Let $(\Omega,U,P)$ ia a probability space
$\Omega = [-1,1] , U=\mathscr B ([-1,1])$ (Borel algebra on that interval) and $P(A)= \frac {interval length of A}{2} $
$\mathcal X : \Omega \to \mathbb R$
$\omega \to \mathcal X (\omega)=\omega$ random variable is given
Calculate probability $P_{\mathcal X} [0,1/4]$
I know it is so basic but I don’t know why I couldn’t write.
What did I write :
Let $a \in \mathbb R$ If $a\lt -1$ then $(\mathcal X \leq a) = \{ \omega : \omega \leq a\} = \emptyset$
I cannot continue. I am not sure about should I write $-1 \leq a \lt 1 $ and $a \geq 1$ cases or only $-1 \leq a \leq 1 $ case.
If I write $-1 \leq a \lt 1 $ and $a \geq 1$ set will be $\Omega$ for second case. But I cannot think the other cases.
I need any help. Thanks in advance
My trouble is writing $(\mathcal X \leq a)$ according to situations of $a$
If $l(A)$ denotes the length of $A$ then $P_{\chi} [0,\frac 1 4] =\frac {l([0,\frac 1 4])} 2=\frac {\frac 1 4} 2=\frac 1 8$. That is it! $P\{\chi \leq a\} =\frac {l([-1,1]\cap (-\infty,a])} 2= 0$ if $a \leq -1$, $\frac {a+1} 2$ if $-1\leq a \leq 1$ and $1$ if $a >1$.