm
C. Exponential
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
74
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
A. quadratic
B. Logarithm
C. Exponential
Natural Logarithm
• Constant Elasticity Model
log(y) = 0 + 1log(x)
• Elasticity of y with respect to x can be written as
y x %y

x y %x
• The slope or elasticity is approximately equal to
– It is the percentage change in y when x increases by 1%
1 
• If y is a linear function, y = 0 + 1x, then the elasticity is y x
x
x
 1  1
x y
 0  1 x
y
 log( y )
 log( x)
• It is the elasticity of y with respect to x
– It depends on the value of x.
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Example A.5: Constant Elasticity Demand Function A. quadratic
B. Logarithm
C. Exponential
Natural Logarithm
•
• Let q
p
Log‐level model
log(y) log(y) 100log(y) %y
quantity demanded (unit)
price (dollar)
log(q) = 4.7 – 1.25log(p)
•
%y
 100  1
x
• 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
•
77
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Example A.6: Logarithmic Wage Equation
= 0 + 1x
= 1x
= 1001x
= (1001)x
The slope or the semi‐elasticity of y with respect to x is
– The price elasticity of demand = ‐1.25
– Interpret
IV. Some Special Functions and Their Properties
A. quadratic
B. Logarithm
C. Exponential
A. quadratic
B. Logarithm
C. Exponential
The semi‐elasticity is the percentage change in y when x increases by one unit. It is equal to 1001.
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Natural Logarithm
A. quadratic
B. Logarithm
C. Exponential
• Level‐log function
y = 0 + 1log(x)
• Hourly wage and years of education are related by
y = 1 log(x) 100y = 1100 log(x) log(wage) = 2.78 + .094 educ
• Using approximation, 100y = 1%x
y = (1/100)(%x)
%(wage) = 100(.094)educ
%(wage) = 9.4educ
• Interpret:
• Interpret
– One more year education increases hourly wage by about 9.4%
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
– 1/100 is the unit change in y when x increases by 1%
79
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
80
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Example A.7: Labor Supply Function
• Let hours
wage
A. quadratic
B. Logarithm
C. Exponential
A. quadratic
B. Logarithm
C. Exponential
Summary
hours worked per week
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.
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
81
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Exponential Function A. quadratic
B. Logarithm
C. Exponential
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
82
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Exponential Function
A. quadratic
B. Logarithm
C. Exponential
• We write the exponential function as
y = exp(x)
• The exponential function is related to the log function. – Other notation can be written as
• For example, given log(y) that is a linear function of x, y = ex
– How to find y itself as a function of x?
• Facts
–
–
–
–
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
83
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
exp(0) =1; exp(1) = 2.7183 exp(x) is defined for any value of x.
exp(x) is always greater than 0
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
84
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Graph: Exponential Function
A. quadratic
B. Logarithm
C. Exponential
Exponential Function
A. quadratic
B. Logarithm
C. Exponential
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
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
Exponential Function
•
V. Calculus
85
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
A. quadratic
B. Logarithm
C. Exponential
In other words, the log “undoes” the exponential, and vice versa.

The exponential function is sometimes called the anti‐log function.
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Summary: Logarithm and Exponential
Given a function,
Logarithm
log (y)= 0 + 1x
•

It is equivalent to the function
l.1 log(x1x2) = log(x1) + log(x2)
y = exp(0 + 1x)
•
If 1>0, then x has an increasing marginal effect on y.
•
Some algebraic facts:
e.1) exp(x1+ x2) = exp(x1)∙exp(x2) or
e.2) exp(x1‐ x2) = exp(x1)/exp(x2) or
e.3) exp[clog(x)] = xc
or
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
Exponential
l.2 log(x1/x2) = log(x1) – log(x2) e.2 exp(x1‐ x2) = exp(x1)/exp(x2)
l.3 log(xc) = clog(x)
ex1+ x2 = ex1∙ex2
ex1‐ x2 = ex1/ex2
eclog(x) = xc
IV. SpecFunc
e.1 exp(x1+ x2) = exp(x1)∙exp(x2)
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
e.3 exp[clog(x)] = xc
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Problem A.6
Solution A.6 (i)
A.6 Suppose that Person A earns $35,000 per year and Person B earns $42,000.
(i)
100[42,000 – 35,000)/35,000] = 20%.
(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
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
89
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
90
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Solution A.6 (ii)
Problem A.7
(ii) • The approximate proportionate change is • 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(42,000) – log(35,000) =.182, so the approximate percentage change is %18.2. log(salary) = 10.6 + .027exper.
• [Note: log() denotes the natural log.]
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
(i) What is salary when exper = 0? when exper =
5? (Hint: You will need to exponentiate.) [ans.]
V. Calculus
91
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
92
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Problem A.7 continue … Solution A.7 (i)
(ii) Use equation (A.28) to approximate the percentage increase in salary when exper increases by five years. [ans.]
(i) • When exper = 0, log(salary) = 10.6; therefore, exp(10.6) = $40,134.84. (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.]
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
log(salary) = 10.6 + .027exper
• When exper = 5, salary = exp[10.6 +.027(5)]
= $45,935.80.
93
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
Solution A.7 (iii)
(ii) 100log(y)= 1001x
(iii) log(salary) = 10.6 + .027exper
exper=0 log(salary) = 10.6 + .027exper
salary = exp(10.6) = $40,134.84. (A.28)
• Exact percentage increase
100[(45,935.80‐40,134.84)/40,134.84)
= 14.5%, so the approximate percentage change is 13.5%.
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
94
exper=5 log(salary) = 10.6 + .027exper
salary = exp[10.6 +.027(5)] = $45,935.80.
• The approximate proportionate increase is .027(5) = .135, I. Summation
V. Calculus
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Solution A.7 (ii)
IV. Some Special Functions and Their Properties
salary = so the exact percentage increase is about one percentage point higher.
95
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
96
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Problem A.8
Solution 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
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.
grthemp = .043 –.78salestax. [ans.]
I. Summation
II. Linear
IV. Some Special Functions and Their Properties
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
III. Prop&Perc
IV. SpecFunc
V. Calculus
98
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
• Example
y = log(x)
Let y = f(x) for some function f

II. Linear
Differential Calculus
V. Differential Calculus

I. Summation
IV. Some Special Functions and Their Properties
For small changes in x
dy 1

dx x
• For a small change, evaluated at the initial point x0,

df/dx is the derivative of the function f, evaluated at the
initial point x0.
y 
1
x
x0
 log( x) 
x , x0
which is the approximation of the proportionate change in x.
I. Summation
V. Differential Calculus
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
I. Summation
V. Differential Calculus
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
100
Basic Mathematical Tools . Intensive Course in Mathematics and Statistics . Chairat Aemkulwat
Differential Calculus
Differential Calculus
•
•
Other Important Rules
Let y = f(x)
DC.5 Let y =cf(x) DC.1 Let f(x) = c
= 0
or = 0
or = = c
– The derivative of a constant times any function is that same constant times the derivative of the function, – Example: y =cf(x) = xc
– The derivative of a constant c is zero.
= c = cxc‐1
DC.2 Let f(x) = log(x)
= – The derivative of the log function of x is one over x.
DC.3 Let f(x) = exp(x)
= exp(x)
or DC.6 Let y = f(x)+g(x)
I. Summation
= cxc‐1
II. Linear
or III. Prop&Perc
DC.7 Let y = z(f(x)) = cxc‐1
IV. SpecFunc
= – Chain Rule
V. Calculus
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+ – The derivative of the sum of two functions is the sum of the derivatives
= exp(x)
– The derivative of the exponential function of x is the exponential function of x
DC.4 Let f(x) = xc
= Chain Rule
I. Summation
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
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V. Differential Calculus
Differential Calculus
= DC.7 Let y = f(z(x)) Find the derivatives of the following functions:
= – y = 0 + 1x+ 2x2
Example:
y = exp(0 + 1x)
z = 0 + 1x
= – y = 0 + 1/x
= = exp
= exp
– y = 0 + 1
x
– y = 0 + 1log(x)
x
– y = exp(0 + 1x)
I. Summation
V. Differential Calculus
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
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I. Summation
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Summary: Calculus DC.1 Let f(x) = c
= 0
Differential Calculus
= DC.3 Let f(x) = exp(x)
= exp(x) DC.7 Let y = z(f(x)) DC.4 Let f(x) = xc
= cxc‐1
= DC.6 Let y = f(x)+g(x)
= I. Summation
II. Linear
III. Prop&Perc
exp
= • Two partial derivatives
1) The partial derivative of y with respect to x1 =
(where x2 is treated as a constant)
= exp
IV. SpecFunc
V. Calculus
2) The partial derivative of y with respect to x2 =
(where x1 is treated as a constant)
x
I. Summation
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V. Differential Calculus
II. Linear
III. Prop&Perc
y
x1
y
x2
IV. SpecFunc
V. Calculus
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V. Differential Calculus
Differential Calculus
Differential Calculus
•
Suppose that
y = f(x1, x2)
x
= •
+ Example DC.7: y = exp(0 + 1x)
= Notion of a partial derivative
= c
DC.5 Let y =cf(x) DC.2 Let f(x) = log(x)
•
We can approximate the change in y as
• y = f(x1,…, xn)
x2,…, xn fixed
•
Example
Example:
log(
y = 0 + 1x1 + 2x2
log(
1) What is the partial derivative of y with respect to x1 ?
) = 0 + 1educ + 2exper + 3tenure + 4age
) = .514 + .078educ +.002exper +.0088tenure +.0033age
1) What is the partial derivative of y with respect to x1 ?
2) What is the partial derivative of y with respect to x2?
2) What is the partial derivative of y with respect to x4?
I. Summation
V. Differential Calculus
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Example A.8 wage function with interaction
Differential Calculus
wage = 3.1 + .41educ + .19exper ‐ .004exper2 + .007educexper
• Example
•
Find the partial effect of exper on wage:
wage/exper = .19 ‐ .008exper +.007educ
•
At the initial values (exper0=5, educ0=12), the approximate change in wage is
wage/exper = 23.4 cents per hour
•
Given exper0=5, educ0=12; and exper1=6, educ1=12, the exact change in wage is
wage = 23 cents per hour
1) What is the partial derivative of y with respect to x1 ?
2) What is the partial derivative of y with respect to x2?
I. Summation
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
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V. Differential Calculus
I. Summation
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
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Basic Mathematical Tools
Problem A.9
Differential Calculus
•
Minimizing and maximizing functions. Let f(x1, x2, …, xk) is the differentiable function of k variables.
•
A necessary condition for x1*, x2*, …, xk* to optimize f over all possible values of xj is
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.]
• Notes
1) All of the partial derivatives of f must be zero when they are evaluated at the xh*.
2) These are called the first order conditions.
3) We hope to solve above equations for the xh*.
I. Summation
V. Differential Calculus
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
fertilizer
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V. Differential Calculus
II. Linear
III. Prop&Perc
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V. Calculus
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Solution A.9 (i)
Solution A.9 (ii)
(i) • The relationship between yield and fertilizer is graphed below.
yield
(ii) • Compared with a linear function, the function
yield = .120 + .19
fertilizer
has a diminishing effect.
• The slope approaches zero as fertilizer gets large. 122
121
– The initial pound of fertilizer has the largest effect, and each additional pound has an effect smaller than the previous pound.
120
0
50
100
fertilizer
I. Summation
V. Differential Calculus
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
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I. Summation
V. Differential Calculus
II. Linear
III. Prop&Perc
IV. SpecFunc
V. Calculus
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