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Lecture 7 Multiple Regression & Matrix NotationPowerPoint Presentation

Lecture 7 Multiple Regression & Matrix Notation

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Lecture 7 Multiple Regression & Matrix Notation. Quantitative Methods 2 Edmund Malesky, Ph.D., UCSD. Order of Presentation. 1. Review of Variance of Beta Hat 2. Review of T-Tests 3. Review of Quadratic Equations 4. Introduction to Multiple Regression 5. The Role of Control Variables

Lecture 7 Multiple Regression & Matrix Notation

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Lecture 7Multiple Regression & Matrix Notation

Quantitative Methods 2

Edmund Malesky, Ph.D., UCSD

1. Review of Variance of Beta Hat

2. Review of T-Tests

3. Review of Quadratic Equations

4. Introduction to Multiple Regression

5. The Role of Control Variables

6. Interpreting Regression Output

- Remember, we are working with a sample of the true population.
- We are using that sample as a way to estimate the true relationships between variables (regression parameters in the population.
- As in QM1, we must always remember that our estimates will be slightly different each time we sample from the population.
- We know that the mean of repeated sampling, we equal the population parameter, but we still might want to have some sense of the variance.
- The smaller the variance, the more efficient the estimate
- As a result, we need some sense of the range that would occur after repeated sampling.
- The confidence interval, derived from the standard error (SE) of the regression parameter is the way we estimate that range.

Remember, how we did this is STATA in Lecture 4. We divided the Root Mean Squared Error of the Model by the Standard Deviation of the Independent Variable. This gave us the Standard Error (SE) or the variance of Beta Hat.

- has nice intuitive qualities
- As the size of the errors decreases,
decreases

- The line fits tightly through the data. Few other lines could fit as well

- As the variation in x increases, decreases
- Few lines will fit without large errors for extreme values of x

- Because the variance of the estimated errors has n in the denominator, as n increases, the variance of β1hat decreases
- The more data points we must fit to the line, the smaller the number of lines that fit with few errors
- We have more information about where the line must go.

- T-test – tests that individual coefficients are not zero.
- This is the central task for testing most policy theories

- In general, our theories give us hypotheses that β0 >0 or β1 <0, etc.
- We can estimate β1hat , but we need a way to assess the validity of statements that β1 is positive or negative, etc.
- We can rely on our estimate of β1hat and its variance to use probability theory to test such statements.

- We know that β1hat ~ N(β1 , σβ)
- Subtracting β1 from both sides, we can see that (β1hat - β1 ) ~ N( 0 , σβ )
- Then, if we divide by the standard deviation we can see that:
(β1hat - β1 ) / β1hat ~ N( 0 , 1 )

- To test the “Null Hypothesis that β1 =0, we can see that: β1hat / σβ~ N( 0 , 1 )

- This variable is a “z-score” based on the standard normal distribution.
- 95% of cases are within 1.96 standard deviations of the mean.
- If β1hat / σβ > 1.96 then in a series of random draws there is a 95% chance that β1 >0
- The key problem is that we don’t actually know σβ, the true population parameter.

- Obvious solution is to substitute
in place of σβ

- Problem: β1hat / is the ratio of two random variables, and this will not be normally distributed
- Fortunately, an employee of Guinness Brewery figured out this distribution in 1919

- The statistic is called “Student’s t,” and the t-distribution looks similar to a normal distribution
- Thus β1hat / ~ t(n-2) for bivariate regression.
- More generally β1hat / ~ t(n-k)
- where k is the # of parameters estimated

- Note the addition of a “degrees of freedom” constraint
- Thus the more data points we have relative to the number of parameters we are trying to estimate, the more the t distribution looks like the z distribution.
- When n>100 the difference is negligible

- Results often illustrative rather than precise
- Only tests “not zero” hypothesis – does not measure the importance of the variable (look at confidence interval)
- Generally reflects confidence that results are robust across multiple samples

As the degrees of freedom increase, the t-distribution approaches the normal distribution.

T-distribution:The Statistical Workhorse

df=6

df=4

df=2

-3

3

0

- In STATA, the null hypothesis for a two-tailed t-test is:H0: βj=0

- To test the hypothesis, I need to have a rejection rule. That is, I will reject the null hypothesis if, t is greater than some critical value (c) of the t distribution. You may know this in excel lingo as tcrit. c is up to me to some extent, I must determine what level of significance I am willing to accept. For instance, if my t-value is 1.85 with 40 df and I was willing to reject only at the 5% level, my c would equal 2.021 and I would not reject the null. On the other hand, if I was willing to reject at the 10% level, my c would be 1.684, and I would reject the null hypotheses.

t-distribution:5 % rejection rule for the that H0: βj=0 with 25 degrees of freedom

Looking at table G-2, I find the critical value for a two-tailed test is 2.06

Rejection Region

Area=.025

Rejection Region

Area=.025

-2.06

2.06

0

- But this operation hides some very useful information.
- STATA has decided that it is more useful to provide what is the smallest level of significance at which the null hypothesis would be rejected. This is known as the p-value.
- In the previous example, we know that .05<p<.10.
- To calculate the p, STATA computes the area under the probability density function.

P-value=P(|T|>t)

In this case, P(|T|>1.85)=

2P(T>1.85)=2(.0359)

=.0718

Area=.9282

Rejection Region

Area=.0359

Rejection Region

Area=.0359

0

- reg approval cpi
- Source | SS df MS Number of obs = 148
- ---------+------------------------------ F( 1, 146) = 9.76
- Model | 1719.69082 1 1719.69082 Prob > F = 0.0022
- Residual | 25731.4061 146 176.242507 R-squared = 0.0626
- ---------+------------------------------ Adj R-squared = 0.0562
- Total | 27451.0969 147 186.742156 Root MSE = 13.276
- ------------------------------------------------------------------------------
- approval | Coef. Std. Err. t P>|t| [95% Conf. Interval]
- ---------+--------------------------------------------------------------------
- cpi | -.1348399 .0431667 -3.124 0.002 -.2201522 -.0495277
- _cons | 60.95396 2.283144 26.697 0.000 56.44168 65.46624
- ------------------------------------------------------------------------------
- . sum cpi
- Variable | Obs Mean Std. Dev. Min Max
- ---------+-----------------------------------------------------
- cpi | 148 46.45878 25.36577 23.5 109

- . reg approval unemrate
- Source | SS df MS Number of obs = 148
- ---------+------------------------------ F( 1, 146) = 0.85
- Model | 159.716707 1 159.716707 Prob > F = 0.3568
- Residual | 27291.3802 146 186.927262 R-squared = 0.0058
- ---------+------------------------------ Adj R-squared = -0.0010
- Total | 27451.0969 147 186.742156 Root MSE = 13.672
- ------------------------------------------------------------------------------
- approval | Coef. Std. Err. t P>|t| [95% Conf. Interval]
- ---------+--------------------------------------------------------------------
- unemrate | -.5973806 .6462674 -0.924 0.357 -1.874628 .6798672
- _cons | 58.05901 3.814606 15.220 0.000 50.52003 65.59799
- ------------------------------------------------------------------------------
- . sum unemrate
- Variable | Obs Mean Std. Dev. Min Max
- ---------+-----------------------------------------------------
- unemrate | 148 5.640541 1.744879 2.6 10.7

y

x1

0 1

- β0hat is the intercept as in the linear equation
- β1hat is the slope when x is 0 to the first unit of x.
- β2hat is used to calculate the slope at other points on the line.
- A positive coefficient on β2hatmeans the curve turns upward.
- A negative coefficient on β2hatmeans the curve turns downward

- Use equation 1 to get predicted value for each point on the line.
- Use equation 2 to get the slope for each point on the curve.
- Use equation 3 to isolate the point where the slope is equal to 0

Figure 1: Kuznets Predictions and Actual Relationship between Growth and Inequality

- Multiple regression to address multiple causes

- We have found an estimator for the relationship between x and y
- We have developed methods to use the estimator to test hypotheses derived from theories about x and y.
- But we have only 1 x (and only 1 β)
- The world is more complicated than that!

- We can make a simple extension of the bivariate model to the multivariate case
- Instead of a two dimensional space (x and y axes) we move into multi-dimensional space
- If we have x1 and x2, then we are fitting a two dimensional plane through points in space.

y: Votes for Candidate A in 2004

x1: Expenditures of Candidate A in $1000s (2000-2003)

x2: 100s of new jobs created (2000-2003)

x1: Expenditures of Candidate A in $1000s (2000-2003)

y: Votes for Candidate A in 2004

Because multiple linear regression includes more than a single independent variable, the result of an analysis is best visualized as a plane rather than as the line of a bivariate regression analysis.

This plane is defined by a series of slopes and a y-intercept value, and oriented such that deviations between the observed data points and the plane are minimized in the direction of the dependent variable.

x2: 100s of new jobs created (2000-2003)

x1: Expenditures of Candidate A in $1000s (2000-2003)

y: Votes for Candidate A in 2004

Q1. If we modeled only an equation with expenditures, where would the impact of Job Growth show up in our results?

A1. Correct. It would show up in a larger residual size.

A2. β1hat is the ceteris paribus effect of expenditures on vote changes, controlling for job growth.

Q2. How do I interpret the coefficient β1hat in my STATA output?

A3. It means that the effect of Job Growth is held fixed or constant. ▲Job Growth = 0

Q3. What does “controlling for” mean?

In other words, I regress x1 on x2.

Basically, β1hat measures the sample relation between x1 and y, after x2 has been partialled out.

Covariance between x1 and x2

- Coefficient β1hat is calculated based on area in yellow circle that overlaps with blue, but NOT with red.

x1

y

x2

Center area discarded –

We can’t say which variable accounts

for it

- Zero conditional mean, before we summarized GM4 as “the population error (u) has an expected value of 0 for any value of the explanatory variable (x).” Essentially, this meant that other factors having a direct impact on y (i.e. changes in votes) are unrelated on average to x (expenditures).

The equation also implies that we have correctly specified the functional form between the independent and dependent variables!

- Now, GM4 becomes “The population error (u) has an expected value of 0 for any combination of x1 & x2Other factors having a direct impact on y (i.e. changes in votes) are unrelated on average to x1 (expenditures) and x2 (job growth).

- Before, GM3 was that there must be sample variation in explanatory variables. xi’s are not all the same value

Essentially, if any one of our x’s is perfectly explained by the others, it will drop out of our model.

- Now GM3 reads, none of the independent variables is constant and there are no exact linear relationships among the dependent variables.

- Above 3 dimensions MR becomes difficult to visualize.
- Logic of the process is the same. We are fitting SETS of x’s to each point on a y dimension.
- β0hatremains the intercept and β1hat,β2hat.. are called slope estimates.
- Though in a quadratic function, slope estimates for both coefficients is slightly incorrect. Why?
- The basic equation of the true population model in scalar notation is:

- In scalar terms the equation for βhat of variable k becomes:
- where = the linear prediction of xik based on the other x’s
- Similar to the bivariate estimator. Then we used because we lacked any better expectation about x

Please note that Wooldridge indexes observations by “t” instead of “i” in the matrix algebra discussions. We use “i” to maintain the analogy with scalar algebra.

- Writing out these terms and multiplying them in scalar notation is clumsy.
- Represented in simpler terms through linear (matrix) algebra
- The basic equation becomes:

- The vectors and matrices in are represented by
- Note that we post-multiply X byβsince this order makes them conformable.

- To derive our vector of coefficients βhat, we will need to do some math with matrices
- Multiplying matrices
- Taking the transpose of a matrix
- Inverting a matrix

- Multiplication
- Where

- Taking the transpose is an operation that creates a new matrix based on an existing one.
- The rows of A = the columns of A'
- Hold upper left and lower right corners and rotate 180 degrees.

- Thus IS the sum of squared residuals (SSR)
- In matrix terminology, we want to pick the vector of βhats’s that minimizes
- As in scalar notation, the vector of βhats’s is a function of the X matrix and the y vector

- For an nxn matrix A, there may be a B such that AB = I = BA.
- The inverse is analogous to a reciprocal
- A matrix which has an inverse is called “nonsingular”.
- A matrix which does not have an inverse is “singular”.
- An inverse exists only if the determinant of A does not equal 0.
- This is true only if columns of the matrix are not linearly dependent (i.e. 2*column1=colum2).

How to find inverse matrixes? determinants? and more?

- If and |A| 0

- Begin again with minimizing squared errors
- Take the partial derivative with respect to each element of β hat

- Setting the derivative equal to 0 for minimization yields:
- Notice that this statement implies k separate equations with k unknowns
- Where k=# of parameters in the β-hat vector

- Rearranging terms, this becomes:
- We want to divide both sides by X’X, which means multiplying by its inverse (just like with scalar fractions)
- Thus, we must take the inverse of X’X

- (X’X)-1 will exist only if X’X is non-singular (is of full rank)
- That is, none of the x’s can be a perfect linear function of the other x’s
- If this is true, then:

Essentially, the same as scalar notation. The covariance of the X matrix and the y vector over the variance-covariance matrix of X.

- Each element of the vector βhat is a slope coefficient for one of the x’s
- Same as in bivariate context except that βhat1 is the expected change in y for a 1 unit increase in x1, while holding x2…xn constant
- Thus βhat1 represents the direct effect of x1 on y, controlling for x2…xn

- Calculation of each element of the βhat vector includes information from observations on all the independent variables and the dependent variable
- The computation includes information about covariation among the x’s
- It also includes information about the relationship between other x’s and y

=

1. A Matrix of Independent Variables: X and a Vector Dependent Variable y

- X[5,3]
- x1 x2 x3
- r1 1 -6.5837102 -12.336432
- r2 1 -17.004963 3.0143378
- r3 1 -1.9127336 -15.459048
- r4 1 12.721842 1.3890865
- r5 1 .54984921 11.332677
- y[5,1]
- Y
- r1 -2.9141634
- r2 -53.135086
- r3 12.545163
- r4 38.019218
- r5 -3.9364302

- Note that x4=2*x3
- X[5,4]
- x1 x2 x3 x4
- r1 1 -6.5837102 -12.336432 -24.672865
- r2 1 -17.004963 3.0143378 6.0286756
- r3 1 -1.9127336 -15.459048 -30.918097
- r4 1 12.721842 1.3890865 2.778173
- r5 1 .54984921 11.332677 22.665354
- . matrix XtX = X'*X
- . display det(XtX)
- 0
- Determinant of X’X = 0
- Thus X’X is not of full rank

- Create the TRANSPOSE of X
- . matrix Xt = X'
- Xt[3,5]
- r1 r2 r3 r4 r5
- X1 1 1 1 1 1
- X2 -6.5837102 -17.004963 -1.9127336 12.721842 .54984921
- X3 -12.336432 3.0143378 -15.459048 1.3890865 11.332677
- Create X'X Matrix - Variation & Covariation of X's
- . matrix XtX = X'*X
- symmetric XtX[3,3]
- X1 X2 X3
- X1 5
- X2 -12.229716 498.32015
- X3 -12.05938 83.432836 530.6151

- Note Matrix is of FULL RANK - CAN BE INVERTED
- . display det(XtX)
- 1160053.6
- . matrix B = inv(X'*X)*X'*y
- B[3,1]
- Y
- X1 3.1000365
- X2 3.0111852
- X3 -.98715343

- . reg Y X2 X3
- Source | SS df MS Number of obs = 5
- ---------+------------------------------ F( 2, 2) = 256.94
- Model | 4415.23139 2 2207.61569 Prob > F = 0.0039
- Residual | 17.1837301 2 8.59186507 R-squared = 0.9961
- ---------+------------------------------ Adj R-squared = 0.9922
- Total | 4432.41512 4 1108.10378 Root MSE = 2.9312
- ------------------------------------------------------------------------------
- Y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
- ---------+--------------------------------------------------------------------
- X2 | 3.011185 .1362818 22.095 0.002 2.424812 3.597558
- X3 | -.9871534 .1317047 -7.495 0.017 -1.553833 -.4204737
- _cons | 3.100036 1.380879 2.245 0.154 -2.841404 9.041477
- ------------------------------------------------------------------------------

- . matrix b = inv(X'*X)*X'*y
- matrix has missing values
- r(504);
- . reg Y X2 X3 X4
- Source | SS df MS Number of obs = 5
- ---------+------------------------------ F( 2, 2) = 256.94
- Model | 4415.23139 2 2207.61569 Prob > F = 0.0039
- Residual | 17.1837301 2 8.59186507 R-squared = 0.9961
- ---------+------------------------------ Adj R-squared = 0.9922
- Total | 4432.41512 4 1108.10378 Root MSE = 2.9312
- ------------------------------------------------------------------------------
- Y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
- ---------+--------------------------------------------------------------------
- X2 | 3.011185 .1362818 22.095 0.002 2.424812 3.597558
- X3 | (dropped)
- X4 | -.4935767 .0658524 -7.495 0.017 -.7769166 -.2102369
- _cons | 3.100036 1.380879 2.245 0.154 -2.841404 9.041477
- ------------------------------------------------------------------------------

Optional!

- Estimator β# is a function of y and some set of weights C
- Set of weights C can be rewritten as our OLS estimator plus some matrix of weights D

Note, this is also a linear estimator of β

Set of Weights

- Take expectations of both sides and substitute Xβ+u for y, as this is the population definition of y.
- Separate out (X’X)-1X’ and D terms to yield:

- Take expectation of each term
- Recall that (X’X)-1X’X β =(1)* β= β, and E(u)=0
- Pull out the identity matrix (I), in order to separate β

- Unbiased: Since DX β is non-stochastic, then if β# is Unbiased, meaning E(β#) = β , then DX must be equal to 0. In other words, the weights must equal 0 for β# to be unbiased.
- Efficiency: If DX=0, then we can write β# as:

F.O.I.L

=(1)*β

- Now, subtract β to calculate the distance of β# from β as:
- To get variance, we square the errors (the right side of the equation above

F.O.I.L

- Since E(uu’)=σ2I
- Recall that DX=0 and therefore X’D’=0

- First term of this equation is the variance of β, plus the squared weights = β#
- We are concerned with variance – the diagonal elements of the matrix
- Diagonal elements of DD’ must be positive – they are sums of squares

- If diagonal elements of DD’ are positive, then the variance of β#>variance of β
- Unless D=0, but then β#= β

- Thus among the class of linear estimators
- β#= Cy

- That are also unbiased linear estimators
- E(β#)= β

- OLS has the least variance
- Thus OLS is the Best Linear Unbiased Estimator
- Its BLUE!

- We needed a number of assumptions to justify the claim that OLS is BLUE
- Next time, we will review the assumptions underlying our use of OLS

- Then we will discuss how to assess the performance of regression models
- Then we begin addressing the violation of our central assumptions and how to repair our estimator of the vector βhat