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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:

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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.

enter image description here

enter image description here

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'?

enter image description here

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

enter image description here

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@Learner, done. –  power Dec 23 '12 at 14:15
    
I have started a bounty. If the question gets more up-votes, I will increase the bounty because I will have more reputation to give. Each up-vote gives me 5 points and the next bounty increment is an increase by 50 points. Therefore, if I get ~10 up votes, I will increase the bounty to 100 points. With even more up-votes, I will make a greater bounty. –  power Dec 26 '12 at 2:06

2 Answers 2

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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 $t \left( T \left( x \right) \right)$.

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I deem this as the best answer so far. Thanks. I am still trying to understand the "relation between two densities" reference. I will try again tomorrow with a fresh head. I will also try to find it on Wolfram Mathworld - I have previously come across misinformation on with Wikipedia. –  power Dec 27 '12 at 17:00
1  
@power I think it is a good investment to study transformation of random variables in some good statistics textbook. Chapter 2 in Casella and Berger's "statistical inference" is a good candidate. –  Learner Dec 28 '12 at 0:47
    
okay I'll try that one. Thank you. –  power Dec 28 '12 at 7:12
    
thanks. I have been using trail and error for the Latex. –  power Dec 30 '12 at 11:18
1  
Yes, and you can see it looks much simpler! –  Learner Dec 30 '12 at 11:20

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 $f_X(x) = \frac{f_Y(y)}{|T^{-1}{\prime}(y)|}$

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@Learner, I actually carefully read your answer and it seems you claim the fourth formulae is wrong. I was not sure to understand that claim, especially if you rewrite $det(T^{-1}'(y))$ as $T^{-1}'(y)$ in R. For the formatting, I have tried to edit it repeatedly; I have no idea why the latex is not compiled. –  saradi Dec 27 '12 at 11:20
1  
Yes it is wrong and you are writing the correct answer. $T^{-1}'(y)$ is the actually the derivate of the inverse function. This will be equal to one over the derivative of $T$, which I denote by $t$. If you look at the fourth formula, there is an additional function composition $t(T(x))$ and not just $t(x)$ as you and I wrote. (One quick sign that adding this additional composition is wrong is that it would truncate the support from $(0,\infty)$ to $(0,1)$, $T$ being a CDF. –  Learner Dec 27 '12 at 11:30
    
Has saradi's answer been voted down? Why? –  power Dec 27 '12 at 16:53

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