If $\hat{F}(x)$ is a consistent estimator of $F(x)$ where $F(x)$ is a cdf, can we state that $\hat{F}^{-1}(x)$ is also a consistent estimator of $F^{-1}(x)$? Is that straightforward? Why? You can make any assumption you want if that is necessary. I am looking for a discussion.

  • $\begingroup$ not exactly clear here. We are estimating $F(X)$ where $X$ is some random variable (in which case it doesn't make sense $\hat F(x)$ depend on $x$?) . Or we are estimating the CDF itself (in which case it doesn't make sense to call it an estimator of $F(X).$ It is an estimator of the indicator $I(X \le x)$ for each $x$)? $\endgroup$ – spaceisdarkgreen Feb 20 '17 at 3:53
  • $\begingroup$ Why should $\hat{F}^{-1}(X)$ even make sense? The domain of $\hat{F}^{-1}$ is $(0,1)$ (maybe including the endpoints depending on convention). $\endgroup$ – Ian Feb 20 '17 at 4:04
  • $\begingroup$ I corrected it. We are estimating the cumulative distribution function itself. In order to estimate the cdf we use data that come from a continuous random variable X. For example one might consider $\hat{F}(x)$ to be the empirical estimator, but I guess there are other consistent estimators of $F(x)$ such as kernel based. I was wondering if one has consistency for the cdf then is it a direct result that the inverse estimator is also a consistent estimator of the true inverse cdf. IntuitivelyI would say it is since graphic-wise you just reverse the axes, but I would a more solid justification. $\endgroup$ – leo Feb 20 '17 at 4:05
  • $\begingroup$ A rephrasing of your question: you want to know whether for each $x \in (0,1)$, $E[\hat{F}^{-1}(x)]=F^{-1}(x)$. Note that this $x$ is not random! $\endgroup$ – Ian Feb 20 '17 at 4:08
  • $\begingroup$ Did you want to write $E(\hat{F}^{-1}(x))=F^{-1}(x)$? which implies unbiasedness? (yes I agree that $x$ is not random and it is an argument of the function). $\endgroup$ – leo Feb 20 '17 at 4:11

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