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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?

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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: $y = \sinh(\theta y')/\theta$.

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  • $\begingroup$ 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 Aug 7 '17 at 10:08

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