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I'm taking a course on probability theory. This is how my teacher has defined a uniform random variable:

Let $X$ be a random variable. We say $X$ is uniformly distributed in $(0,1)$ if for every $a,b \in\mathbb R$,

$P(a < X < b) =\lambda[(a,b)\cap (0,1)]$, where $\lambda$ is Lebesgue measure

I'm having trouble in understanding and applying this definition.

For example, I was doing the following question:

Let $U$ be a uniform random variable on $[0,1]$. Let $X=U^2$. Find distribution function of $X$.

What I tried:

$$\{X \leq x\} = \{U^2 \leq x\} = \{-\sqrt x \leq U \leq \sqrt x\}$$

So, $F(x)= P\{X \leq x\} = \lambda([-\sqrt x ,\sqrt x\,] \cap [0,1])$

Is this right?

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  • $\begingroup$ You're in the right direction, but not finished. What is $[-\sqrt x,\sqrt x] \cap [0,1]$? What is the measure of that set? So which function is $F$? $\endgroup$ – Mees de Vries Sep 5 '16 at 19:47
  • $\begingroup$ $\sqrt x$ if 0<x<1 0 if x<0 or x>1 $\endgroup$ – Sahiba Arora Sep 5 '16 at 19:51
  • $\begingroup$ Almost. Note that $F$ must be increasing, so we can't have $F(x) = 0$ for $x > 1$. $\endgroup$ – Mees de Vries Sep 5 '16 at 19:56
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    $\begingroup$ Oh yes, my bad. F(x)=1 for x>1 $\endgroup$ – Sahiba Arora Sep 5 '16 at 19:57
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It's ok, but since you know that $\Pr(U>0)=1$, I'd just write $0\le U \le \sqrt x$. Then of course, simplify the expression $\lambda([0,\sqrt x\,] \cap [0,1])$.

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