How to transform leptokurtic distribution to normality?

Question

Suppose I have a leptokurtic variable that I would like to transform to normality. What transformations can accomplish this task? I am well aware that transforming data may not always be desirable, but as an academic pursuit, suppose I want to "hammer" the data into normality. Additionally, as you can tell from the plot, all values are strictly positive.

I have tried a variety of transformations (pretty much anything I have seen used before, including $\frac 1 X,\sqrt X,\text{asinh}(X)$, etc.), but none of them work particularly well. Are there well-known transformations for making leptokurtic distributions more normal?

See the example Normal Q-Q plot below:

Answer 1

I use heavy tail Lambert W x F distributions to describe and transform leptokurtic data. See (my) following posts for more details and references:

• Transformation to increase kurtosis and skewness of normal r.v: this shows some illustrations of how the densities vary when varying $\delta$.
• What's the distribution of these data?: an application example of how to use this to estimate model parameters and Gaussianize your data.

Here is a reproducible example using the LambertW R package.

library(LambertW)
set.seed(1)
theta.tmp <- list(beta = c(2000, 400), delta = 0.2)
yy <- rLambertW(n = 100, distname = "normal", 
                theta = theta.tmp)

test_norm(yy)

## $seed
## [1] 267509
## 
## $shapiro.wilk
## 
##  Shapiro-Wilk normality test
## 
## data:  data.test
## W = 1, p-value = 0.008
## 
## 
## $shapiro.francia
## 
## 	Shapiro-Francia normality test
## 
## data:  data.test
## W = 1, p-value = 0.003
## 
## 
## $anderson.darling
## 
##  Anderson-Darling normality test
## 
## data:  data
## A = 1, p-value = 0.01

The qqplot of yy is very close to your qqplot in the original post and the data is indeed slightly leptokurtic with a kurtosis of 5. Hence your data can be well described by a Lambert W $\times$ Gaussian distribution with input $X \sim N (2000, 400)$ and a tail parameter of $\delta = 0.2$ (which implies that only moments up to order $\leq 5$ exist).

Now back to your question: how to make this leptokurtic data normal again? Well, we can estimate the parameters of the distribution using MLE (or for methods of moments use IGMM()),

mod.Lh <- MLE_LambertW(yy, distname = "normal", type = "h")
summary(mod.Lh)

## Call: MLE_LambertW(y = yy, distname = "normal", type = "h")
## Estimation method: MLE
## Input distribution: normal
## 
##  Parameter estimates:
##        Estimate  Std. Error  t value Pr(>|t|)    
## mu     2.05e+03    4.03e+01    50.88   <2e-16 ***
## sigma  3.64e+02    4.36e+01     8.37   <2e-16 ***
## delta  1.64e-01    7.84e-02     2.09    0.037 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## -------------------------------------------------------------- 
## 
## Given these input parameter estimates the moments of the output random variable are 
##   (assuming Gaussian input): 
##  mu_y = 2052; sigma_y = 491; skewness = 0; kurtosis = 13.

and then use the bijective inverse transformation (based on W_delta()) to backtransform the data to the input $X$, which -- by design -- should be very close to a normal.

# get_input() handles does the right transformations automatically based on
# estimates in mod.Lh
xx <- get_input(mod.Lh)
test_norm(xx)

## $seed
## [1] 218646
## 
## $shapiro.wilk
## 
##  Shapiro-Wilk normality test
## 
## data:  data.test
## W = 1, p-value = 1
## 
## 
## $shapiro.francia
## 
## 	Shapiro-Francia normality test
## 
## data:  data.test
## W = 1, p-value = 1
## 
## 
## $anderson.darling
## 
##  Anderson-Darling normality test
## 
## data:  data
## A = 0.1, p-value = 1

Voila!

Answer 2

Credit for this answer goes to @NickCox's suggestion in the comments section of the original question. He suggested I subtract the median of the data and apply the transformation to the deviations. For instance, $\text{sign(.)}\cdot\text{abs(.)}^{\frac 1 3}$, with $Y-\text{median}(Y)$ as the argument.

Although the cube root transformation didn't work out well, it turns out the square root and the more obscure three-quarters root work well.

Here was the original kernel density plot corresponding to the Q-Q plot of the leptokurtic variable in the original question:

---

After applying the square root transformation to the deviations, the Q-Q plot looks like this:

Better, but it can be closer.

---

Hammering some more, applying the three-quarters root transformation to the deviations gives:

---

And the final kernel density of this transformed variable looks like this:

Looks close to me.

Answer 3

In many cases, there may simply be no simple-form monotonic transformation that will produce a close-to-normal result.

For example, imagine that we have a distribution which is a finite mixture of lognormal distributions of various parameters. A log transform would transform any of the components of the mixture to normality, but the mixture of normals in the transformed data leaves you with something that's not normal.

Or there may be relatively nice transform, but not of one of the forms you'd think to try -- if you don't know the distribution of the data, you may not find it. For example, if the data were gamma-distributed, you won't even find the exact transform to normality (which certainly exists) unless I tell you exactly what the distribution is (though you might stumble upon the cube-root transformation that in this case would make it pretty close to normal as long as the shape parameter isn't too small).

There are myriad ways in which the data can look reasonably amenable to being transformed but which doesn't look great on any of a list of obvious transformations.

If you can give us access to the data, it may well be that we can either spot a transformation that does okay -- or that we can show you why you won't find one.

Just from the visual impression there, it looks rather like a mixture of two normals with different scales. There's only a slight hint of asymmetry, which you could easily observe by chance. Here's an example of a sample from a mixture of two normals with common mean - as you see it looks quite a bit like your plot (but other samples may look heavier or lighter tailed - at this sample size there's a lot of variation in the order statistics outside 1 sd either side of the mean).

In fact here are yours and mine superimposed:

$\quad\quad\quad $