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$ f(x) = \begin{cases} 1 & \text{if $0\le x \le 1$ } \\ 0 & \text{otherwise }\end{cases}$

Find the PDF of Random variable $Y=\dfrac{X}{1+X}$

$F(Y\le y)=F\bigg(\dfrac{X}{1+X}\le y\bigg)=F\bigg(X\le \dfrac{y}{1-y}\bigg)$

$U=F_x\bigg(\dfrac{y}{1-y}\bigg)$

$\dfrac{dU}{dy}=f(y)\bigg(\dfrac{1}{(1-y)^2}\bigg)$

$f(y)=\bigg(\dfrac{1}{(1-y)^2}\bigg)$

Please check if I did any mistake. I am not sure how to figure out domain after transformation. Heres my try.

$0\le x \le 1$

$1\le x+1\le2$

$1\ge \dfrac{1}{x+1}\ge\dfrac{1}{2}$

$x\ge \dfrac{x}{x+1}\ge\dfrac{x}{2}$

$x\ge y\ge\dfrac{x}{2}$

Is this a correct way to approach result?

Deepesh Meena
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DSR
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  • Editing question after looking at someone's answer is not allowed on this site – Deepesh Meena Sep 02 '18 at 17:12
  • @James really ? It's $1$ I know. The only mistake was typo. I wrote the whole pdf see man. – DSR Sep 02 '18 at 17:12
  • @James Write down whole pdf then match may be. – DSR Sep 02 '18 at 17:14
  • Yes you can not keep editing the question like that you asked for the mistake I pointed it out and you edited it than what is the meaning o my answer if a new person comes and read my answer what he will think he will think you have not done any mistake I am wrong – Deepesh Meena Sep 02 '18 at 17:15
  • @James No james write the whole pdf in your answer FINAL ANSWER. – DSR Sep 02 '18 at 17:17
  • so you are asking me too write down the whole answer?? – Deepesh Meena Sep 02 '18 at 17:18
  • @James Is it fine now? – DSR Sep 02 '18 at 17:18
  • @James I am saying substitute the value of $f(\frac{y}{1-y})$ – DSR Sep 02 '18 at 17:19
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    As an alternative to the CDF method, you can simply solve $x$ in terms of $y$ and use the transformation formula (If your $Y$ was defined with higher powers of $X$, this would have been more straightforward. For the problem at hand, it doesn't matter)

    $$y=\frac{x}{1+x} \implies x=\frac{y}{1-y}$$

    Naturally, $$0<x<1\implies 0<y<1/2$$

    So the density of $Y$ is

    \begin{align} g(y)&=f\left(\frac{y}{1-y}\right)\left|\frac{dx}{dy}\right| \&=\frac{1}{(1-y)^2}\mathbf1_{0<y<1/2} \end{align}

     – StubbornAtom Sep 02 '18 at 18:06
  • @StubbornAtom I just want to ask. When I tried to calculate the domain of $y$. What was wrong in my method? Why it didn't come out to be $0\le y\le\frac{1}{2}$ – DSR Sep 02 '18 at 18:31
  • @Damn1o1 Don't just rely on mechanical manipulations. When $x$ is $0$, $y$ is $0$ and when $x=1$, $y=1/2$; and $y$ is continuous throughout. You could draw a picture to convince yourself. – StubbornAtom Sep 02 '18 at 18:47
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    @StubbornAtom I got it but I want to know mistake where it didn't go well. Can I say the way I calculated never works all the time ? – DSR Sep 02 '18 at 18:52
  • @StubbornAtom Are you studying from ISI? – DSR Sep 02 '18 at 18:53
  • @StubbornAtom You are very good in statistics man and I am struggling here pretty hard. Which book do you recommend? – DSR Sep 02 '18 at 18:59

2 Answers2

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There are simple and practical reasons why one writes $F_X(x)$ and not $F_x(X)$ or $F_x(x)$ or $F_X(X).$ If one does not know about this, how would one even understand something like $\Pr(X\le x)\text{ ?}$ \begin{align} \text{For } 0\le y \le \frac 1 2,\text{ we have } & \frac X {1+X} = 1 - \frac 1 {1+X} \le y \\[10pt] \text{iff } & \frac 1 {1+X} \ge 1 - y \\[10pt] \text{iff } & 1 + X \le \frac 1 {1-y} \\[10pt] \text{iff } & X \le \frac 1 {1-y} - 1 = \frac y {1-y}. \\[10pt] \text{Therefore for } 0 \le y \le \frac 1 2, \quad & \Pr\left( \frac X{1+X} \le y \right) = \Pr\left( X\le \frac y {1-y} \right) \\[10pt] = {} & \int_0^{y/(1-y)} 1\, dx = \frac y {1-y}. \\[12pt] \text{Hence } & f_Y(y) = \frac d {dy}\, \frac y {1-y} = \cdots. \end{align}

Michael Hardy
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So you are asking for the pdf

$$F(Y\le y)=F(X\le \frac{y}{1-y})$$ so you can write

$$F(Y\le y)=\int_0^{\frac{y}{1-y}} 1 \, dx=\frac{y}{1-y}$$

$$f_Y(y)=\frac{1}{(1-y)^2}$$

$f_Y(y)$ is your PDF

$\frac{1}{0}$ is not defined thus $$(1-y)^2\ne 0$$ $$1-y \ne 0$$ $$y \ne 1$$

also, you want $$0\le\frac{y}{1-y}\le 1$$ Solve it to get your domain

Deepesh Meena
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