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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 $\text{sign}()$:

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

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  • $\begingroup$ The function is an odd function, so you can do it just for $x>0$, whence we just have $y = -\frac 12\ln(1-x^2)$.(The square root , when eventually taken to determine $x$ will be positive). $\endgroup$ Sep 7, 2020 at 7:00
  • $\begingroup$ the derivation without the sign multiplier is easy and well-known, so I would like to learn how to do it for the function shown instead ($x\in \mathbb{R}$ not just $x>0$) $\endgroup$
    – develarist
    Sep 7, 2020 at 7:09
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    $\begingroup$ The domain of $y$ is $x \in (-1,1)$,since otherwise $\ln (1-x^2)$ would be undefined. My comment still stands : break $y$ into a piecewise function by breaking the $\mbox{sign}$ into one, then find the piecewise inverses and put them together. In a follow up question you have asked for the derivative : you can get it everywhere except at $0$ where you need to create the differential quotient. The sign multiplier is handled by breaking the function into pieces, so that on each piece we know what to do. $\endgroup$ Sep 7, 2020 at 14:18
  • $\begingroup$ is it possible to isolate $x$ without resorting to piece-wise functions, I meant to add that to the question and will do so now $\endgroup$
    – develarist
    Sep 7, 2020 at 14:23
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    $\begingroup$ Thanks for the edits, I now see what you want to do. Here's something : note that $sign(x) = sign(y(x))$ for all $x$ (proved by noting that $\frac{sign(x)}{x} = \frac 1{|x|}$ is always positive for $x \neq 0$, so the quotient $\frac{y}{x}$ is also always positive for $x \neq 0$), so you can replace $sign(x)$ with $sign(y)$. This should help. Let us discuss this. $\endgroup$ Sep 7, 2020 at 16:23

3 Answers 3

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

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

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

therefore

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

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  • $\begingroup$ this omits all the actual content of the function and is incomplete since $x$ is not isolated, could you show the line-by-line operations without masking with chain rule $\endgroup$
    – develarist
    Sep 7, 2020 at 7:25
  • $\begingroup$ what's the problem? The $sign$ function is $1$ for x >0 and $-1$ for $x <0$ $\endgroup$
    – Tortar
    Sep 7, 2020 at 15:47
  • $\begingroup$ @Tortar what does that mean for the first derivative of $\text{sign}(x)$? $\endgroup$
    – develarist
    Sep 7, 2020 at 16:43
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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

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

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  • $\begingroup$ doesn't differenting the equation cause an unwanted called $y'$? I would like to retain $y$ as is while isolating $x$ $\endgroup$
    – develarist
    Sep 7, 2020 at 16:44
  • $\begingroup$ you singled out a single component of the equation while ignoring two others: $y$ and $\text{sign}(x)$. if i take the derivative of the entire equation, i have to deal with a new quantity called $y'$, which is the first derivative of variable $y$. I would like to retain $y$ as is, while isolating $x$ $\endgroup$
    – develarist
    Sep 7, 2020 at 16:49
  • $\begingroup$ read the question again, especially the first two formulas $\endgroup$
    – develarist
    Sep 7, 2020 at 16:51
  • $\begingroup$ @develarist: oops, sorry, because of the other answer, I was on the way of a differentiation question. That was silly ! $\endgroup$
    – user65203
    Sep 7, 2020 at 16:52

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