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Suppose i have a large data pool with a particular PDF, $F(x)$, interval $[x,y]$ estimated from KDE of the datapool. I drew $N$ samples at random from that data pool and saw that their distribution is also represented quite well by $F(x)$. let this draw be $D_{bef}\sim F(x)$

Now I want another distribution $G(x)$, on same interval, such that, if i draw another $N$ sample from $G(x)$, $D_{aft}\sim G(x)$, then total $2N$ samples follow $(D_{bef} + D_{aft}) \sim Uniform(x)$.

Is it possible?? A lot of questions here want to generate uniform from PDF, but I want to draw from a PDF which when combined with my original draw, will convert them uniform.

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    $\begingroup$ I do not understand the question: what is the meaning of the interval $[x,y]$? What is the role of $N$? How do you get $2N$ samples $D_{bef}+D_{aft}$? What is the meaning of Uniform$(x)$ and is this $x$ related with the $x$ in $[x,y]$? What is the connection between the title of the question and the contents of the question? $\endgroup$
    – Xi'an
    Commented Jun 7, 2020 at 14:06

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It's not always possible: suppose that $F$ samples almost entirely from a subinterval $[a,b]$. The combined sample will be almost 50% from that subinterval, so if $b-a <\frac{1}{2}(y-x)$ you can't get uniformity on $[x-y]$.

This shows what you need: for every subset of $[x,y]$, $F$ must put no more than twice as much probability on it than the uniform distribution does. That is, you need the density $f(s)$ of $F$ to be less than twice the uniform density $1/(y-x)$ everywhere on the interval. You can then sample from the distribution $G$ with density $$g(s) = \frac{2}{y-x}-f(s)$$ to get up to uniformity.

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  • $\begingroup$ Thank you. That was informative. I understood the limitation of uniform and sampling bias as you said. Suppose if I remove the $N$ new samples limit, and say that the new pdf $G$ shall approach uniform as N approaches infinity, will that change the answer? $\endgroup$
    – ipcamit
    Commented Jun 7, 2020 at 11:29
  • $\begingroup$ If the second round can go to $\infty$ then you can handle any $F$ that has a density. But if the interval is [0,1] and $F$ is the Bernoulli CDF you are still out of luck. $\endgroup$
    – Thomas Lumley
    Commented Jun 7, 2020 at 21:34

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