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Let $X$ be a Gaussian distributed random variable and $f: \mathbb{R} \to \mathbb{R}^+$ a transformation function. Looking for $f$ that makes $f(x)$ exponentially (or Laplace) distributed.

For instance if $f=x^2$, $f(x)$ is Chi-square distributed, if $f=e^x$, $f(x)$ becomes lognormally distributed, etc.

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If $X$ is a random variable with continuous, strictly increasing CDF $F$, then $F(X)$ is uniformly distributed on $[0,1]$ (this is a really good exercise if you don't already know it). Conversely, if $U$ is uniformly distributed on $[0,1]$, then $F^{-1}(U)$ has the same distribution as $X$. This latter fact can be generalised, but then you need to define $F^{-1}$ properly.

Combining these two observations, if $Z\sim \mathcal{N}(0,1)$, then $\Phi(Z)$ is uniformly distributed on $[0,1]$, where $\Phi$ denotes the standard Gaussian CDF. Note that the CDF of the standard Laplace distribution is $$ F(x)= \begin{cases} \frac{1}{2} e^{x} & x<0 \\ \frac{1}{2}(2-e^{-x}) & x\geq 0 \end{cases} $$ Thus, $$ F^{-1}(u)=\begin{cases} \log(2u) & 0<u< \frac{1}{2} \\ -\log(2(1-u)) & \frac{1}{2}\leq u< 1 \end{cases} $$ and $F^{-1}(\Phi(Z))$ is Laplace distributed. You can find a transformation which yields an exponential distribution in a similar way.

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  • $\begingroup$ This does not answer the question, unless I am missing something. The probability integral transform maps to a uniform for all distribution. How to go from a Normal dist to an exponential/laplace? $\endgroup$
    – Nero
    Commented Jul 3, 2020 at 11:33
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    $\begingroup$ This does answer the question. $F^{-1}\circ \Phi$ above is an explicit map that transforms a standard Gaussian into a Laplace distributed variable. Note the bit starting with "conversely". $\endgroup$ Commented Jul 3, 2020 at 11:37
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    $\begingroup$ The requirement that $F$ is strictly increasing is not needed. That it must be continuous is, or course, crucial. $\endgroup$ Commented Jul 3, 2020 at 11:44
  • $\begingroup$ @kimchi Sure... I guess surjectivity is the only necessary part. And of course, $F^{-1}(U)$ having the same distribution as $X$ is always true if $F^{-1}$ is the unique choice of right continuous "inverse" of $F$. $\endgroup$ Commented Jul 3, 2020 at 11:45
  • $\begingroup$ I see I can use the erf to map to [0,1] then invert according to whatever distribution on needs (sort of Box-Mueller). I was confused by ref to the probability integral transform. Actually I did that initially with an Ito Process. $\endgroup$
    – Nero
    Commented Jul 3, 2020 at 12:09

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