let $X = exp(1)$, and let $Y = \frac{x+1}{x}$ how to find density function of Y?

Question

let X be an exponential random variable with parameter 1. $X = exp(1)$, and let Y be equal to $$Y = \frac{x+1}{x}$$ I need to find density function of Y. $f_y(x)$.

I've done as follows: $$F_y(Y) = P(Y \leq y) = P(\frac{x+1}{x} \leq y)$$ but here's the problem, because I don't know how to get $P(X <= y)$ from $P(\frac{x+1}{x} \leq y)$. Meaning, I need to obtain X in the left hand side, and the rest on the right hand side, because this is the way to obtain $F_x(y)$. When I have distribution function I can take the first derivative and obtain the density function. But the problem is that I don't know how to proceed further. For now, I'm stuck in this phase: $$P(\frac{x+1}{x} \leq y)$$

EDIT: I think I can simplify the fraction as follows: $$\frac{x}{x} + \frac{1}{x}$$ and I'd get: $$x >= \frac{1}{y-1}$$ and I get: $$1-F_x( \frac{1}{y-1})$$ and the other steps are about finding the derivatives and substitute what I got with the exponential distribution. (I think that's the correct answer).

Answer 1

Because $P(X > 0) = 1$ and $Y = 1 + \frac{1}{X}$, we have $P(Y > 1) = 1$. Hence, $$ F_Y(y) = 0, \qquad y \le 1. $$ Next, for y > 1 it follows that $$ P(Y \le y) = P(\tfrac{X+1}{X} \le y) = P(X+1 \le Xy) = P\bigl(X(1-y) \le -1\bigr) = P(X \ge \tfrac{1}{y-1})\\ = 1 - F_Y(\tfrac{1}{y-1}). $$ Finally $$ F_Y(y) = \begin{cases} 0, & y \le 1,\\ 1 - F_X(\tfrac{1}{y-1}), & y > 1. \end{cases} $$

Answer 2

I use the definition:

$$\begin{align} F_Y(y)&=Pr(Y\le y)=Pr(\frac{X+1}{X}\le y)=Pr(X+1\le yX)\\ \\ &=Pr(1\le yX-X)=Pr(1\le(y-1)X)=Pr(X\ge \frac{1}{y-1})\\ \\ &=1-Pr(X\le \frac{1}{y-1})=1-\int_0^{\frac{1}{y-1}} e^{-x}dx \end{align}$$

Probability density function: $$f_Y(y)=\frac{dF_Y(y)}{dy}=-e^{-\frac{1}{y-1}}\cdot\left(\frac{-1}{(y-1)^2}\right)=\frac{1}{(y-1)^2}\cdot e^{-\frac{1}{y-1}}~, ~~~y\in(1,\infty)$$