-2
$\begingroup$

Suppose I can observe $x_1,...,x_n$ as the realization of the random variables $X_1,..,X_n$. Using $x_1,...,x_n$, I can estimate the empirical cumulative distribution function (CDF), $F_n(x)=\sum_{i=1}^n\frac{I(x_i\leq x)}{n}$. Now, with a given $\lambda$, I can transform this CDF by using the Wang transform which is $F^*(x)=\Phi\big[\Phi^{-1}(F_n(x))-\lambda\big]$, where $\Phi(.)$ is the cdf of standard normal distribution.

Question: How can I estimate the $f^*(x)$ (i.e. the probability density function under the new transformation) using $F^*(x)$? Is there any package in R to do that?

$\endgroup$

closed as not a real question by whuber Mar 10 '13 at 4:00

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Well the obvious (perhaps naive) estimate would be the $\hat{F^*}$ obtained by applying $F^*$ to $\hat{F}$; obviously that doesn't work for places where $\hat{F}$ is 0 or 1, though the usual methods of dealing with percentiles/percentile ranks in QQ plots and such could be applied (the $(i-\alpha)/(n+1-2\alpha)$ kinds of things in place of $\hat{F}$). $\endgroup$ – Glen_b Mar 10 '13 at 2:41
  • $\begingroup$ I really don't get what you mean. $\endgroup$ – Stat Mar 10 '13 at 2:56
  • $\begingroup$ I think part of the issue is you're talking about ecdfs then saying you want to estimate pdfs. I was confused by that and focused on estimating $F^*$, rather than $f^*$ there. Now ecdfs are discrete estimates of $F$, pdfs aren't either of those things. My comment is not an adequate answer and should probably be ignored. $\endgroup$ – Glen_b Mar 10 '13 at 3:01
1
$\begingroup$

Since $F_n$ is discrete, you cannot talk about a density. $F^*$ is the cdf of a discrete distribution.

I would suggest using a continuous distribution estimator, such a kernel estimators, instead of the EDCF. Using this, you can simply differentiate the expression of interest. You will get an expression that will require the estimation of a PDF and a CDF. Both can be done using kernel estimators. They can be implemented in R using the package kerdiest and the default packages.

$\endgroup$
  • $\begingroup$ No buddy, that's not true. $F^{*}$ is a distribution function of a new random variable that has a density. I need to estimate this density. $\endgroup$ – Stat Mar 10 '13 at 2:51
  • $\begingroup$ Nope, it depends on a discrete estimator: the ECDF (hint: its jumps at each observation). See my recommendation. $\endgroup$ – Filth Mar 10 '13 at 2:51
  • $\begingroup$ I am not arguing with you. Good luck. $\endgroup$ – Filth Mar 10 '13 at 3:16
0
$\begingroup$

I originally came to post something like this answer but got distracted by the details of the way the question was asked. Apologies.

If you take a kernel estimate of $f$, then the density of a transformation ($W=T(X)$) of $X$, $f_W(w)$ can be estimated from the Jacobian and $f(T^{-1}(x))$. That is, you can transform a kernel density estimate just as you would any other density. [I use such a trick when dealing with nearly lognormal random variables (transforming data by taking logs, doing a standard KDE and then transforming the estimate back via the above trick) - it often - but not always - results in density estimates that behave nicely, and the need for wider bandwidths at the right than at the left is automatically handled.]

e.g. see p3 here (the reference to theorem 11 of p200 of MGB is presumably a reference to 'Mood, Graybill and Boes, Introduction To The Theory Of Statistics')

$\endgroup$

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