How to calculate the inverse function of $y=-\frac{1}{2} \ln(1-x^2) \times \text{sign}(x)$? 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 $\text{sign}()$:
$$ -2 y = \ln(1-x^2) \times \text{sign}(x)$$
$$ \exp(-2y) = (1-x^2) \times \exp(\text{sign}(x))$$
 A: We can use chain rule noting that for $x\neq 0$
$$(\text{sign}(x))'=0$$
therefore
$$(f(x)\cdot \text{sign}(x))'=f'(x)\cdot \text{sign}(x)$$
A: 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
$$x=\text{sgn}(y)\sqrt{1-e^{-2|y|}}.$$
A: There are 2 cases:

*

*$x \geq 0$:
\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*}

*$x < 0$:
\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:
$$x=\text{sign}(y)\sqrt{1-e^{-2|y|}}$$
