multicollinearity, howto

[terance]

Could you explaine me, how can I test my regression on multicollinearity in eviews?

[startz]

I can't imagine why you would want to do this, but look under View/Coefficient diagnostics/Variance inflation factors

[irulishka]

Sorry? but I do`n undirstand haw to calculated the Variance Inflation Factors?

I have the equation of regression: y=c0+c1x1+c2x2+c3x3

In help says: The uncentered VIF is the ratio of the variance of the coefficient estimate from the original equation divided by the variance from a coefficient estimate from an equation with only one regressor (and no constant) =>

for x1 I must to take the equation y=x1 and the variance of the coefficient estimate from this equation.
and divided variance of the coefficient estimate (x1) on variance of the coefficient estimate from auxiliary equation, but i have different result. :( :( :(

Please, help me...

[EViews Gareth]

The manual isn't quite right in the description, since the calculations hold sigmasq constant between the two regressions. The uncentered VIF is actually the variance of the coefficient from the original regression, divided by sigmasq (from original regression) / inner_product(Xi).

As an example, you can run the following program that calculates it manually:

Code: Select all

create u 100 series y=nrnd series x1=nrnd series x2=nrnd equation eq1.ls() y x1 x2 c freeze(vif) eq1.varinf scalar vif1 = eq1.@coefcov(1,1) / (eq1.@se^2 / @inner(x1)) scalar vif2 = eq1.@coefcov(2,2) / (eq1.@se^2 / @inner(x2))

[irulishka]

Thanks :D , I'll try this option

[irulishka]

Hello, i'm realy soory, but second test on multicollinearity also fails(((

in help written, V is a matrix whose columns are equal to the corresponding eigenvectors and mu(j) is the j-th eigenvalue, and v(ij) is the (i,j)-th element of V, but I calculat eigenvalue and eigen vector in Eviews 7 (eigenvalue decomposition) and get other numbers!!!

i have matrix
6
69 983
26 278 152
31 359 130 199

in Coefficient Variance Decomposition
Eigenvalues 275.1193 2.574688 1.316740 0.088263

Eigenvectors
Variable 1 2 3 4
C -0.995697 -0.001729 -0.065226 0.065798
X1 0.030188 -0.098771 0.441103 0.891494
X2 0.065934 -0.644378 -0.709543 0.277450
X3 0.057690 0.758300 -0.545640 0.352038

and in
eigenvalue decomposition
Eigenvalues:
1 1216.375
2 81.53567
3 41.69875
4 0.390235

Eigenvectors (loadings):
Variable PC 1 PC 2 PC 3 PC 4
C1 0.065798 0.065226 -0.001729 0.995697
C2 0.891494 -0.441103 -0.098771 -0.030188
C3 0.277450 0.709543 -0.644378 -0.065934
C4 0.352038 0.545640 0.758300 -0.057690

please explain to me the paradox ...

[EViews Gareth]

I can't follow what you've done there, but for an equation, you can match the eigenvalues reported in the coefficient variance decomposition, by using this command:

Code: Select all

=@eigenvalues(eq1.@coefcov)

where eq1 is the name of the equation.


If you're trying to decipher the technical details behind these type of diagnostic statistics, I highly recommend getting the Belsley, Kuh and Welsch (2004) book. It is a great book.

[coolsaad]

can u tell whether we should look ''centered vif'' or ''uncentered vif' nd y'???plz help me... :( :(