1
$\begingroup$

Suppose we have a sample $X_1, \ldots, X_n$ of i.i.d. real-valued random variables with an (unknown cumulative) distribution $F$.

The goal is to estimate the value of $F$ in multiple points. That is, given a set of $k$ numbers $x_1 < x_2 < \ldots < x_k$, we want to derive a set of pairs $\{ (l_i, u_i) \}_{i \in [1,k]}$ such that $\mathbb{P}[\forall i \in [1,k] : l_i \le F(x_i) \le u_i] \ge 1 - \alpha,$ where $1 - \alpha$ is the required coverage probability.

If $k = 1$, then one can use a binomial proportion confidence interval (to determine the proportion of elements that are smaller than $x_k$). If $k$ is large, then one can use Dvoretzky–Kiefer–Wolfowitz inequality (DKW).

However, what if $k$ is reasonably small? Let's say that $k \in [2, 5]$.

  • I know that in this case DKW inequality may be needlessly wide. So, probably, it is not the best approach.
  • At the same time, we cannot use (at least in theory) a bunch of binomial proportion confidence intervals (one interval for each $x_i$), since the estimates would be correlated and thus $\prod_{i \in [1,k]} \mathbb{P}[l_i \le F(x_i) \le u_i] \not= \mathbb{P}[\forall i \in [1,k] : l_i \le F(x_i) \le u_i].$
  • I was thinking about applying simultaneous confidence intervals for multinomial proportions. However, note that they give estimates for proportion of elements falling into $[x_i, x_{i+1})$, whereas I need an estimation of proportion of elements falling into $(-\infty, x_{i})$ for each $i$.

  1. What is the right approach here? Are there any papers on this topic?
  2. Does the problem changes if $X_1, \ldots, X_n$ is a sample of i.i.d. integer-valued random variables?
$\endgroup$
2
  • 2
    $\begingroup$ You seem to be asking about a generalization of the Kolmogorov-Smirnov statistic. Are you familiar with the theory of that test? $\endgroup$ – whuber Sep 28 '20 at 23:29
  • $\begingroup$ No, I'm not familiar with this test. Is your idea to view the input sample as interval-censored data and then to apply some method to derive CDF? $\endgroup$ – Sergey Bozhko Sep 30 '20 at 21:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.