Finding and sketching CDF for $Y=g(X)$ random variable

Question

The question asks me to find an expression for $f_Y(y)$ if $Y = g(X)$ is defined as in (1). And also sketching $F_Y (y)$ if $F_X(x)$ is given by Fig. 1

image here

I got the derivative of $g(x)$. And I used the formula of $\sum_{i=0}^n f_X(x_i)/g'(x_i)$ . But I couldnt see how to find $f_X(x)$. And sketching was whole another story. I couldnt even imagine. Can anyone help?

Answer 1

Hi Osman ağabey I think we don't need to re-sketch the $F_Y(y)$, it's just shifted version of $F_X(x)$. I found the expressions like these: image here

Answer 2

For $y<0$ you have $$F_Y(y)=P(Y\leq y)=P(X+c\leq y, X\leq-c)=P(X\leq y-c)=F_X(y-c).$$ For $y>0$ you have $$F_Y(y)=F_X(y+c)$$ by a similar calculation. Lastly, $$P(Y=0)=P(X=-c)+P(-c<X\leq c)=P(X=-c)+F_X(c)-F_X(-c)=F_X(c)-F_X(-c),$$ where $P(X=-c)=0$ comes from the observation that $F_X$ does not jump as $-c$.

So what does the graph of $F_Y$ look like? On $(-\infty,0)$ you see the graph of $F_X$ on the interval $(-\infty,-c)$. At $0$ there is a jump of size $F_X(c)-F_X(-c)$. On $(0,\infty)$ you see the graph of $F_X$ on the interval $(c,\infty)$.