# Distributions of random variables

Let $(\omega, F, P)$ be Lebesgue measeure on$[0,1]$, and set

$X(\omega) = 1$ if $0 \le \omega < \frac{1}{4}$

$X(\omega) = 2\omega^2$ if $\frac{1}{4} \le \omega < \frac{3}{4}$

$X(\omega) = \omega^2$ if $\frac{3}{4} \le \omega \le 1$

Compute $P(X \in A)$ where

(A) $A = [0,\frac{3}{4}]$

(B) $A = [\frac{1}{2},1]$

From a previous question I know that:

(A) $A = [0,1] = \frac{\sqrt{2}}{2} + \frac{1}{4}$

(B) $A = [\frac{1}{2},1] = \frac{\sqrt{2}}{2}$

However for this problem I need to solve it via one of these equations I'm assuming: $\mu(B) = P(X \in B) = P(X^{-1}(B)), B \in B$

$\int_\Omega f(X(\omega))P(d\omega) = \int^\infty_{-\infty} f(t)\mu(dt)$

I've been looking around on Google for anything that may be helpful and so far I've found this: http://terrytao.wordpress.com/2010/09/19/245a-notes-2-the-lebesgue-integral/

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You need to find the measure of those values of $\omega$ which give $X(\omega)$ in the required interval
For (A) you need to calculate $\mu\left(\left[0,\frac14\right)\right)+\mu\left(\left[\frac14,\sqrt{\frac{1}{2}}\right)\right) +\mu\left(\left[\frac34,1\right]\right)$
For (B) you need to calculate $\mu\left(\left[0,\frac14\right)\right)+\mu\left(\left[\frac{1}{2},\sqrt{\frac{1}{2}}\right)\right) +\mu\left(\left[\frac34,1\right]\right)$
Could you give me an example of how you might solve: $\mu\left(\left[\frac14,\sqrt{\frac{1}{2}}\right)\right)$ I'm having trouble understanding how this is calculated. –  dmcqu314 Dec 3 '12 at 0:29
It is just $\sqrt{\frac{1}{2}} - \frac14$ or as you might write it $\frac{\sqrt{2}}{2} - \frac{1}{4}$ –  Henry Dec 3 '12 at 7:38
Or perhaps you meant "Why is this expression here?" If $\frac{1}{4} \le \omega \lt \frac{3}{4}$ and $0 \le 2\omega^2 \le 1$ then $\frac{1}{4} \le \omega \lt \frac{1}{\sqrt{2}}$. Perhaps I should have written $\mu\left(\left[\frac14,\sqrt{\frac{1}{2}}\right]\right)$ with a right square bracket but in this case it makes little difference. –  Henry Dec 3 '12 at 7:57