Suppose X$\sim$ Cauchy(0,1). Then what will be the distribution of $\frac{1-X}{1+X}$?

In order to find distribution of $$\frac{1-X}{1+X}$$ below approach I followed,

Let, \begin{align} Y = \frac{1-X}{1+X} \end{align} Then, cdf of Y is \begin{align} F_{Y}(y) = P(Y \leq y) \end{align} \begin{align} = P\left(\frac{1-X}{1+X} \leq y\right) \end{align} \begin{align} = 1 - P\left(X < \frac{1-y}{1+y}\right) \end{align} \begin{align} = 1 - \int_{-\infty}^{\frac{1-y}{1+y}} f(x) \,dx \end{align} \begin{align} = 1 - \int_{-\infty}^{\frac{1-y}{1+y}} \frac{1}{\pi}\cdot \frac{1}{1+x^2} \,dx \end{align} \begin{align} = 1 - \frac{1}{\pi}\cdot \left[tan^{-1}x\right]_{-\infty}^{\frac{1-y}{1+y}} \end{align} \begin{align} F_{Y}(y) = \frac{1}{\pi}\cdot \left[-\frac{\pi}{2}+tan^{-1}\left({\frac{1-y}{1+y}}\right)\right] = \frac{1}{2} -\frac{1}{\pi}.tan^{-1}\left({\frac{1-y}{1+y}}\right) \end{align} and then \begin{align} f_{Y}(y) = \frac{d F_{Y}(y)}{dy} = \frac{1}{\pi}\cdot \frac{1}{1+y^2} \end{align}

But I have a little confusion here how to find range of Y from X? And CDF of Y is doesn't looks like cdf of a cauchy distribution.

• You got exactly the same distribution you started with, so the range is all real line. Feb 6, 2022 at 17:38
• Hi, sorry to bother but I really can't understand how did you conclude $1 - P\left(X < \frac{1-y}{1+y}\right)$ from $P\left(\frac{1-X}{1+X} \leq y\right)$? The support of $x$ is $x\in (-\infty,\infty)$, $1+x$ might be negative! Feb 6, 2022 at 18:34

This fact is useful in thinking about your problem: If $$U,V\sim N(0,1)$$ are independent standard normals, their ratio $$X=U/V$$ has the standard Cauchy distribution. Then $$Y=(1-X)/(1+X) = (V-U)/(V+U) = S/T$$ where $$S=(V-U)/\sqrt 2$$ and $$T=(V+U)/\sqrt 2$$. But $$S$$ and $$T$$ are also independent standard normals, so $$Y$$ has the same distribution as $$X$$.

Equivalently, one can represent the iid variables $$U,V\sim N(0,1)$$ as $$U=R\cos\Theta, V=R\sin\Theta$$, where $$R,\Theta$$ are independent, with $$R$$ Rayleigh distributed and $$\Theta$$ uniform on $$[0,2\pi)$$, so $$X=\tan\Theta$$ and $$Y=(1-X)/(1+X)=\tan(\pi/4-\Theta)$$. Since $$\Theta$$ is uniform, so is $$\pi/4-\Theta$$, so $$Y$$ has the same distribution as $$X$$.

Let $$g(x) = (1-x)/(1+x)$$. Then $$g$$ is self-inverse; i.e., $$g^{-1} = g$$, but it is not everywhere monotone. Thus, consider $$\Pr[g(X) \le y]$$ for the case $$y \le -1$$ versus $$y > -1$$ separately. In the first case, $$g$$ is monotone decreasing on $$X \in (-\infty, -1)$$ and we have $$\Pr[g(X) \le y] = \Pr[g^{-1}(y) \le X < -1] = \Pr[g(y) \le X < -1],$$ with no issues (the inequality reverses because $$g$$ is order-reversing). You can check this with a numerical example; e.g., for $$y = -2$$, $$\frac{1-x}{1+x} \le -2 \iff -3 \le x \le -1.$$

However, for $$y > -1$$, we see that the inequality is compound: \begin{align} \Pr[g(X) \le y] &= \Pr[-\infty < X < -1] + \Pr[g^{-1}(y) \le X < \infty] \\ &= 1 - \Pr[-1 < X \le g(y)]. \end{align}

Hence we have $$F_Y(y) = \int_{x=g(y)}^{-1} f_X(x) \, dx$$ when $$y \le -1$$ and $$F_Y(y) = \int_{x = -\infty}^{-1} f_X(x) \, dx + \int_{x=g(y)}^\infty f_X(x) \, dx = 1 - \int_{x=-1}^{g(y)} f_X(x) \, dx$$ when $$y > -1$$. I leave the rest of the computation as an exercise.

• Why are you considering for y=-1 if g(x) is not defined at -1? Feb 8, 2022 at 3:48
• @DaeHyun Because $(1-x)/(1+x) \le -1$ is still a valid inequality even if the function $(1-x)/(1+x)$ is not well-defined at $x = -1$. Feb 8, 2022 at 4:39

$$P((1-X)/(X+1)\leq t)=\begin{cases} P(\{X\leq -1\}\cup\{X\geq (1-t)/(1+t)\})&t>-1\\ P(\{(1-t)/(1+t)\leq X\leq -1\}& t<-1 \end{cases}$$ So for $$t>-1$$ $$P((1-X)/(X+1)\leq t)=\frac{1}{2}+\frac{1}{\pi}\arctan(-1)+\frac{1}{2}-\frac{1}{\pi}\arctan{\frac{1-t}{1+t}}=\\\frac{3}{4}-\frac{1}{\pi}\arctan{\frac{1-t}{1+t}}$$ while for $$t<-1$$ $$P((1-X)/(X+1)\leq t)=\frac{1}{2}+\frac{1}{\pi}\arctan(-1)-\frac{1}{2}-\frac{1}{\pi}\arctan{\frac{1-t}{1+t}}=\\-\frac{1}{4}-\frac{1}{\pi}\arctan\frac{1-t}{1+t}$$

• How can we ignore t=-1 for cdf since it is a continuous function? Feb 8, 2022 at 3:46
• @DaeHyun it's a removable discontinuity Feb 8, 2022 at 10:41

Remark (Theorem of transformation for one random variable): Let be $$X$$ a continuous r.v. with pdf given by $$f_X$$ and let be Y=g(X) with g a diffeomorphism, then the pdf $$f_Y$$ is given by:
$$f_Y(y)=f_X(g^{-1}(y))\cdot|{{d\over{dy}}g^{-1}(y)}|$$.

If $$Y=g(X)={{1+X}\over{1-X}}$$ then $$X=g^{-1}(Y)={{1-Y}\over{1+Y}}$$. We find that $${dx\over{dy}}=-{{2\over{1+y^2}}}$$. Now:
$$f_Y(y)=f_X({{1-y}\over{1+y}})\cdot{2\over{1+y^2}}$$.
I'll let you do the math, then you get:
$$f_Y(y)={1\over{\pi{(1+y^2)}}}$$.
So $$Y\sim{Cauchy(0,1)}$$.

• The function $g(x) = (1-x)/(1+x)$ is not differentiable at $x = -1$. Feb 6, 2022 at 19:48