Kernel density estimation for heavy-tailed distributions using the champernowne transformation I am trying to follow this paper to estimate the density for a heavy-tailed distributions using the champernowne transformation.
Alternative link to the paper
Another alternative link to the paper
However, I do not understand the final step to transform the kernel density estimate of the transformed data back to the untransformed data set.
An outline of the procedure is below:
Firstly, the data, X, is transformed:

Where T() is a modified Champernowne CDF. The parameter alpha, M and c have already been estimated.
Then a Kernel Density Estimate, with a Gaussian kernel is done on the transformed data. However, the data must lie in the interval (0,1), so we only take the that part of the estimated density and then divide by the integral of that part of the density.


The final step, which I don't understand is the formula below. What does the denominator mean?
I understand that the numerator is the estimate of the transformed data set.
I can also see the transformered data set in the denominator, T(), but what is T'?

The authors of the paper then write the following expression for the density estimator of the untransformed dataset:

 A: There is a mistake in the fourth formula, the one you are trying to
understand (that is apparent from the last formula where that mistake
disappears). Precisely, I mean it should be written $$\hat{f} \left( x \right) =
\frac{\hat{f}_{\text{trans}} \left(
T_{\hat{\alpha}, \hat{M}, \hat{c}} \left( x \right) \right)}{\left| \left(
T^{- 1}_{\hat{\alpha}, \hat{M}, \hat{c}}  \right)' \left( x
\right) \right|}$$ and not $$\hat{f} \left( x \right) =
\frac{\hat{f}_{\text{}} \left(
T_{\hat{\alpha}, \hat{M}, \hat{c}} \left( x \right) \right)}{\left| \left(
T^{- 1}_{\hat{\alpha}, \hat{M}, \hat{c}}  \right)' \left(
T_{\hat{\alpha}, \hat{M}, \hat{c}} \left( x \right) \right) \right|}$$
The notation in these formulas are clumsy and not very intuitive, but I will explain how that formula is derived and where the mistake occurs.
The relation between the two random variables $Y$ and $X$ is given by $Y = T
\left( X \right)$ (I will denote by $T$ the function $T_{\hat{\alpha},
\hat{M}, \hat{c}}$ to simplify notation). The transformation $T$ is the
cumulative distribution function of an absolutely continuous random variable
and thus is strictly montonically increasing with unique inverse $T^{- 1}$.
Let $t \left( x \right) = T' \left( x \right) = \frac{\partial T \left( x
\right)}{\partial x}$ be the density corresponding to $T$. Denote the by $f_X$
and $f_Y$ the densities of $X$ and $Y$. The relation between the two
densities is
$$ f_X \left( x \right) = f_Y \left( T \left( x \right) \right) t \left(
   \left. x \right) \right. $$
$$ f_Y \left( y \right) = f_X \left( T^{- 1} \left( y \right) \right)
   \frac{1}{t \left( T^{- 1} \left( y \right) \right)} $$
This is clear because the jacobian of the transformation $X = T^{- 1} \left( y
\right)$ is $t \left( x \right)$. (and not $t \left( T \left( x \right)
\right)$ as is assumed in the fourth formula). The term $\left|
\frac{1}{\left( T^{- 1} \right)' T \left( x \right)} \right|$ appears mistakenly in the
fourth formula because
$$ t \left( T \left( x \right) \right) = \left| t \left( T \left( x \right)
   \right) \right| = \left| T' \left( T \left( x \right) \right) \right| =
   \left| \frac{1}{\left( T^{- 1} \right)' T \left( x \right)} \right| $$
(Notice that the derivative of the inverse function is the inverse of the
derivative of the original function. Also there is no need for the absolute
values here because densities are positive. To quickly check the error notice that $T:(0,\infty)\rightarrow (0,1)$ and $t:(0,\infty)\rightarrow (0,\infty)$). In the last formula, the last
term is $t \left( x \right)$ and not the incorrect one $t \left( T \left( x \right) \right)$.
A: The general formulae for a density of a transformed variable Y in $R^d$ is:$$f_Y(y) = |det(T^{-1}{\prime}(y))|f_X(T^{-1}(y))$$
In R, the jacobian collapses to the derivative of the inverse. 
Then $f_X(x) = \frac{f_Y(y)}{|T^{-1}{\prime}(y)|}$
