How do you find the conditional PDF of X, if it is known that X was in a certain interval?

Question

Given random variable X with PDF: $f_X(x)=\frac{1}{2}sin(x)$ on the interval $[0,\pi]$

If we know that $X\in(\frac{\pi}{3},\frac{2\pi}{3}]$

What is the conditional probability density function of X? $f(x|X\in(\frac{\pi}{3},\frac{2\pi}{3}])=?$

Answer 1

For $x\leq \pi/3$ and $x\geq 2\pi/3,$ $F(x|X\in(\pi/3, 2\pi/3])$ equals $0$ and $1$ respectively. If $x\in (\pi/3, 2\pi/3),$ we have $$P(X\leq x| X\in (\pi/3, 2\pi/3]) = \frac{P(X\leq x, X\in (\pi/3, 2\pi/3])}{P(X\in (\pi/3, 2\pi/3])} = \frac{P(X\in (\pi/3, x])}{P(X\in (\pi/3, 2\pi/3])}.$$

Can you express the RHS in terms of $F_X$?