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I have the following problem:

Let $U$ be a $\text{Unif}(−1,1)$ random variable on the interval $(−1,1)$.

Find the CDF and PDF of $U^2$. Is the distribution of $U^2$ Uniform on $(0, 1)$?

The solution is as follows:

Let $X = U^2$, for $0 < x < 1$, $P(X \le x) = P(−\sqrt{x} \le U \le \sqrt{x}) = P(U \le \sqrt{x}) - P(U \le -\sqrt{x}) = \dfrac{\sqrt{x} + 1}{2} - \dfrac{-\sqrt{x} + 1}{2} = \sqrt{x}$ (Note that $P(U \le u) = \dfrac{u + 1}{2}$ for $-1 \le u \le 1$. The density is then given by $f_X(x) = \dfrac{d}{dx} P(X \le x) = \dfrac{d}{dx}x^{1/2} = \dfrac{1}{2} x^{-1/2}$. The distribution of $X = U^2$ is not Unif$(0, 1)$ on the interval $(0, 1)$ as the PDF is not a constant on this interval.

The first fact that I am confused about is how the author got that the interval is now $0 < x < 1$ instead of $-1 < x < 1$.

The second fact that I am confused about, which it seems is related to the first, is how the author calculated that $P(U \le \sqrt{x}) - P(U \le -\sqrt{x}) = \dfrac{\sqrt{x} + 1}{2} - \dfrac{-\sqrt{x} + 1}{2}$. Did they just insert the values into the formula for the CDF (since the formula for the CDF is $\dfrac{x - a}{b - a}$), or did they otherwise somehow calculate the CDF? I ask because I'm unsure if it's just a matter of memorization, or whether there is some mathematical understanding here that I am missing.

I would greatly appreciate it if people could please take the time to clarify this.

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  • $\begingroup$ Isn't it $X = |U|$? In this case for $0<x<1$ we have $P(X \leq x) = P(|U| \leq x) = P(-x \leq U \leq x)$ $\endgroup$
    – G. Gare
    Commented Nov 1, 2019 at 15:23
  • $\begingroup$ @G.Gare My apologies!!! I made a transcription error; it should be $U^2$. I'm really sorry!!! $\endgroup$
    – The Pointer
    Commented Nov 1, 2019 at 15:25
  • $\begingroup$ Then the square roots make more sense. There's another edit you missed: $P(-x\leq U^2 \leq x)$. $\endgroup$
    – G. Gare
    Commented Nov 1, 2019 at 15:26
  • $\begingroup$ @G.Gare How's that? $\endgroup$
    – The Pointer
    Commented Nov 1, 2019 at 15:28
  • $\begingroup$ It's ok now! Don't worry $\endgroup$
    – G. Gare
    Commented Nov 1, 2019 at 15:28

1 Answer 1

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Explaining fact 1: while $U$ has support $[-1,\,1]$, $X=U^2$ has support $[0,\,1]$.

Explaining fact 2: yes, it was just use of the CDF.

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    $\begingroup$ Thanks for the answer. Can you please explain the reasoning for how we go from support $[-1, 1]$ to $[0, 1]$? I ask because it is very simple to naively just reason that, since $X = U^2$, we have $(-1)^2$ and $1^2$, and so the new support is $[1, 1]$ (which is obviously nonsense). I recognize that this would be due to a misunderstanding of what "support" is, but I'm curious what explanation you would use to describe the reasoning behind how we get $[0, 1]$, so that it is clear in my mind as a person who is new to this concept. $\endgroup$
    – The Pointer
    Commented Nov 1, 2019 at 15:32
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    $\begingroup$ @ThePointer The support is the set of allowed values. Note that $\{u^2|u\in[-1,\,1]\}=[0,\,1]$. (For example, the graph $x=u^2$ has $x$-coordinates $\in[0,\,1]$ for $u\in[-1,\,1]$.) $\endgroup$
    – J.G.
    Commented Nov 1, 2019 at 15:46

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