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

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.

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 $$\ [0,1]\$$.
• if $$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}
• 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. Apr 26, 2020 at 1:49
• @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 Apr 26, 2020 at 2:05