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I have started off by this:

$F_Y(Y)=P(Y\leq y)=P(X^2 \leq y).$ Now, I have been told that $P(X^2 \leq y) = P(- \sqrt y \leq X \leq \sqrt y) $. I don't quite understand why this is and any help would be appreciated!

Kim
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    $X^2 \leq y = |X| \leq \sqrt{y} = -\sqrt{y} \leq X \leq \sqrt{y}$ What is the exact confustion that you are having here? – TenaliRaman Oct 19 '15 at 08:55

1 Answers1

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Follow TenaliRaman's comment

$X^2\le y\Leftrightarrow |X|\le\sqrt{y}\Leftrightarrow -\sqrt{y}\le X \le \sqrt{y}$ and $y \in [0,1]$

$$F(y)=P(Y<y)=P(X^2<y)=P(-\sqrt{y}\le X \le \sqrt{y})=\int_{-\sqrt{y}}^{\sqrt{y}}|x|dx=\int_{-\sqrt{y}}^{0}-xdx+\int_{0}^{\sqrt{y}}xdx=-\frac{1}{2}x^2\mid_{-\sqrt{y}}^{0}+\frac{1}{2}x^2\mid_{0}^{\sqrt{y}}=y$$

You can see $F(y)=y$ and $y \in [0,1]$

This shows $Y\sim U(0,1)$

Deep North
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    This being a self-study question, providing the complete answer does not help the OP to figure out a working solution. – Xi'an Oct 19 '15 at 11:06
  • I agree with you. To fully answer Op's question, one may need start from definition of a random variable which is a set function then go to $\sigma$ field and Borel $\sigma$ field and also need to explain the probability defined on a set which may be more difficult to for me explain. So I take a easier way to solve the problem. – Deep North Oct 19 '15 at 11:40
  • No, there is no any irony in my comments or answer. I respect you since you have given me very good suggests for my questions. – Deep North Oct 19 '15 at 11:50
  • Also, when I have questions I really think if I post the problem to SE then people such as Xi'an, Wuber, Glen, etc. would give me good suggestions. – Deep North Oct 19 '15 at 12:03
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    Ok, good to hear! And sorry for the paranoia! What I meant is that we should help Kim by going stepwise in the answer so that she or he can spot his or her jamming blocks. For instance, starting only at $X^2-\sqrt{y}$. – Xi'an Oct 19 '15 at 12:05