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I have observation data $x_1, \dots, x_n$ with associated positive weights $w_1, \dots, w_n$ such that $\sum_i w_i = 1$, representing a "weighted" empirical distribution with CDF $$ \hat{F}(x) = \sum_{i=1}^n w_i I(x_i \le x)$$ where $I$ is the indicator function. This generalizes the usual empirical distribution, which is the special case $w_i = 1/n$. Is there a closed-form expression for a confidence interval for the median in this case?

I'm aware that this question has been answered before for the case $w_i = 1/n$, for which the confidence interval can be obtain from the binomial distribution, but it's not obvious to me how to treat general case. I realize that bootstrap estimators are possible, but I'm interested in deriving an analytical expression, if possible.

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  • $\begingroup$ You need to specify the distribution of the $x_i$'s to hope for a solution. $\endgroup$
    – Xi'an
    Commented Feb 5, 2019 at 17:34
  • $\begingroup$ @Xi'an Possibly, but can you show why? Given that there is a distribution-free solution for the case $w_i = 1/n$, why isn't there a solution for arbitrary $w_i$ ? $\endgroup$
    – Roland
    Commented Feb 5, 2019 at 17:39
  • $\begingroup$ Because the weights need be connected with the distribution of the $x_i$'s. $\endgroup$
    – Xi'an
    Commented Feb 5, 2019 at 21:08

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