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Dec 20, 2018 at 21:18 comment added Iosif Pinelis Oops, I missed it that the distribution on $[0,1]$ may arbitrary. This does not change the answer much, though.
Dec 20, 2018 at 21:16 history edited Iosif Pinelis CC BY-SA 4.0
deleted 235 characters in body
Dec 20, 2018 at 20:19 comment added dohmatob The formula you're proposing is valid for the uniform distribution on $[0,1]$. My problem is more general. The distribution is only required to supported on $[0, 1]$, not necessarily uniform thereupon. Agreed ? Concerning the remark about my use of "empirical processé", it's in fact not a very interesting process. Fixed.
Dec 20, 2018 at 19:58 history answered Iosif Pinelis CC BY-SA 4.0