Political Science 6000: Beginnings and Mini Math Boot Camp

Transcription

1 Political Science 6000: Beginnings and Mini Math Boot Camp January 20, 2010

2 First things first Syllabus This is the most important course you will take. 1. You need to understand these concepts in order to read and evaluate the work in your field. 2. You need to understand these concepts in order to do a basic level of quantitative research. 3. You need to understand these concepts in order to learn more advanced methods. But you can probably say these things about many of your other classes. Given the above reasons this class is the most important you ll take is because you probably cannot teach yourself these things.

3 Reading assignments Syllabus I. Required and optional texts Wooldridge is required; Kennedy helpful; Chiang a good sourcebook. II. Doing the work: A. Complete reading before class so you can perform calculations in class. 1. Read each chapter quickly first time, more thoroughly the second. B. Reading should be done with a pen(cil) and pad of paper at hand. I strongly encourage you to replicate the textbook s examples. C. You should expect to have weekly problem sets. D. Additional readings may be assigned to provide examples of how techniques are used.

4 Expectations and Grading Standards I. Problem sets: Both mathematical and computer exercises geared to understanding topics. You may discuss problems with each other, but work should be your own. II. Exams: Exams involve a combination of demonstrating concepts mathematically, recognizing the presence or absence of particular properties in hypothetical problems, and interpreting results of data analysis. III. Term paper: You need to develop an idea for a paper quickly and work on the paper as we cover new topics. IV. Attendance is required.

5 s Learning Objectives I. Understand Regression as a specific approach to research design. II. Review some of the basis mathematical tools that you will need for the course

6 Regression as an Approach to Research Design I. A research design involves: (a) a theory (b) hypotheses (c) data selection or collection (d) means of analysis II. If your theory is clear, you should be able to write it down in the form of a regression equation. III. Your regression equation should be written in a parsimonious way that focuses upon your main and confounding hypotheses. It should not include everything. IV. A good regression equation should direct you to the data that you need to collect in order to test your theory so that you have sufficient degrees of freedom and variation on your independent variables. V. A good regression equation should tell you if you have to consider things like selection effects and reciprocal causation. These are all things you need to consider even if you are doing qualitative work.

7 Basic arithmatic operations = 13/ ( 2 3 )( 1 4 )( 3 2 ) = 1/ = 13/ ( ) 1 = 1/9. You should understand these key ideas: 1. The least common denominator in addition of fractions. 2. Multiplication of the numerator and denominator in fractions. 3. The idea that x x = 1 4. The inverse function: x 1 = 1 x, x 2 = 1 x 2, etc. 5. The order of operations: (1) anything in parentheses (2) exponents (3) multiplication/division (4) addition/subtraction.

8 Basic arithmatic properties 1. Commutative property of addition: a + b = b + a 2. and multiplication: ab = ba 3. Associative property of addition: (a + b) + c = a + (b + c) 4. and multiplication: (ab)c = a(bc) 5. Distributive property of multiplication: a(b + c) = ab + ac

9 The Summation Operator 5. 6 i=1 x i where x i = i Answer: 21 xi = (x 1 + x 2 + x x n ) (1) 1. The summation operator is a short-hand for writing the sum of a series of terms. 2. Until you are fully comfortable with the idea, just write out a series of terms and do manipulations on the basis of something more familiar

10 The Summation Operator 6. 6 i=1 (ax i + b) where x i = i, a = 3, b = 2 Answer: 75. n (ax i + b) = (ax 1 + b + ax 2 + b + ax 3 + b... + ax n + b) i=1 = ax 1 + ax 2 + ax ax n + nb = a(x 1 + x 2 + x x n ) + nb = n n n a x i + nb = a x i + b (2) i=1 i=1 So, 3 x i + 5(2) = 3(21) + 12 = 75 Note: Pay attention to parentheses. Ex: xi + b (x i + b) and the presence/absence of subscripts. i=1

11 The Summation Operator 7. x i y i where x = {3, 4, 2, 7, 9} and y = {2, 3, 1, 4, 2} Answer: -26 n x i y i = (x 1 y 1 + x 2 y 2 + x 3 y x n y n ) (3) i=1 So, n i=1 x iy i = ( ) Remember x i y i x i ( y i ), which you can verify using the above example.

12 Basic Linear Equations 8. What is the intercept of y = β 0 + β 1 x? β 0 The intercept is simply the point where a line crosses the y-axis that is, where x = For y = β 0 + β 1 x, what is the change in y given a 2-unit change in x? 2β 1 β 1 reflects the slope of the line which gives the magnitude and direction of the relationship between y and x. y/ x (rise over run).

13 Graphical Illustration of a Line Copyright 2009 South-Western/Cengage Learning 1

14 Basic Algebra Syllabus 10. Solve y = β 0 + β 1 x for x. x = (y β 0 )/β Solve x 2 + 5x + 6 = 0 for x. x = 3, 2 Key ideas: 1. Know how to solve equations. 2. Ability to see factors. 3. Consider the characterisics of a parabola. If x 2 is multiplied by a positive number, the parabola opens upward. If x 2 is multiplied by a negative number, the parabola opens downward. For our purposes, quadratic terms can be useful for specifying relations between variables that are not linear.

15 A Quick Check on Some Odds and Ends 12. Calculate x where x = {3, 4, 2, 7, 9} Answer: 5 x is often used to denote the mean. 13. If taking the average of 10 numbers produces x = 5, then what is 10 i=1 x i? Answer: 50 It is useful to remember that n x = x. 14. Determine the median of x = {3, 4, 2, 7, 9} Answer: 4 You should be able to distinguish the median (the point with 50% higher and 50% lower) from the mean. 15. Calculate: exp(0) (that is, e 0 ). Answer: 1 You should recognize that anything to the zeroth power equals Calculate: ln exp(βx ) (that is, ln e βx ). Answer: βx You should recognize that the natural log is the inverse function of the exponential function.

16 Graphical Illustration of the Exponential Function Copyright 2009 South-Western/Cengage Learning 1 The exponential function applied to any number produces a positive number.

17 Graphical Illustration of the Natural Log Copyright 2009 South-Western/Cengage Learning 1 You can only take the log of a positive number. This makes sense given that the exponential function always produces a positive number.

18 Basic Calculus Syllabus 17. Calculate dy dx for y = β 0 + β 1 x β 1. The derivative is the instantaneous rate of change in a function. The basic formula for the derivative of a variable is: d dx ax b = abx b 1 For a line, the slope is the rate of change. For a line, the rate of change is constant across the entire line.

19 Basic Calculus Syllabus 18. Calculate dy dx for y = β 0 + β 1 x + β 2 x 2. 2xβ 2 + β 1 The same formula applies for this function as the previous one. You just have additional terms to apply that formula to. In this case, the rate of change depends where you are on the curve. In other words, the rate of change is not constant.

20 Basic Calculus Syllabus 19. What value of x minimizes the function: y = β 0 + β 1 x + β 2 x 2 for β 0 = 3, β 1 = 5, β 2 = 1? -5/2 To find the value that minimizes or maximizes a function, take the derivative, set it equal to 0, and solve for (in this case) x. β 1 + 2β 2 x = x = 0 2x = 5 x = 5 2 (4) If you plug in different values of x above and below -5/2, you will be able to verify this result.

21 Basic Calculus Syllabus 20. What value of x maximizes the function: y = β 0 + β 1 x + β 2 x 2 for β 0 = 3, β 1 = 5, β 2 = 1? To find whether a value minimizes or maximizes a function, take the second derivative (the derivative of the first derivative). If the second derivative is positive, the value is a minimum; if the second derivative is negative, the value is a maximum. d dx β 1 + 2β 2 x = 2β 2 > 0 (5) since β 2 is positive. So, the value above, -5/2, is a minimum. As x becomes larger in absolute value, y will continue to increase.

22 Expectation Syllabus If X is a discrete random variable that takes on the values of -1, 0, and 2 with respective probabilities of 1/8, 1/2, and 3/8, what is the value of: 1. E(X )? 5/8 Take the case of a discrete variable, the expectation is the probability-weighted average of the variable: E(X ) = k x j f (x j ) (6) where f (x j ) is the pdf (probability density function) for X and x j are the different values taken by X. This means: j=1 E(X ) = 1(1/8) + 0(1/2) + 2(3/8) = 1/8 + 6/8 = 5/8 (7)

23 Properties of Expectations: cx The expectation of cx, where c is a constant, is simply ce(x ). Again, take the case of a discrete variable, the expectation is: k E(X ) = x j f (x j ) (8) j=1 and multiply every value of x J by c: k k E(cX ) = cx j f (x j ) = c[x j f (x j )] = c So E(8X )=8E(X )=8 5 8 =5 j=1 j=1 k x j f (x j ) = ce(x ) (9) j=1 Also note, this shows that E(c)=c.

24 Properties of Expectations: cx + b The expectation of cx + b, where c and b are constants, is simply ce(x ) + b. Again, take the case of a discrete variable, the expectation is: E(X ) = k x j f (x j ) (10) j=1 and multiply every value of x J by c and add b: E(cX + b) = = c k (cx j + b)f (x j ) = j=1 k x j f (x j ) + b j=1 k {c[x j f (x j )] + bf (x j )} j=1 k f (x j ) = ce(x ) + b (11) j=1 So E(8X +2)=8E(X )+2= =7 Political Science 6000: Beginnings and Mini Math Boot Camp

25 Properties of the Expectation If X is a discrete random variable that takes on the values of -1, 0, and 2 with respective probabilities of 1/8, 1/2, and 3/8, what is the value of: E(X 2 )? 13/8 E(X 2 ) [E(X )] 2. In this case: E(X 2 ) = 1 2 (1/8) (1/2) (3/8) = 1(1/8) (3/8) = 1/8 + 12/8 = 13/8 (12)

26 The Variance Syllabus If X is a discrete random variable that takes on the values of -1, 0, and 2 with respective probabilities of 1/8, 1/2, and 3/8, what is the value of: Var(X )? This is actually simple to determine given what you ve done. Letting µ = E(X ) (a constant), Var(X ) = E[(X µ) 2 ] = E[X 2 2µX + µ 2 ] = E(X 2 ) 2µE(X ) + µ 2 = E(X 2 ) [E(X )] 2 (13) So with earlier work, Var(X ) = 13/8 [5/8] 2 = 104/64 25/64 = 79/64

27 Properties of the Variance, cx The variance of cx, where c is a constant, is simply c 2 Var(X ). Var(cX ) = E[(cX ) 2 ] [E(cX )] 2 = E[c 2 X 2 ] [ce(x )] 2 = c 2 E(X 2 ) c 2 [E(X )] 2 = c 2 {E(x 2 ) [E(X )] 2 } = c 2 Var(X ) (14) So with earlier work, Var(8X ) = 64(79/64) = 79

28 Properties of the Variance, X + b The variance of X + b, where b is a constant, is simply Var(X ). Var(X + b) = E[(X + b) 2 ] [E(X + b)] 2 = E(X 2 + 2bX + b 2 ) [E(X ) + b] 2 = E(X 2 ) + 2bE(X ) + b 2 {[E(X )] 2 + 2bE(X ) + b 2 } = E(X 2 ) + 2bE(X ) + b 2 [E(X )] 2 2bE(X ) b 2 = E(X 2 ) [E(X )] 2 = Var(X ) (15) So Var(8X +2) = Var(8X ) = 79 This result makes sense because whereas adding a constant to a random variable shifts the mean, it does not affect the spread of the variable s distribution.

29 Standard Deviation Syllabus The standard deviation is simply the square root of the variance. So, sd(x )=[Var(X )] 1/2 = (79/64) 1/2

30 Covariance Syllabus Covariance refers to the amount of linear dependence between two variables. Note, linear excludes quadratic and higher-order relationships. So, let µ x = E(X ) and µ y = E(Y ) then: Cov(X, Y ) = E[(X µ x )(Y µ y )] = E[XY Y µ x X µ y + µ x µ y ] = E(XY ) µ x E(Y ) µ y E(X ) + E(X )E(Y ) = E(XY ) E(X )E(Y ) (16) Note, that if X = Y, then we have the same formula as we had for the variance.

31 Properties of the Covariance 1. If X and Y are independent, then E(XY )=E(X )E(Y ) and the Cov(X, Y )=0. 2. Cov(cX, by ) = cbcov(x, Y ). This follows from the observation that Cov(X, Y ) has to follow the form of Var(X ) if Y = X. 3. By similar logic Cov(X + c, Y + b)=cov(x, Y ). 4. Finally, Cov(X, Y ) sd(x )sd(y ). They are equal if y = f (x) (i.e. Y is a linear function of X, or even clearer, Y = X ), making Cov(X, Y ) a maximum or X or Y is a constant, in which case the Cov(X, Y )=0 as is one of the standard deviations. But in between, Cov(X, Y ) could equal zero even if neither X nor Y is a constant (i.e. sds are non-zero), but if y f (x), then Cov(X, Y ) will be less than sd(x )sd(y ).

32 Variance of Sums of Random Variables Var(cX + by ) = c 2 Var(X ) + b 2 Var(Y ) + 2cbCov(X, Y ) Var(cX + by ) = E[(cX + by ) 2 ] [E(cX + by )] 2 = E[c 2 X 2 + 2cbXY + b 2 Y 2 ] [ce(x ) + be(y )] 2 = c 2 E(X 2 ) + 2cbE(XY ) + b 2 E(Y 2 ) [c 2 [E(X )] 2 + 2cbE(X )E(Y ) + b 2 [E(Y )] 2 ] = c 2 E(X 2 ) c 2 [E(X )] 2 + b 2 E(Y 2 ) b 2 [E(Y )] 2 + 2cb[E(XY ) E(X )E(Y )] = c 2 Var(X ) + b 2 Var(Y ) + 2cbCov(X, Y ) (17) So, Var(X + Y ) = = 6. Change the appropriate signs above and Var(X Y ) = = 4.

33 Conditional Expectation Conditional Expectation is the expectation of a variable for a specific value of another variable: E(Y X = x). So if y = {4, 4, 2, 5, 6, 9, 3, 5, 2, 8} and x = {1, 0, 0, 0, 1, 0, 1, 1, 1, 0}, you determine E(Y x = 1) by taking the mean of Y for only those values of Y where the corresponding value of x = 1. In this case, E(Y x = 1)=4. And E(Y x = 0)=5.6. If Y is independent of X, then E(Y X = x)=e(y ). You can see this because if there is no relationship between Y and X, then the expectation of Y should be the same for all values of X.