How to calculate the inverse function of $y=-\frac{1}{2} \ln(1-x^2) \times \text{sign}(x)$?

Question

How can the continuous random variable $x$ by isolated by itself on one side of the following equation

$$y = -\frac{1}{2} \ln(1-x^2) \times \text{sign}(x)$$

without resorting to a piece-wise equation?

$$ x = ?$$

Below is my initial, incomplete and probably wrong attempt since I don't know the exponential of a product or the exponential of $\text{sign}()$:

$$ -2 y = \ln(1-x^2) \times \text{sign}(x)$$ $$ \exp(-2y) = (1-x^2) \times \exp(\text{sign}(x))$$

Answer 1

There are 2 cases:

1. $x \geq 0$: \begin{align*} y&=-\frac{1}{2}ln(1-x^2)\\ -2y &= ln(1-x^2)\\ e^{-2y} &= 1-x^2\\ x^2 &= 1-e^{-2y}\\ x &= +\sqrt{1-e^{-2y}} \qquad :\text{since } x \geq 0 \end{align*}
2. $x < 0$: \begin{align*} y&=\frac{1}{2}ln(1-x^2)\\ 2y &= ln(1-x^2)\\ e^{2y} &= 1-x^2\\ x^2 &= 1-e^{2y}\\ x &= -\sqrt{1-e^{2y}} \qquad :\text{since } x < 0 \end{align*}

Now, as stated in the comments, you notice that $y(x)$ has the same sign as $x$, i.e: $$\text{sign}(x) = \text{sign}(y(x))$$ So, the different formulas for $x$ can be unified using $\text{sign}(y)$, as follows: $$x=\text{sign}(y)\sqrt{1-e^{-2y.\text{sign}(y)}}$$ Also, since $y.\text{sign}(y)=|y|$, we can write $x$ as: $$x=\text{sign}(y)\sqrt{1-e^{-2|y|}}$$

Answer 2

We can use chain rule noting that for $x\neq 0$

$$(\text{sign}(x))'=0$$

therefore

$$(f(x)\cdot \text{sign}(x))'=f'(x)\cdot \text{sign}(x)$$

Answer 3

From

$$y=-\frac12\ln(1-x^2)\text{ sgn}(x)$$ we can draw

$$x=\text{sgn}(x)\sqrt{1-e^{-2y\text{ sgn(x)}}}$$ because the square root is a positive number. But the function is odd, $\text{sgn}(x)=\text{sgn}(y)$, and

$$x=\text{sgn}(y)\sqrt{1-e^{-2|y|}}.$$