# Transforming a Gaussian r.v. into an exponential/Laplace

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.

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, $$F^{-1}(u)=\begin{cases} \log(2u) & 0 and $$F^{-1}(\Phi(Z))$$ is Laplace distributed. You can find a transformation which yields an exponential distribution in a similar way.

• 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?
– Nero
Jul 3, 2020 at 11:33
• 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". Jul 3, 2020 at 11:37
• The requirement that $F$ is strictly increasing is not needed. That it must be continuous is, or course, crucial. Jul 3, 2020 at 11:44
• @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 $F$. Jul 3, 2020 at 11:45
• 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.
– Nero
Jul 3, 2020 at 12:09