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$X\sim U[0,1]$ and $Y\sim U[-1,1]$ are two uniform-distributed R.V.'s. Are $X^2$ and $Y^2$ still uniform? Do they have explicit probability density funtion?

wiskundeliefhebber
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JFK
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2 Answers2

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They are not uniform distribution.$$F_{X^2}(x) = \mathbb{P}(X^2 \leq x) = \mathbb{P}(X \in [0,\sqrt{x}]) = \sqrt{x}$$ Hence, $$f_{X^2}(x) = \dfrac{dF_{X^2}(x)}{dx} = \dfrac1{2\sqrt{x}}$$ You can do the same thing for $Y \mapsto Y^2$. $$F_{Y^2}(y) = \mathbb{P}(Y^2 \leq y) = \mathbb{P}(Y \in [-\sqrt{y}, \sqrt{y}]) = \sqrt{y}$$ Hence, $$f_{Y^2}(y) = \dfrac{dF_{Y^2}(y)}{dy} = \dfrac1{2\sqrt{y}}$$

  • Since the support of $X$ and $Y$ are different, shouldn't their distributions be different as well? Namely, isn't $F_{Y^2}(y) = 2\sqrt{y}$ and thus $f_{Y^2}(y)=1/\sqrt{y}$? – Jason Morgan Sep 23 '13 at 21:02
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    @JasonMorgan No. Note that $F_{Y^2}(1)$ should be $1$, which is not the case, with your answer. –  Sep 23 '13 at 21:10
  • Right. Thank you. I should have noticed that. – Jason Morgan Sep 23 '13 at 21:15
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    I don't understand a step of your proof: with $Y$ being uniform between $[-1,1]$ shouldn't the probability of an interval $[a,b]$ be $\frac{1}{b-a}$ and therefore the $\mathbb{P}(Y \in [-\sqrt{y}, \sqrt{y}]) = \frac{1}{2\sqrt{y}}$ – Matteo Nov 08 '13 at 18:29
  • @Matteo no, be careful about the integrand – aflous Apr 24 '14 at 13:48
  • @aflous - is it because y is under a square root and thus needs to be positive (so between 0 and 1)? – Matteo Apr 25 '14 at 16:12
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    No, $ \mathbb{P}(Y \in [-\sqrt{y}, \sqrt{y}]) = \frac{\sqrt{y} -(-\sqrt{y})}{b-a}=\frac{2\sqrt{y}}{1-(-1)}= \frac{2\sqrt{y} }{2}=\sqrt{y}$ – aflous Apr 28 '14 at 07:26
  • why doesn't this work? $E(y^2)=\int_{-1}^{1}y^2\frac{1}{2}dy=\int_0^1y^2dy?$ – Dave ddd Apr 08 '16 at 14:22
  • What are the domain of $y$, is it $y \in [0,1]$? – Logan May 15 '17 at 11:48
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First, find the CDF of the transformed RV. Let $S = X^2$.

$P(S \leq s) = P(X^2 \leq s) = P(X \leq \sqrt s) = \sqrt s$, for $0 < s < 1$ and $0$ otherwise. Where the last equality is true since $X$ ~ $\text{Unif}(0,1).$ Note, in general, the support is not the same under the transform.

Next, differentiate to find the PDF. $\frac{d}{dx} \sqrt{s} = \frac{1}{2 \sqrt s}$ for $0 \leq s \leq 1$ and $0$ otherwise.

The PDF of $Y^2$ follows similarly. Also, in general, uniformity is not preserved under non-linear transforms.

Jake Wren
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  • The PDF is not defined at 0 or 1, although it can be set to 1/2 at 1 by convention. – Alan Feb 12 '22 at 16:24