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

Let $$X$$ be an rv with pdf: $$$$f(x)=1,0\leq x\leq\ 1$$$$ or being $$0$$ otherwise. Let $$Y=\frac{X}{X+1}$$. What is the pdf of $$X$$?

I have tried to solve this problem several ways, however I feel that there is something that I'm missing. This is my latest attempt: $$$$P[Y\leq y]=P[X\leq \frac{y}{1-y}]=\frac{y}{1-y}$$$$ This is what I always get for the CFD. Then, when I differentiate, I get $$\frac{1}{(1-y)^2}$$ which, correct if I'm wrong, cannot be a pdf. Can you please help me? Thanks.

PS: Note that I've ommited the step of finding the inverse of the $$g$$ functiont that gives us $$Y$$. Also the CFD of $$X$$ is $$F(x)=x$$ for $$0\leq x\leq\ 1$$ and $$0$$ otherwise.

• Well, one initial problem is that a function that is $1$ for all $x\ge 0$ cannot be a PDF either. Apr 12 at 21:23
• sorry I already corrected the post. It was a mistake on my part. Apr 12 at 21:24

The first step is to think about the support of $$Y$$. If $$X \in [0,1]$$, then what is the range of $$Y = f(X) = X/(X+1)$$? When $$X = 0$$, we have $$Y = 0$$. But when $$X = 1$$, then $$Y = 1/2$$. Intuition tells us that we cannot have $$Y > 1/2$$. We can prove this by observing that $$X/(X+1) > 1/2$$ implies $$2X > X+1$$, or $$X > 1$$. So this suggests the support of $$Y$$ is $$Y \in [0,1/2]$$. Now $$\Pr[Y \le y] = \Pr\left[\frac{X}{X+1} \le y\right] = \Pr\left[0 \le X \le \frac{y}{1-y}\right] = \frac{y}{1-y}, \quad 0 \le y \le 1/2.$$ Hence $$f_Y(y) = \frac{1}{(1-y)^2}, \quad 0 \le y \le 1/2.$$ The restricted support is the part of your computation that was missing.