After all we calculate the VIF by $1/(1-R_j^2)$. A VIF of $5$ corresponds to an $R_J^2$ of $0.8$. To me, the information given by $R_j^2$ just becomes more obscure when I apply the VIF formula. Why can't I just use $R_j^2$ to detect multicollinearity?

  • 3
    $\begingroup$ You certainly can, there is one-to-one correspondence between VIF values and $R^2$ values. $\endgroup$
    – JohnK
    Apr 18, 2016 at 12:50
  • $\begingroup$ @JohnK But why do we bother to calculate Vif Then in the First place? $\endgroup$
    – user105833
    Apr 18, 2016 at 12:56
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    $\begingroup$ (+1) The VIF has an interpretation: it is closely related to the condition number of the Normal Equations. $\endgroup$
    – whuber
    Apr 18, 2016 at 13:49

2 Answers 2


You make a good point. I'd like to point out that one thing we like to use VIF for is its relationship to the standard error of the beta coefficient estimates. We can say that, the standard error is a function of MSE (the total variability around the model), $s^2\left\{X_k\right\}$ (the variability of the kth variable), and the VIF for the kth variable. It would be weird to say, a function of the inverse of 1 minus the coefficient of partial determination. i.e.

$$s^2\left\{b_k\right\} = \frac{MSE}{(n-1)s^2\left\{X_k\right\}}(VIF_k)$$


When I learned it, I was told the the larger numbers made it easier to identify to the naked eye. My instructor also used 10 as the cut off and not 5. So if you had many VIF calculations in a matrix of some sort, you would round to the digit and then numbers with 2 digits = multicolinearity.

Also I think the VIF intuition is that we are changing the values to grow exponentially rather than linearly: 0.8, 0.85, 0.90, 0.95 vs 5, 6.66, 10, 20 again making it easier to identify.

That being said, since there is a 1-1 mapping, you can use $R^2$ and nothing changes


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