# Interpreting regression coefficients when the outcome variable is an inverse hyperbolic sine function

I just learned that when there are zero or negative values, a good alternative to using a logged function is the inverse hyperbolic sine function.

When using log transformation on the dependent variable, as in log(y)=βx+ε, the interpretation of β is 1 unit change in x = (100β)% change in y.

Likewise, what would the interpretation of β be if the regression is asinh(y)=βx+ε?

• Is x continuous or a dummy? – Dimitriy V. Masterov Mar 8 at 0:48
• @DimitriyV.Masterov, yes it is a continuous variable. – Samyam Shrestha Mar 8 at 0:51
• On why asinh() is better notation than arcsinh() please see comments under @DimitriyV.Masterov's helpful answer. – Nick Cox Mar 8 at 10:22

The interpretation is not as easy as with the log-linear semi-elasticity case, but there is a nice working paper by Marc Bellemare and Casey Wichman on Elasticities and the Inverse Hyperbolic Sine Transformation that covers some of this in Section 2.2.

In a model like

$$\DeclareMathOperator{\asinh}{asinh} \asinh ( y)= \alpha + \beta \cdot x+ \varepsilon,$$

you can recover $$y$$ as

$$\DeclareMathOperator{\sinh}{sinh} y= \sinh ( \alpha +\beta \cdot x+ \varepsilon).$$

Then taking the derivative with respect to $$x$$, $$\frac{\partial y}{\partial x} = \beta\cdot \cosh(\alpha +\beta \cdot x+ \varepsilon) = \beta\cdot \cosh(\asinh(y))= \beta \cdot \sqrt{y^2+1}.$$

Multiplying that by $$\frac{x}{y}$$, we get the standard elasticity of $$y$$ with respect to $$x$$

$$\epsilon=\frac{\partial y}{\partial x}\cdot\frac{x}{y}=\beta \cdot x \cdot \frac{\sqrt{y^2+1}}{y}=\beta \cdot x \cdot \sqrt{\frac{{y^2+1}}{y^2}}=\beta \cdot x \cdot \sqrt{1+\frac{1}{y^2}}.$$

This looks messy, but when $$y$$ gets large in absolute value, $$\sqrt{1+\frac{1}{y^2}} \rightarrow 1,$$ so you can ignore the last term and just focus on $$\epsilon \approx \beta \cdot x.$$ You can plug the average $$x$$ into that or evaluate it as function of $$x$$ for some interesting values of $$x$$. Since this is just a linear transformation of $$\beta$$, calculating the SEs is pretty easy.

You can also calculate the average elasticity:

$$\epsilon=\sum_i^N \hat \beta \cdot x_i \cdot \frac{\sqrt{\hat y_i^2+1}}{\hat y_i},$$

or evaluate it at interesting values of $$x$$ and other covariates. This has the interpretation of a percent change in expected $$y$$ for a 1% change in $$x$$. Since this is a non-linear function, SEs are more challenging.

If you want to retain the semi-elasticity interpretation, the

$$\epsilon^*=\frac{\partial y}{\partial x}\cdot\frac{1}{y}=\beta \cdot \sqrt{1+\frac{1}{y^2}}.$$

You may want to multiply that by 100 to get the expected % change in $$y$$ for a 1 unit increase in $$x$$.

• The notation arcsinh() is quite common, but unfortunate: use of the word "arc" is based on an improper analogy with terminology for inverse trigonometric functions, such as arcsine. In the latter case, "arc" means the arc of a circle, or equivalently angle, as is proper for circular functions. But the inverse of a hyperbolic function is geometrically interpreted not as an arc but as an area. This point is argued in detail by Saler, B. M. 1985. Inverse hyperbolic functions as areas. College Mathematics Journal 16: 129-131. – Nick Cox Mar 8 at 10:20
• Various minor misunderstandings in this territory may arise partly from the linguistic coincidence that arc, area, and argument have the same initial characters. The notation arsinh() is possible, but infelictious on other grounds. Hence I recommend asinh() -- which often matches names in software -- and have edited accordingly. – Nick Cox Mar 8 at 10:20
• That should be "infelicitous". – Nick Cox Mar 8 at 10:35
• Hi Dimitriy, thanks for the detailed answer. I am still trying to grasp everything you wrote with no prior knowledge of asinh() function. Anyway, it seems like beta is totally dependent on each x and their corresponding y variable, right? This makes the interpretation pretty complication compared to a simple log-level or log-log model. – Samyam Shrestha Mar 9 at 21:30
• @SamyamShrestha The elasticities are only a function of x (since expected y is itself a function of x). The approaches I suggested do return a single number, so you get a kind of typical elasticity. – Dimitriy V. Masterov Mar 9 at 21:34