L8 - The Multivariate Regression Model Flashcards

1
Q

How do you work out the residual sum of squares?

A
  • Same as a bivariate model

Min RSS = Σt=1N(ui)2

So for example less take a 3 variable model of:

Yi= β12Xi23Xi3+ui

Min RSS = Σt=1N(Yi- β1(hat) -β2(hat)Xi2 - β3(hat)Xi3)2

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

How do we work out the RSS of a 3 variable regression model?

A
  • By differentiating
  • FOC conditions for this problem are:

dRSS/d(β1(hat)) = -2Σ(Yi- β1(hat) -β2(hat)Xi2 - β3(hat)Xi3)=0

dRSS/d(β2(hat)) = -2ΣXi2*(Yi- β1(hat) -β2(hat)Xi23(hat)Xi3)=0

dRSS/d(β3(hat)) =-2ΣXi3*(Yi- β1(hat) -β2(hat)Xi2 - β3(hat)Xi3)=0

This gives us a system of three equations in three unknownvariables. More generally, if:

Yi- β1j=2kjXij)+ui

then the first order conditions will yield a system of k equationsin k unknown variables.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

What are the Least-Square normal equations of the three variable regression model?

A

Yi= β12Xi23Xi3+ui

has least-squares normal equations of the form:

β1(hat)N +β2(hat)ΣXi2 + β3(hat)ΣXi3=ΣYi

β1(hat)ΣXi22(hat)ΣXi22 + β3(hat)ΣXi2Xi3=ΣXi2Yi

β1(hat)ΣXi32(hat)ΣXi3Xi2 + β3(hat)ΣXi32=ΣXi3Yi

This is a system of three equations in three unknowns. Solving these yields the OLS estimates.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

What happens when you divide through the OLS estimator by N?

A

β1(hat)N +β2(hat)ΣXi23(hat)ΣXi3=ΣYi

β1(hat) +β2(hat)(ΣXi2/N)+β3(hat)(ΣXi3/N)=(ΣYi/N)

β1(hat) +β2(hat)Xi2(bar) +β3(hat)Xi3(bar)=Yi(bar)

  • this is the regression line that passes through the mean of the data just like a bivariate regression model
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

What is the general equation to solve multivariate OLS estimators?

A

In general, let X be a matrix whose columns contain the data for the explanatory variables and let y be a vector containing the data for the dependent variable. The multivariate OLS estimator can be calculated as:

β(hat)=(X(Ubar)TX(Ubar)-1*X*(Ubar)Ty(Ubar)

This is k x 1 vector where k is the number of explanatory variables (including the intercept)

In order to calculate the OLS estimator in this way we must assume:

  1. k < N, where N is the number of observations
  2. The X variables are linearly independent.
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

What is a example of solving for an OLS estimator of a multivariate regression model?

A

variable(Ubar) = vector

  • the X variable is derived from –> X(Ubar)TX(Ubar) = X(Ubar)-1
  • X(Ubar)TY(Ubar) = the y matrix
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

When it possible to solve multivariate regression models with matrices?

A

only if X(Ubar)TX(Ubar) is invertible

This occurs when:

  1. The number of colums is < than the number of rows (K < N) where N is the number of observation and K is number of parameters
  2. Columns of X(Ubar) are linearly independent (there cant be perfect correlation between any of the X variables)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

What does R-squared measure?

A

the fraction of the variance of the endogenous variable which is explained by the model

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

How can the regression model be wrote in matrix form?

A

Note that we can also write the regression model itself in matrix form.

y(Ubar)=X(Ubar)β + u

where β is a k x 1 vector of unknown population parameters and u is an N x 1 vector of unoberserable random errors

We can substitute for y in the expression for the estimator to obtain:

β(hat)= (X(Ubar)TX(Ubar))-1X(Ubar)T(X(Ubar)β + u)
= β+(X(Ubar)TX(Ubar))-1X(Ubar)Tu

The OLS estimator is therefore a linear combination of the random errors and is itself therefore a random variable whose distribution will depend on the properties of the errors

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

What are the Guass- Markov assumption in matrix form?

A
  1. E(ui)=0; i=1,…,N –> expect value of the error term is equal to zero for every value of i
  2. E(uiuj)=0; i≠j –> the expected value of uiuj=0, the errors are independent
  3. E(ui2)= σu2; i= 1,…,N –> the expected value of the squared error term is a constant σu2 –> the variance of the error
  4. E(X’u)=X’E(u)=0 –> if X is not random and is fixed, we can take it out of the expectation operation

and finally:

  1. the errors follow a normal distribution

These assumptions are essentially the same as for the bivariate regression model

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

What does assumption 2 and 3 tell you about the variance-covariance matrix of the errors?

A

IN –> identity matrix

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

What happens to a multivariate regression model when the X variables are correlated?

A

Yi= β12Xi23Xi3+ui

ρXi2=Xi3

This means we could write:

Yi= β12Xi23ρXi2+ui

Yi= β1+(β2 +ρβ3)Xi2+ui

The equation only contains one genuine independent variable and we therefore cannot estimate separate effects from the two variables

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

What happens to a multivariate regression model when we have only partial collinearity?

A

The previous example was one in which there was perfect collinearity between two variables.

In cases like this the least-squares method will not allow us to estimate separate coefficients for the two right-hand side variables.

More generally we might have less than perfect collinearity:

ρXi2i=Xi3

where V(εi) ≠0 but is small relative to V(Xi2)

  • a software package here will treat them both as two completely two different variables

If this is the case then it is possible to estimate separate effects using least-squares but the estimates may be very inaccurate.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly