Exponential distribution and density

Question

Let X be a positive random variable with density $e^{x}\, 1_{(0,\infty)}(x)$. What is the density of $1/(1 + X)$?

What I am getting is $P(1/1+X < x) = 1-P(X>1/x - 1) = 1- e^{(x^{-1})-1}$ which turns into $F'(x)=e^{x^{-1}-1}$

Does this seem correct?

Answer 1

Note that the random variable $\frac{1}{1+X}$ "lives" in the interval $(0,1)$, and outside that interval the density function is $0$.

Let $Y=\frac{1}{1+X}$, and let $0\lt y\lt 1$. We have $$F_Y(y)=\Pr(Y\le y)=\Pr\left(\frac{1}{1+X}\le y\right)=\Pr\left(X \ge \frac{1}{y}-1\right)=e^{1-1/y}.$$ For the density function, differentiate. We get $f_Y(y)=\frac{1}{y^2}e^{1-1/y}$ (on the interval $(0,1)$, and $0$ elsewhere).