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Let $U$ be a uniform random variable on $[-1,1]$. Let $X=U^2$. Find the distribution of $X$.

This is what I have right now.

$F_X(x)=P(X \leq x) = P(U^2 \leq x) \rightarrow P(-\sqrt{x} \leq U \leq \sqrt{x})$.

$$F(x)= \begin{cases} 0, & \text{if } x <0, \\ \sqrt{x}, & \text{if } 0 \leq x < 1\\ 1, & \text{if } x \geq 1 \end{cases}$$

Am I doing this wrong?

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    $\sqrt{x}$ is not defined for negative values – 2017-02-27
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    then that is why it is 0 if x is less than 1 – 2017-02-27

1 Answers 1

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You got $\sqrt{x}$ right. But, the intervals are wrong.

$F_x(x)=P(U^2 \leq x)=P(-\sqrt{x} \leq U\leq\sqrt{x})=2\int_{0}^{\sqrt{x}}dt$

$F_x(x)=2\int_{0}^{\sqrt{x}}U(t)dt=2\int_{0}^{\sqrt{x}}\frac{1}{2}dt=\sqrt{x}$

$$F_x(x)=\begin{cases} 0, & \text{if } x < 0, \\ \sqrt{x}, & \text{if } 0 \leq x < 1\\ 1, & \text{if } x \geq 1 \end{cases}$$

Hope it is clear.

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    How did you get $P(\sqrt{x} \leq U \leq \sqrt{x})$ shouldn't it be $\pm\sqrt{x}$? – 2017-02-27
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    "Hope it is clear" Clearly marred with typos, yes. Please revise carefully your post. – 2017-02-27
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    @Killercamin, you are right, I made a mistake. – 2017-02-27
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    One last question, why was [-1,1] important then? – 2017-02-27
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    With the new random variable, we do not have negative values, as we have mapped them all to the interval $[0,1]$. Does it answer your question? – 2017-02-27
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    Yes, thank you now it makes sense – 2017-02-27
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    I made an edit. Sorry I was being clumsy – 2017-02-27
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    Now, it is very nice (+1). I would still suggest that you spell out for future readers that the $U(t)=1/2$ inside the integral is the pdf of the uniform $U(-1,1)$, or $f_X=\frac{1}{1-(-1)}.$ – 2017-02-27