If $X$ is a continuously distributed random variable and $F$ is the c.d.f. of its distribution, then $F(X)$ is uniformly distributed in the interval $(0,1).$
While i'm clear with the mathematical proof, I'm looking for an intuitive answer to this.
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1$\begingroup$ This must be a duplicate, but cannot find it $\endgroup$– kjetil b halvorsenCommented Aug 21, 2017 at 22:29
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1$\begingroup$ Please give a tiny bit of hint about what you don't find intuitive. Are you comfortable with CDF's in general but uniform RVs somehow baffle you, or the opposite, or both? $\endgroup$– kimchi loverCommented Aug 21, 2017 at 22:41
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$\begingroup$ This only applies to continuous cumulative distribution functions, i.e. for continuous random variables $\endgroup$– HenryCommented Aug 22, 2017 at 0:14
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2$\begingroup$ It's really unclear what this is about. To ask for an intuitive explanation, it helps to give a clear statement of the mathematical fact you are interested in--the definition, formula, whatever, saying what all the parts of it are; explain the way in which you already understand it (in this case, what is the mathematical proof?), and then ask for a more intuitive explanation. If you want to try this, use the "edit" link just below the question text to insert this information into the question; don't try to explain in comments. $\endgroup$– David KCommented Aug 22, 2017 at 0:31
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1$\begingroup$ @MichaelHardy Yes, the language was so poorly organized that I completely missed the point. Now that you've helped OP out by fixing it, I think this is distinct from the linked question because this one asks for intuition rather than a proof. I'm voting to reopen. $\endgroup$– David KCommented Aug 22, 2017 at 5:07
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1 Answer
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Assuming we are dealing with continuous random variables
the CDF $F(X)$ is a continuous increasing function of $X$ where the probability that $F(X) \le p$ for $p \in (0,1)$ is the probability that $X \le q$ when $F(q)=p$, which is $p$
so $F(X)$ has a uniform distribution on the unit interval, since the probability it is less than or equal to a given $p \in (0,1)$ is $p$
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