Regarding the following problem:
Given $ X \sim U[-1, 1]$ find the PDF of $Y=\frac{1}{1+X}$
So:
$F_Y(t)=P(Y\leq t)=P(\frac{1}{1+X}\leq t)=P(X> \frac{1-t}{t}) = 1 - P(X\leq \frac{1-t}{t})$
Now I need to seprate the different cases for t - and this is the point I always fail to do. I tried to plot the function:
And to extract information from there, and the only ranges I could think of were $t<0$ and $t\geq 0$ which is probably incorrect.
Could someone please explain me how to approach such problems? Thanks
Observe that $1+X \in [0,2]$ so that $Y \in [1/2,+\infty)$ almost surely. This means that for $t \leq 1/2$ you have $F_Y(t) = 0$. The expression you wrote holds for $t>1/2$.
In general, your procedure is correct. You just have to take care of where your random variable takes values. If, for example, $Z \in [a,b]$ then you'll have $F_Z(x) = 0$ for $x < a$ and $F_Z(x) = 1$ for $x \geq b$.
Also, remind to take care of what happens when you divide. In this case, there's a point where you divide by $t$ but that would change the sign of the inequality. However, we are sure that $t \geq 1/2$ for the previous discussion, so there's no need to make this distinction now.
Side note: how did I calculate the support of $Y$? I calculated the image of $$f(x) = \frac{1}{1+x}, \quad x \in [-1,1]$$ Since $f$ is decreasing and continuous, it suffices to observe that $$f(1)=1/2$$ $$\lim_{x \to -1^+} f(x) = + \infty$$ to get the desired image.