# How to interpret regression coefficients when outcome variable was transformed to $1 / \sqrt{Y}$?

In OLS regression, what does a one unit change in X represent in terms of a beta-related change in Y when the outcome variable has been transformed to $1/\sqrt(Y)$?

(Transformation was done to ensure linearity in the model. My predictor variables are 0-1 dummies.)

I've consulted the following useful posts, but I still don't know what is the answer:

https://stats.stackexchange.com/a/2148

Back-transformation of regression coefficients

How to interpret regression coefficients when response was transformed by the 4th root?

• One of those links you reference shows how to deal with the fourth root. What part of it are you having trouble adapting to the reciprocal square root?
– whuber
Oct 5 '12 at 19:12
• Thanks for your reply, whuber. This is the part where my math skills are wanting:
– Argi
Oct 5 '12 at 22:35
• Sorry, here is the quote: "These changes are the derivatives dY/dX i , which the Chain Rule tells us are equal to 4β i Y 3 . We would plug in the estimates, then, and say something like The regression estimates that a unit change in X i will be associated with a change in Y of 4b i Y ˆ 3 = 4b i (b 0 +b 1 X 1 +b 2 X 2 ) 3 ."
– Argi
Oct 5 '12 at 22:35
• I'm not sure what "the Chain Rule" would tell us for the case of the inverse square root transformation.
– Argi
Oct 5 '12 at 22:36

The model is

$$\frac{1}{\sqrt{Y}} = \beta_0 + \beta_1 X_1 + \cdots + \beta_p X_p + \varepsilon$$

where $$Y$$ is the original outcome, the $$X_i$$ are the explanatory variables, the $$\beta_i$$ are the coefficients, and $$\varepsilon$$ are iid, mean-zero error terms. Writing $$b_i$$ for the estimated value of $$\beta_i$$, we see that a one-unit change in $$X_i$$ adds $$b_i$$ to the right hand side. Starting from any baseline set of values $$(x_1, \ldots, x_p)$$, this induces a change in predicted values from $$\widehat{1/\sqrt{y}} = b_0 + b_1 x_1 + \cdots + b_p x_p$$ to $$\widehat{1/\sqrt{y'}} = b_0 + b_1 x_1 + \cdots + b_p x_p + b_i$$. Subtracting the first equation from the second gives

$$\frac{1}{\sqrt{\hat{y'}}} - \frac{1}{\sqrt{\hat{y}}} = b_i.$$

Solving for $$\hat{y'}$$ gives

$$\hat{y'} = \frac{\hat{y}}{1 + 2b_i\sqrt{\hat{y}} + b_i^2 \hat{y}}.$$

One may stop here, but often we seek simpler expressions: the behavior of this one might not be any easier to understand than the original model. Simplification can be achieved provided $$b_i$$ is very small. If necessary, we can contemplate a tiny change in $$X_i$$, say by an amount $$\delta$$, which would replace $$b_i$$ in the preceding equation by $$\delta b_i$$. Using a sufficiently small value of $$\delta$$ will assure the denominator is close to $$1$$. When it is,

$$\frac{\hat{y}}{1 + 2\delta b_i\sqrt{\hat{y}} + \delta^2 b_i^2 \hat{y}} \approx \hat{y}(1 - 2\delta b_i\sqrt{\hat{y}} - \delta^2 b_i^2 \hat{y}),$$

whence the change in predicted values is

$$\hat{y'} - \hat{y} \approx -\delta (2b_i\sqrt{\hat{y}} + \delta b_i^2 \hat{y}).$$

Taking $$\delta$$ to be so small that $$\delta b_i^2 \hat{y} \ll 2 b_i\sqrt{\hat{y}}$$ allows us to drop the second term in the right hand side. That is, for very tiny changes, the predicted outcome changes by $$-(2b_i\sqrt{\hat{y}})$$ times the amount of change in $$x_i$$.

The appearance of the negative sign indicates that an increase in $$X_i$$ will decrease $$Y$$ when $$b_i$$ is positive and increase $$Y$$ when $$b_i$$ is negative. Normally, we avoid this (potentially confusing) sign reversal by using $$-1/\sqrt{Y}$$ instead of $$1/\sqrt{Y}$$ when making a reciprocal square root transformation (or any other transformation that reverses the order of numbers).

This solution method is always applicable no matter how $$Y$$ is re-expressed, but it can lead to complicated algebra for other transformations of $$Y$$. Those who know the basics of differential calculus will recognize that all we're doing here is approximating the change in $$\hat{y}$$ to first order using its derivative with respect to $$x_i$$, so they will be able to avoid most of the algebraic manipulations.