I have a regression equation that uses covariates of the following form

asinh(y) = b0 + b1 asinh(x) + b2 (asinh(x))^2 + error

and am wondering how to intepret the coefficients b1 and b2. Does anybody have an idea?

Is it still right to conjecture about the form of the relationship (concave, convex) by means of the sign on the linear and the quadratic coefficient?

For logarithmic transformation the turning point may be found at e^(-a/b) where a and b are the coefficients on the linear and the quadratic term. Is there a similar shortcut for the case of asinh?

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    $\begingroup$ I hate to be pessimistic, but unless there was a scientific rationale for including those terms in the first place (which would immediately lend itself to their interpretation), I don't think you will be able to ad hoc any interpretation to the coefficient values. $\endgroup$ – AdamO Jun 11 '18 at 16:13
  • $\begingroup$ The turning point of predictions in terms of say $X_1 := \text{asinh}\ x$ and $X_2 := X_1^2$ can be got in the usual way by elementary calculus. Then read off the corresponding value of $x$ given that $\text{asinh}\ x$ is monotonic in $x$. $\endgroup$ – Nick Cox Jun 11 '18 at 16:23
  • $\begingroup$ Well, then thank you even more for your reply. The rationale is to use the asinh-transformation instead of log due to its desirable property of keeping the zeros. An I am running a quadratic regression because I suspected a U-shaped functional form of the logarithmized (or asinh-tized) relationship between y and x. $\endgroup$ – Papayapap Jun 11 '18 at 16:26
  • $\begingroup$ Thank you for the explanation. Just to be sure I am getting this right, so I calculate with the ususal quadratic formula the turning point and then I transform this turning point back (using the sinh function)? $\endgroup$ – Papayapap Jun 11 '18 at 16:36
  • $\begingroup$ Yes, that's right. Also, you should see it on a graph of $b_1 X_1 + b_2 X_2$ as a function of $X_1$. . $\endgroup$ – Nick Cox Jun 11 '18 at 17:03

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