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Let $X$ be an rv with pdf: \begin{equation}f(x)=1,0\leq x\leq\ 1 \end{equation} 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: \begin{equation}P[Y\leq y]=P[X\leq \frac{y}{1-y}]=\frac{y}{1-y} \end{equation} 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.

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

1 Answer 1

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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.

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  • $\begingroup$ Thank you very much! I completely forgot about checking the support of Y! $\endgroup$
    – Dinis P.
    Commented Apr 12 at 21:42

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