Put a number on the variance inflation

Question

You regress returns on two signals whose correlation with each other is $0.90$ .

Compute the variance inflation factor, the resulting standard-error inflation, and find the correlation at which VIF crosses the usual alarm threshold of $10$ .

Accepted Answer

For regressor j, the sampling variance carries a factor \frac{1}{1 - Rj^2}, the variance inflation factor (VIF), where Rj^2 is the R^2 from regressing xj on the other regressors. With only two regressors, Rj^2 = 	ext{corr}^2. VIF. Here Rj^2 = 0.90^2 = 0.81, so 	ext{VIF} = \frac{1}{1 - 0.81} = \frac{1}{0.19} \approx 5.26. Standard-error inflation. Standard errors scale with \sqrt{	ext{VIF}}, so they are inflated by \sqrt{5.26} \approx 2.29	imes versus orthogonal regressors. Alarm threshold. 	ext{VIF} 10 \iff 1 - Rj^2 < 0.1 \iff Rj^2 0.9 \iff 	ext{corr} \sqrt{0.9} \approx 0.95. So the rul…

Put a number on the variance inflation

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