1
$\begingroup$

Suppose I have a highly skewed real-valued variable that takes on negative, zero, and positive values. One of the seemingly efficient ways to transform such a variable so that it's distribution becomes more normal-like and thus more suitable for further analyses is to use the Inverse Hyperbolic Sine (IHS) transformation (discussed in this Stackexchange thread, and originally described in this paper).

In its general form, IHS has a parameter $\theta$:

$f(y, \theta) = \sinh^{-1}(\theta y)/\theta = \log[\theta y + (\theta^2 y^2 + 1)^{1/2}]/\theta$

Let's assume that I've been able to estimate $\theta$ from my data and apply the transformation as specified in the above formula. As my trigonometry is extremely rusty, I'd appreciate any help with the following question: how do I back-transform to the original scale after applying such an IHS transformation?

| cite | improve this question | |
$\endgroup$
1
$\begingroup$

As it turned out after some additional digging (see this Stackexchange question), the answer is very simple. If $y' = \sinh^{-1}(\theta y)/\theta$ is the transformed variable, then the reverse transformation would be as follows: $y = \sinh(\theta y')/\theta$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, that's correct - the inverse of "inverse hyperbolic sine" is indeed "hyperbolic sine". But it's probably easier to think of it flipped about -- as "the inverse of hyperbolic sine is inverse hyperbolic sine". $\endgroup$ – Glen_b -Reinstate Monica Aug 7 '17 at 10:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.