Find the pdf of $Y=\frac{1}{1+X}$

Question

I know that $X$ has an exponential distribution and from there I must calculate the pdf of $Y=\frac{1}{1+X}$. My work so far: $$F_{Y}(y)=P(Y\leq y)=P\left (\frac{1}{1+X} \leq y \right)=1-P\left(X \leq \frac{1}{y} -1 \right) = 1-F_{X}\left(\frac{1}{y} -1 \right)$$ Now I find the derivative on both sides of the equality: $$f_{Y}(y)=\frac{1}{y^2}·f_{X}\left(\frac{1}{y} -1 \right)$$ And I get that: $$f_{Y}(y)=\frac{1}{y^2}·\lambda e^{-\lambda ·\left(\frac{1}{y}-1\right)}$$ if $y\leq1$ and $f_{Y}(y) = 0$ in other cases. When I integrate this, the resulting integral is divergent, when it should be 1. Can anyone tell me where is my mistake please?

Answer 1

No it's not infinite.

Indeed, a primitive function is

$$\phi(y)=e^{-\lambda ·\left(\frac{1}{y}-1\right)}$$

with $\phi(1)=1$ and $\phi(0_+)=0.$ Therefore

$$\int f_Y(y)dy=\phi(1)-\phi(0)=1.$$

Here is, on the same figure, in the case $\lambda=1$, the curve of $f_Y$, the pdf (in red), and the histogram (in blue) obtained as a result of a simulation of random variable $ 1/(1-\ln(U))$ where $U$ is a uniformly distributed random variable on $[0,1]$ (let us recall that $X=-\ln U$ follows the standard exponential law with parameter $\lambda=1$).

Fig. 1 : one can check that the area under the curve is equal to one.

Answer 2

Since $Y$ ranges from a minimum of $0$ at $X=\infty$ to a maximum of $1$ at $X=0$, $f_Y$ is nonzero only if $y\in[0,\,1]$. So the integral that needs to equal $1$ is $\int_0^1\frac{1}{y^2}\lambda e^{-\lambda\left(\tfrac1y-1\right)}dy$, not $\int_{-\infty}^1\frac{1}{y^2}\lambda e^{-\lambda\left(\tfrac1y-1\right)}dy$.