Given $ X \sim \exp(\lambda) $ find the density function of $Y=\frac{1}{1-X}$
So I tried doing the following and gut stuck in the way:
$ F_Y(y)=P(Y\leq y)=P(\frac{1}{1-x}\leq y)$
Now I though about isolating X, which gives me $ F_x(\text{something}) $ and by deriving it, get the desired density of Y, but I couldn't figure out how to handle the inequality signs (because I can't simply multiple both sides by $1-x$).
What is so wrong with my way of thinking?
Note that $Y$ takes values in $(-\infty, 0) \cup [1, \infty)$. It might be helpful to take each case separately.
If $y < 0$, then $$P(\frac{1}{1-X} \le y) = P(\frac{1}{1-X} \le y, X > 1) = P(1 < X \le 1-\frac{1}{y}).$$
If $y \in [0, 1)$ then $P(\frac{1}{1-X} \le y) = P(\frac{1}{1-X} \le 0) = P(X > 1)$.
If $y \ge 1$, then $$P(\frac{1}{1-X} \le y) = P(\frac{1}{1-X} \le y, X < 1) + P(X > 1) = P(X \le 1 - \frac{1}{y}) + P(X > 1).$$