I have some questions about the following example from my lecture book. I can't really make sense of it, if someone could fill in the blanks that would be great.
Theory: Let $X$ be a continuous random variable with the p.d.f. $f_{X}(x)$. The p.d.f. of $Y=H(X)$, where the domain of definition of the function $H(X)$ can be decomposed into disjoint intervals, where $H(X)$ is strictly monotonous, we have
$$f_{Y}(y)=\Sigma_{i}f_{X}(H(X)^{-1}(y))\cdot|\frac{d}{dy}H(X)^{-1}(y)|\chi_{I_i}(y),$$
where $H_i$ indicates the function $H$ restricted to the respective domain $I_i$ of strict monotonicity, and $\chi_{I_i}$ is the corresponding indicator function.
Example
Let $X\in U(0,2\pi)$ and $Y=\sin(X)$. We want to determine the p.d.f. $f_Y(y)$. The function $H(x)=\sin(x)$ is not strictly monotonous in $(0,2\pi)$, hence we shall find the p.d.f. $f_Y(y)$ by means of the theory above.
We make the decomposition $(0,2\pi)=I_1\cup I_2\cup I_3$, where $I_1=(0,\pi /2)$, $I_2=(\pi /2, 3\pi /2)$ and $I_3=(3\pi /2, 2\pi)$. Then for $i=1,2,3$, $H_i(x)=\sin (x)|I_i$, i.e., the function $\sin (x)$ restricted to $I_i$, is strictly monotonous. In fact,
$$H_1(x)=\sin(x)\quad 0\leq x \leq \pi /2,\\ H_2(x)=H_2(x-\pi /2),\pi /2\leq x\leq 3\pi /2 \iff H_2(t)=\cos(t),\quad 0\leq t\leq \pi \\ H_3(x)=H_3(x-2\pi), \pi /2 \leq x\leq 3\pi /2 \iff H_3(t)=\sin(t),\quad -\pi /2\leq t\leq 0.$$
Then we have two cases (i)-(ii) to consider:
(i) $0\leq y<1$. Then (draw a picture)
$$F_Y(y)=\pmb{P}(Y\leq y)=\pmb{P}(0\leq X \leq H_1^{-1}(y))+\pmb{P}(H_2^{-1}(y)\leq X\leq 3\pi /2)+\\ +\pmb{P}(3\pi /2\leq X \leq 2\pi)=\pmb{P}(0\leq X \leq \arcsin(y))+\pmb{P}(\arccos(y)\leq X\leq 3\pi /2)+1/4 = \frac{\arcsin(y)}{2\pi}+\frac{3\pi /2-\arccos(y)}{2\pi}+\frac{1}{4}$$
\...\
I think I'll stop there. My questions are regarding the $H_i$s, the first $H_1$ I can understand, but the following equations $H_2(x)=H_2(x-\pi /2)$ and $H_3(x)=H_3(x-2\pi)$ I don't understand. What is the author trying to convey by writing that? And how is this anything like what the theory suggests? Furthermore, I don't understand the interval of $H_3$, why is the same as for $H_2$? Typo? Why not just use the same $H(x)=\sin(x)$ for all of these?
On the other hand, what is happening exactly when he is calculating the c.d.f. of $Y$? From $\pmb{P}(Y\leq y)$ he gets three separate expressions containing $X$ (Law of total probability or is this the decomposition?). How is this making any sense, it says $0\leq y <1$, but here it seems as $y$ is less then zero (judging from the third expression). I don't know what to make of it, could someone explain?
Best regards,