Find CDF of $U^2$ where $U\sim{}\text{Unif}(-1,1)$. How am I misapplying Universality of the Uniform?

Question

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).

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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.

Answer 1

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}

Answer 2

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]\ $.