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Let $x_{\alpha} = \inf \{x \in\mathbb{R}: F_X(x) \geq \alpha\}$, $U \sim Uniform(0,1)$ and $Z=x_{U}$. I need to prove that Z has the same distribution as X. Obviously this is true as can easily be shown with a numerical example and the intuition behind it is clear. However I cannot seem to formulate a formal mathematical proof. Could anyone provide me with a hint/paper of how to do this?

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Verify that $Z \leq t$ if and only if $U \leq F_X(t)$. You then get $P\{Z\leq t\} =P\{U\leq F_X(t)\}=F_X(t)$ Hints for the first part: it is immediate from definiton that $Z \leq t$ if $U \leq F_X(t)$. Suppse $Z\leq t$ but $U>F_X(t)$. There exists $a>0$ such that $U>F_X(t+a)$. Then $F_X(s) < U$ for all $s \leq t+a$ so $Z \geq t+a$, a contradiction.

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  • $\begingroup$ I have used right continuity of $F$ to conclude that $F_X(t+a) <U$ for some $a>0$. My proof is valid for all distributions. I did not use the fact that $F_X$ is strictly incraesing. $\endgroup$ – Kavi Rama Murthy Jul 26 '18 at 7:51
  • $\begingroup$ @JeannotvandenBerg If you don't want to study my argument you can visit en.wikipedia.org/wiki/Cumulative_distribution_function where the you can find the same statement. What I have proved in my answer if folklore and Probabilists use it routinely. $\endgroup$ – Kavi Rama Murthy Jul 26 '18 at 8:16

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