Re: independent variables - ln


In other words, if the OP asks the question you took it to be,
it would be a no-brainer because the statistical independence
of random variables is invariant under all non-degenerate
transformations, and ln(X) is just any one of infinitely many
transformations that will leave the resultant transformed
variables statistically independent.

On the other hand, if the "independent variables" are only
linearly independent, as is to be understood in the independent
variables in a regression, then ln(X) transformation MAY leave
the variables inappropriate for a regression on the transformed
variables.


I don't think that the independence of non-degenerate transformations of independent random variables is any more or less of a "no-brainer" than the non-"linear independence" of transformations of independent RVs. There is certainly a reason *why* the first result holds (beyond "it's obvious!"), and I was attempting to give some intuition behind it.

-J
.