Showing $\frac{2X}{1-X^2}$ is standard Cauchy when $X$ is standard Cauchy

Question

If $X\sim\mathcal C(0,1)$, find the distribution of $Y=\frac{2X}{1-X^2}$.

We have $F_Y(y)=\mathrm{Pr}(Y\le y)$

$\qquad\qquad\qquad=\mathrm{Pr}\left(\frac{2X}{1-X^2}\le y\right)$

$\qquad\qquad=\begin{cases} \mathrm{Pr}\left(X\in\left(-\infty,\frac{-1-\sqrt{1+y^2}}{y}\right]\right)+\mathrm{Pr}\left(X\in\left(-1,\frac{-1+\sqrt{1+y^2}}{y}\right]\right),\text{if}\quad y>0\\ \mathrm{Pr}\left(X\in\left(-1,\frac{-1+\sqrt{1+y^2}}{y}\right]\right)+\mathrm{Pr}\left(X\in\left(1,\frac{-1-\sqrt{1+y^2}}{y}\right]\right),\text{if}\quad y<0 \end{cases}$

I wonder if the above case distinction is correct or not.

On the other hand, the following seems a simpler method:

We can write $Y=\tan(2\tan^{-1}X)$ using the identity $\frac{2\tan z}{1-\tan^2z}=\tan 2z$

Now, $X\sim\mathcal C(0,1)\implies\tan^{-1}X\sim\mathcal R\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$

$\qquad\qquad\qquad\quad\implies 2\tan^{-1}X\sim\mathcal R(-\pi,\pi)$

$\qquad\qquad\qquad\quad\implies\tan\left(2\tan^{-1}X\right)\sim\mathcal C(0,1)$, the last one being a 2-to-1 transformation.

But if I am asked to derive the distribution of $Y$ from definition, I guess the first method is how I should proceed. The calculation becomes a bit messy, but do I reach the correct conclusion? Any alternate solution is also welcome.

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Continuous Univariate Distributions (Vol.1) by Johnson-Kotz-Balakrishnan has highlighted this property of the Cauchy distribution. As it turns out, this is just a special case of a general result.

Answer 1

An alternative, more simplistic, way to look at it:

standard Cauchy distribution: $$f(x) \text{d}x = \frac{\pi^{-1}}{x^2+1} \text{d}x $$

transformations of variables: $$ u(x) = \frac{2x}{1-x^2} \qquad \text{and} \qquad x_1(u) = \frac{-1 - \sqrt{u^2+1}}{u} \, , \, x_2(u) = \frac{-1 + \sqrt{u^2+1}}{u} $$

transformation of distribution: $$g(u)\text{d}u = \sum_{i=1,2} f(x_i(u)) \left|\frac{\text{d}x_i}{\text{d}u}\right|\text{d}u $$

If you work with that, which does not need to become so messy, then you will get

$$g(u) = \frac{\pi^{-1}}{u^2+1}$$

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graphical representation

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This sort of works like the identity $\frac{2\tan z}{1-\tan^2z}=\tan 2z$, but written more explicitly.

Or like your representation with the split cumulative distribution function $F_Y(y) = Pr(Y \leq y)$ but now for a split in $f_Y(y) = Pr(y-\frac{1}{2}dy \leq Y\leq y+\frac{1}{2}dy)$.

Answer 2

The transformation in the second approach seems lack of motivation (some details in that also need to be filled up with). Here, from the characteristic function calculation, I am trying to back up your "mysterious" transformation.

The characteristic function of $Y$ can be calculated as follows: \begin{align} \varphi_Y(t) = & E[e^{itY}] = \int_{-\infty}^{\infty} e^{it\frac{2x}{1 - x^2}} \frac{1}{\pi(1 + x^2)} dx \\ = & \frac{1}{\pi}\int_{-\infty}^\infty e^{it\frac{2x}{1 - x^2}} d\arctan x, \end{align} which suggests us trying the transformation $u = \arctan x$, which leads to \begin{align} \varphi_Y(t)= \frac{1}{\pi}\int_{-\pi/2}^{\pi/2} e^{it\frac{2\tan u}{1 - \tan^2 u}} du = \frac{1}{\pi}\int_{-\pi/2}^{\pi/2}e^{it\tan(2u)}du. \tag{1} \end{align}

Our goal is to show that the integral in $(1)$ equals to the characteristic function of a standard Cauchy random variable $X$: \begin{align} \varphi_X(t) = & \int_{-\infty}^\infty e^{itx}\frac{1}{\pi(1 + x^2)} dx \\ = & \frac{1}{\pi}\int_{-\pi/2}^{\pi/2} e^{it\tan u} du \tag{2} \end{align}

Why does the integral in $(1)$ equal to the integral in $(2)$? At the first glance, this is a little counter-intuitive. To verify it, we need to treat the monotonicity of the function $\tan(\cdot)$ carefully. Let's continue to work on $(1)$:

\begin{align} \varphi_Y(t) = & \frac{1}{\pi}\int_{-\pi/2}^{\pi/2}e^{it\tan(2u)}du \\ = & \frac{1}{2\pi}\int_{-\pi}^\pi e^{it\tan v} dv \quad (\text{Change of variable } v = 2u) \\ = & \frac{1}{2\pi}\left[\int_{-\pi}^{-\pi/2} + \int_{-\pi/2}^{\pi/2} + \int_{\pi/2}^{\pi}\right]e^{it\tan u} du \\ = & \frac{1}{2}\varphi_X(t) + \frac{1}{2\pi}\int_{-\pi}^{-\pi/2}e^{it\tan v} dv + \frac{1}{2\pi}\int_{\pi/2}^{\pi}e^{it\tan v} dv \quad (3) \\ = & \frac{1}{2}\varphi_X(t) + \frac{1}{2\pi}\int_{-\pi/2}^{0} e^{-it\tan u_1} du_1 + \frac{1}{2\pi}\int_{0}^{\pi/2} e^{-it\tan u_2} du_2 \quad (4) \\ = & \frac{1}{2}\varphi_X(t) + \frac{1}{2\pi}\int_{-\pi/2}^{\pi/2} e^{-it\tan v} dv \\ = & \varphi_X(t) \quad (5) \end{align}

$(3)$: Because the function $u \mapsto \tan(u)$ is not monotone on the interval $(-\pi, \pi)$, I made such division such that each integrand is monotone on the separated interval (which ensures subsequent change of variable formulae valid).

$(4)$: The two change of variable formulae are $u_1 = -\pi - v$ and $u_2 = \pi - v$.

$(5)$: Last change of variable formula $u = -v$.

The steps $(3)$--$(5)$ elaborated the statement "the last one being a 2-to-1 transformation" in OP's question.