A problem on transformation of a random variable.

Question

$ 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?

Answer 1

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}

Answer 2

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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