Re: independent variables - ln

Reef Fish wrote:
J W wrote:
rastompk@xxxxxxxxx wrote:
Hi there,
If x1,...,xn are random independent variables,
it is true that ln(x1),...,ln(xn) are also independent?

rasto

I thought that QUESTION wss so infantile that was not worth
even a one line comment, such as:

What is the meaning of your "independent" in
"random indepdent variables"?
Yes; Independence can be thought of as an *inherent property* of the
random variables in question, and hence unchanged by any mathematical
transformation applied to each individually (combinations like sums, of
course, could be dependent depending on how they're constructed).
Intuitively, you can think of your random variables X1, ..., Xn as being
generated by n "sources" which are completely isolated from each other.
If you now put a filter on each source which applies a ln()
transformation to the random variable it generates, then you can
consider this "source+filter" combination as a new set of n "sources".
Since these new sources are still isolated from each other, your ln(X1),
..., ln(Xn) are still independent.

-J

I think you are at best killing a dead duck is his "independent"
meant the random variables are stochastically independent.

But if his "independent variable" meant the "independent
variables used in a REGRESSION, the observed x's" that
was randomly chosen, then the answer is NO because
the "independent variables" in a linear regression" means
LINEARLY independent but NOT stochastically independent.

Although in this case, of course, they would still be "independent variables" as long as one was using the same meaning of independent for both the x's and the ln(x)'s.

A subject had been covered, and concurrently covered in
several threads.

You reply is, otherwise, WRONG, and missed the boat and
even the dead duck.

I really wish you would explain why posts were wrong, rather than just being abusive: if JW's reply is wrong, then it would help the OP more if you explained why it's wrong. Then both of them would learn something.

Bob
.