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Note: This is the same question posted here, but I am seeking clarification on how my attempt is incorrect (i.e., where I went wrong).


If we let $S\sim{}\text{Unif}(0,1)$, then $U = 2S-1$ and $U^2 = (2S-1)^2$ by location-scale transformation. Letting $X = U^2$ we have $X = (2S-1)^2$.

By universality of the uniform ($X = F^{-1}(S))$, why isn't $F(x) = \frac{\sqrt{x}+1}{2}$ the CDF of $U^2$? I know this is incorrect because $F(x)$ isn't a valid CDF, but I'm a little turned around as to why this logic doesn't work out.

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The universality of the uniform tells you that if $\ X\ $ has a continuous distribution $\ F\ $, then $\ T=F(X)\ $ is uniformly distributed over $\ [0,1]\ $, so when $\ F\ $ is invertible $\ X=F^{-1}(T)\ $. It does not tell you that $\ X=F^{-1}(S)\ $ for any other uniformly distributed random variable $\ S\ $.

In your example, $\ F(x)=\sqrt{x}\ $ so $\ T=\sqrt{X}= |2S-1|\ne S \ $ even though both $\ |2S-1|\ $ and $\ S\ $ are uniformly distributed over $\ [0,1]\ $.

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You are misusing this property of the uniform distribution, which w.r.t. the present problem reads

  • if $F$ is the CDF of some continuous random variable $X=U^2$, then $F(X) = S$ satisfies $S\sim\text{Unif}(0,1)$.

The key point is that the function $F$ must be the CDF of $X=U^2$. In particular, it must satisfy the properties of a continuous cumulative distribution function. Note that the inverse transformation of $X=(2S-1)^2$ is $$ \frac{1\pm\sqrt{X}}2 = S\, , $$ and that many functions which restriction to $(0,1)$ is $x\mapsto\frac12(1+\sqrt{x})$ aren't continuous CDFs! The correct CDF is obtained in the linked post as follows: \begin{aligned} \Bbb P(U^2\leq x) &= \Bbb P(-\sqrt{x}\leq U\leq \sqrt{x}) \\ &= \dots \end{aligned}

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  • $\begingroup$ Thanks for taking the time to answer my question, @EditPiAf (...great SO handle, too :). I'm still a little confused re: "...and that many functions which restrictions...". I thought if I inverted $F^{-1}(S)$, that would yield the CDF of X. I understand the solution provided in the other question...I just don't know how to connect this property of the uniform distribution to the correct solution. $\endgroup$
    – Per48edjes
    Apr 26, 2020 at 1:49
  • $\begingroup$ @Per48edjes Inverting brutally $F(S) = (2S+1)^2$ doesn't even give you a function here. Indeed we'd have $F^{-1}(X) = \tfrac12(1\pm\sqrt X)$ which is multi-valued because of $\pm$... The transformation $U\mapsto U^2$ isn't always one-to-one $\endgroup$
    – EditPiAf
    Apr 26, 2020 at 2:05

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