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:


Back-transformation of regression coefficients

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

  • 2
    $\begingroup$ 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? $\endgroup$
    – whuber
    Oct 5, 2012 at 19:12
  • $\begingroup$ Thanks for your reply, whuber. This is the part where my math skills are wanting: $\endgroup$
    – Argi
    Oct 5, 2012 at 22:35
  • $\begingroup$ 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 ." $\endgroup$
    – Argi
    Oct 5, 2012 at 22:35
  • $\begingroup$ I'm not sure what "the Chain Rule" would tell us for the case of the inverse square root transformation. $\endgroup$
    – Argi
    Oct 5, 2012 at 22:36

1 Answer 1


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.


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