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  • How do you manually calculate regression?

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Week 7: Multiple Regression

Brandon Stewart

1

Princeton

October 22, 24, 20181

These slides are heavily in

uenced by Matt Blackwell, Adam Glynn, Justin Grimmer,

Jens Hainmueller and Erin Hartman.

Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 1 / 140

Where We've Been and Where We're Going...

Last Week

I regression with two variables Iomitted variables, multicollinearity, interactionsThis Week

IMonday:

F matrix form of linear regression Ft-tests, F-tests and general linear hypothesis testsI

Wednesday:

F problems withp-values

Fagnostic regression

Fthe bootstrapNext Week

I break!

Ithen:::diagnosticsLong Run

I probability!inference!regression!causal inference

Questions?

Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 2 / 140

1Matrix Form of Regression

2OLS inference in matrix form

3Standard Hypothesis Tests

4Testing Joint Signicance

5Testing Linear Hypotheses: The General Case

6Fun With(out) Weights

7Appendix: Derivations and Consistency

8The Problems withp-values9Agnostic Regression

10Inference via the Bootstrap

11Fun With Weights

12Appendix: Trickyp-value ExampleStewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 3 / 140

The Linear Model with New Notation

Remember that we wrote the linear model as the following for all i2[1;:::;n]:y i=0+xi1+zi2+uiImagine we had annof 4. We could write out each formula:y

1=0+x11+z12+u1(unit 1)y

2=0+x21+z22+u2(unit 2)y

3=0+x31+z32+u3(unit 3)y

4=0+x41+z42+u4(unit 4)Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 4 / 140

The Linear Model with New Notation

y

1=0+x11+z12+u1(unit 1)

y

2=0+x21+z22+u2(unit 2)

y

3=0+x31+z32+u3(unit 3)

y

4=0+x41+z42+u4(unit 4)We can write this as:

2 6 64y
1 y 2 y 3 y 43
7 75=2
6 641
1 1 13 7 750+2
6 64x
1 x 2 x 3 x 43
7 751+2
6 64z
1 z 2 z 3 z 43
7 752+2
6 64u
1 u 2 u 3 u 43
7

75Outcome is alinea rcombination of the the x,z, anduvectorsStewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 5 / 140

Grouping Things into Matrices

Can we write this in a more compact form?

Yes!

Let Xandbe the following:X

(43)=2 6

641x1z1

1x2z2 1x3z3

1x4z43

7 75
(31)=2 4 0 1 23
5 Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 6 / 140

Back to Regression

Xis then(k+ 1) design matrix of independent variablesbe the (k+ 1)1 column vector of coecients.Xwill ben1:X=0+1x1+2x2++kxkWe can compactly write the linear model as the following:

y (n1)=X (n1)+u(n1)We can also write this at the individual level, wherex0iis theith row ofX: y i=x0i+uiStewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 7 / 140

Multiple Linear Regression in Matrix Form

Let bbe the matrix of estimated regression coecients andbybe the vector of tted values:b =2 6 664b

0b1...

bk3 7 775b
y=XbIt might be helpful to see this again more written out: b y=2 6 664b
y1 b y2... b yn3 7

775=Xb=2

6 6641
b0+x11b1+x12b2++x1Kbk

1b0+x21b1+x22b2++x2Kbk...

1 b0+xn1b1+xn2b2++xnKbk3 7

775Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 8 / 140

Residuals

We can easily write the

residuals in matrix fo rm: b u=yXbOur goal as usual is to minimize the sum of the squared residuals, which we saw earlier we can write:b u0bu= (yXb)0(yXb)Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 9 / 140

OLS Estimator in Matrix Form

Goal: minimize the

sum of the squa redresiduals Take (matrix) derivatives, set equal to 0 (see Appendix)

Resulting rst order conditions:

X

0(yXb) = 0Rearranging:

X

0Xb=X0yIn order to isolate

b, we need to move theX0Xterm to the other side of the equals sign.We've learned about matrix multiplication, but what about matrix \division"? Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 10 / 140

Back to OLS

Let's assume, for now, that the inverse ofX0XexistsThen we can write the OLS estimator as the following:

b = (X0X)1X0y\ex prime ex inverse ex prime y"sear it into your soul. Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 11 / 140

Intuition for the OLS in Matrix Form

b = (X0X)1X0yWhat's the intuition here? \Numerator"X0y: is approximately composed of the covariances between the columns ofXandy\Denominator"X0Xis approximately composed of the sample variances and covariances of variables withinXThus, we have something like: b (variance ofX)1(covariance ofX&y)

i.e. analogous to the simple linear regression case!Disclaimer: the nal equation is exactly true for all non-intercept coecients if you remove the

intercept fromXsuch that^0= Var(X0)1Cov(X0;y). The numerator and denominator are the variances and covariances ifXandyare demeaned and normalized by the sample size minus 1. Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 12 / 140

1Matrix Form of Regression

2OLS inference in matrix form

3Standard Hypothesis Tests

4Testing Joint Signicance

5Testing Linear Hypotheses: The General Case

6Fun With(out) Weights

7Appendix: Derivations and Consistency

8The Problems withp-values9Agnostic Regression

10Inference via the Bootstrap

11Fun With Weights

12Appendix: Trickyp-value ExampleStewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 13 / 140

OLS Assumptions in Matrix Form

1Linearity:y=X+u2Random/iid sample: (yi;x0i) are a iid sample from the population.3No perfect collinearity:Xis ann(k+ 1) matrix with rankk+ 14Zero conditional mean:E[ujX] =05Homoskedasticity: var(ujX) =2uIn6Normality:ujXN(0;2uIn)Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 14 / 140

Assumption 3: No Perfect Collinearity

Denition (Rank)

The rank of a matrix is the maximum nu mberof linea rlyindep endent

columns.In matrix form:Xis ann(k+ 1) matrix with rankk+ 1IfXhas rankk+ 1, then all of its columns are linearly independent...and none of its columns are linearly dependent implies no perfect

collinearityXhas rankk+ 1 and thus (X0X) is invertibleJust like variation inXled us to be able to divide by the variance in

simple OLS Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 15 / 140

Assumption 5: Homoskedasticity

The stated homoskedasticity assumption is: var(ujX) =2uInTo really understand this we need to know what var(ujX) is in full

generality.The variance of a vector is actually a matrix: var[u] = u=2 6

664var(u1) cov(u1;u2):::cov(u1;un)

cov(u2;u1) var(u2):::cov(u2;un) cov(un;u1) cov(un;u2):::var(un)3 7

775This matrix is alwayssymmetric since cov( ui;uj) = cov(uj;ui) by

denition. Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 16 / 140

Assumption 5: The Meaning of Homoskedasticity

What does var(ujX) =2uInmean?I

nis thennidentity matrix,2uis a scalar.Visually: var[u] =2uIn=2 6 664

2u0 0:::0

02u0:::0

0 0 0::: 2u3

7

775In less matrix notation:

I var(ui) =2ufor alli(constant variance)I

cov(ui;uj) = 0 for alli6=j(implied by iid)Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 17 / 140

Unbiasedness of

Is ^still unbiased under assumptions 1-4? DoesE[^] =?^

X0X1X0y(linearity and no collinearity)^

X0X1X0(X+u)^

X0X1X0X+X0X1X0u^

=I+X0X1X0u^ =+X0X1X0uE[^jX]=E[jX] +E[X0X1X0ujX]E[^jX]=+X0X1X0E[ujX]E[^jX]=(zero conditional mean)So, yes! Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 18 / 140

A Much Shorter Proof of Unbiasedness of

A shorter (but less helpful later) proof of unbiasedness,E[^] =E[X0X1X0y] (denition of the estimator)=

X0X1X0X(expectation of y)=Now we know the sampling distribution is centered onwe want to derive

the variance of the sampling distribution conditional onX.Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 19 / 140

Rule: Variance of Linear Function of Random Vector Recall that for a linear transformation of a random variableXwe have

V[aX+b] =a2V[X] with constantsaandb.We will need an analogous rule for linear functions of random vectors.

Denition (Variance of Linear Transformation of Random Vector) Letf(u) =Au+Bbe a linear transformation of a random vectoruwith non-random vectors or matricesAandB. Then the variance of the transformation is given by:

V[f(u)] =V[Au+B] =AV[u]A0=AuA0Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 20 / 140

Conditional Variance of

^=+ (X0X)1X0uandE[^jX] =+E[(X0X)1X0ujX] =so the OLS

estimator is a linear function of the errors. Thus:V[^jX]=V[jX] +V[(X0X)1X0ujX]=V[(X0X)1X0ujX]= (X0X)1X0V[ujX]((X0X)1X0)0(Xis nonrandom givenX)= (X0X)1X0V[ujX]X(X0X)1= (X0X)1X02IX(X0X)1(by homoskedasticity)=2I(X0X)1X0X(X0X)1=2(X0X)1This gives the (k+ 1)(k+ 1)va riance-covariancematrix of ^.To estimateV[^jX], we replace2with its unbiased estimator ^2, which is now

written using matrix notation as: ^2=P i^u2in(k+ 1)=^u0^un(k+ 1)Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 21 / 140

Sampling Variance for

Under assumptions 1-5, the

va riance-covariancematrix of the OLS estimators is given by:

V[^jX] =2X0X1=b

0b1b2bkb

0V[b0] Cov[b0;b1] Cov[b0;b2]Cov[b0;bk]

b1Cov[ b0;b1]V[b1] Cov[b1;b2]Cov[b1;bk] b2Cov[ b0;b2] Cov[b1;b2]V[b2]Cov[b2;bk] bkCov[ b0;bk] Cov[bk;b1] Cov[bk;b2]V[bk] Recall that standard errors are the square root of the diagonals of this matrix. Stewart (Princeton)Week 7: Multiple RegressionOctober 22, 24, 2018 22 / 140

Overview of Inference in the General Setting

Under assumption 1-5 in large samples:

b jjc SE[bj]N(0;1)In small samples, under assumptions 1-6, b jjc

SE[bj]tn(k+1)Estimated standard errors are:

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