Find the distribution of $Y = g(X)$ knowing that $$ g(X) =\begin{cases}X & \text{if}&|X|\geq 1 \\0 & \text{if}&|X| < 1 \end{cases}$$
and
a) let $X$ be a Uniform(−1,1) continuous random variable
b) $p_{\small X} = \frac 14\delta_{-1} + \frac 12\delta_2 + \frac 14f_{\small X}$, where $f_{\small X}(x) = \left(\tfrac{x}{\surd(2\pi)}\right)e^{-{x^2}/{2}}$
so in part b) the probability that I will get -1 is $\frac 14$ and $\frac 12$ for 2 and the rest is the function, idk if that makes it more clear, frankly I dont understand it too well myself and the language barrier doesn't seem to help
I would be immensely grateful if someone could write me a step by step solution to this problem as I struggle quite a bit with this topic and seeing an example solution would help me understand it better (or at all lmao, remote learning kinda sucks :/)
Note that $g:(-\infty..\infty)\mapsto(-\infty..1]\cup\{0\}\cup [1..\infty)$ with, in particular, $g((-1..1))=\{0\}$.
So for (a): since $X\in(-1..1)$ then $Y$ will be degenerate: ie $Y=0$ a.s.. We use the Dirac delta function: $$p_{\small Y}(y)=\delta_0(y)$$
So for (b):
Likewise $Y$ will have a degeneracy at $0$, just as $X$ does at $-1$ and $2$, and the coefficient will be the probability mass at this degeneracy. IE the probability that $\lvert X\rvert<1$
The remaining probability density will only be supported when $y\leq -1$ or $y\geq 1$. We use the indicator function for this.
$$p_{\small Y}(y)=\tfrac 14\delta_{-1}(y)+\tfrac 12\delta_2(y)+\mathsf P(\lvert X\rvert<1)\,\delta_0(y)+f_{\small X}(y)\mathbf 1_{\lvert y\rvert\geq 1}$$
Now you have been given $f_{\small X}$ so may evaluate this probability.