I'm using fourth root (1/4) power transformation on my response variable, as a result of heteroscedasticity. But now I'm not sure how to interpret my regression coefficients.

I assume that I would need to take the coefficients to the fourth power when I back-transform (see below regression output). All of the variables are in units of dollar in millions, but I would like to know the change in dollar in billions.

While holding the other independent variable constant, a billion dollar change in fees, on average, leads to a change of 32 (or 32,000 dollars) in collections. I take 0.000075223 * 1000 (to get to billions) ^ 4 = 0.000032. Now do I multiply this number by 1 million or 1 billion (the original unit of the dependent variable is in millions)?

lm(formula = (Collections^(1/4)) ~ Fees + DIR)

                 Estimate      Std. Error  t value            Pr(>|t|)
(Intercept)   2.094573355     0.112292375   18.653  0.0000000000000151
Fees        **0.000075223   **0.000008411    8.943  0.0000000131878713
DIR           0.000022279     0.000004107    5.425  0.0000221138881913

The best solution is, at the outset, to choose a re-expression that has a meaning in the field of study.

(For instance, when regressing body weights against independent factors, it's likely that either a cube root ($1/3$ power) or square root ($1/2$ power) will be indicated. Noting that weight is a good proxy for volume, the cube root is a length representing a characteristic linear size. This endows it with an intuitive, potentially interpretable meaning. Although the square root itself has no such clear interpretation, it is close to the $2/3$ power, which has dimensions of surface area: it might correspond to total skin area.)

The fourth power is sufficiently close to the logarithm that you ought to consider using the log instead, whose meanings are well understood. But sometimes we really do find that a cube root or square root or some such fractional power does a great job and it has no obvious interpretation. Then, we must do a little arithmetic.

The regression model shown in the question involves a dependent variable $Y$ ("Collections") and two independent variables $X_1$ ("Fees") and $X_2$ ("DIR"). It posits that

$$Y^{1/4} = \beta_0 + \beta_1 X_1 + \beta_2 X_2 +\varepsilon.$$

The code estimates $\beta_0$ as $b_0=2.094573355$, $\beta_1$ as $b_1=0.000075223$, and $\beta_2$ as $b_2=0.000022279$. It also presumes $\varepsilon$ are iid normal with zero mean and it estimates their common variance (not shown). With these estimates, the fitted value of $Y^{1/4}$ is

$$\widehat{Y^{1/4}} = b_0 + b_1 X_1 + b_2 X_2.$$

"Interpreting" regression coefficients normally means determining what change in the dependent variable is suggested by a given change in each independent variable. These changes are the derivatives $dY/dX_i$, which the Chain Rule tells us are equal to $4\beta_iY^3$. 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 $4b_i\widehat{Y}^3$ = $4b_i\left(b_0+b_1X_1+b_2X_2\right)^3$.

The dependence of the interpretation on $X_1$ and $X_2$ is not simply expressed in words, unlike the situations with no transformation of $Y$ (one unit change in $X_i$ is associated with a change of $b_i$ in $Y$) or with the logarithm (one percent change in $X_i$ is associated with $b_i$ percent change in $Y$). However, by keeping the first form of the interpretation, and computing $4b_1$ = $4\times 0.000075223$ = $0.000301$, we might state something like

A unit change in fees is associated with a change in collections of $0.000301$ times the cube of the current collections; for instance, if the current collections are $10$, then a unit increase in fees is associated with an increase of $0.301$ in collections and if the current collections are $20$, then the same unit increase in fees is associated with an increase of $2.41$ in collections.

When taking roots other than the fourth--say, when using $Y^p$ as the response rather than $Y$ itself, with $p$ nonzero--simply replace all appearances of "$4$" in this analysis by "$1/p$".


An alternative to transformation here is to use a generalised linear model with link function power and power 1/4. What error family to use is open, which gives you more flexibility than you have with linear regression and an assumption of conditional normality. One major advantage of this procedure is that predictions are automatically produced on the original measurement scale, so there is no question of back-transforming.


I have seen papers using quartic root regression coefficients in thinking about percentage changes, while avoiding taking logs (and dropping observations).

If we're interested in using quartic roots to calculate percentage changes, we know that:

$\hat{Y} = (\alpha + \hat{\beta}_1 X_1 + \hat{\beta}_2 X_2)^4 \implies \frac{d\hat{Y}}{dX_1} = 4\hat{\beta}_1(\alpha+\hat{\beta}_1 X_1 + \hat{\beta}_2 X_2)^3$

For the equivalent of a log-level regression, in which we're interested in the percentage change in $Y$ resulting from a unit change in $X$, we have to know the levels of all the $X$ variables:

$ \frac{d\hat{Y}/dX_1}{Y} = \frac{4\hat{\beta}_1}{\alpha + \hat{\beta}_1 X_1 + \hat{\beta}_2 X_2} $

For the equivalent of a log-log regression, in which we're interested in the percentage in $Y$ resulting from a percentage change in $X$, we'd have:

$ \frac{d\hat{Y}}{dX_1}\frac{X_1}{\hat{Y}} = \frac{4\hat{\beta}_1 X_1}{\alpha + \hat{\beta}_1 X_1 + \hat{\beta}_2 X_2} $

It doesn't seem especially convenient (I prefer the log transformation), but it can be done, either evaluating the $X$ values at the sample means or at hypothetical values.

I suppose, actually, you could replace the denominator with the sample average value of $Y^{1/4}$, and that would be a bit more convenient.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.