Rectified Normal distribution is a hybrid distribution with the following pdf:

$f(x;\mu ,\sigma ^{2})=\Phi (-{\frac {\mu }{\sigma }})\delta (x)+{\frac {1}{{\sqrt {2\pi \sigma ^{2}}}}}\;e^{{-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}}}{\textrm {U}}(x).$

I want to find the distribution of a sum of independent rectified Normal distributions. I think the best method would be using the characteristic function. But I'm not good with Fourier transforms and cant transform the second term. What is the characteristic function of a rectified Normal distribution?

EDIT: $U$ is the Heaviside step function and $\delta$ is the Dirac delta function

I see that the second term is a unnormalized truncated Normal. If we use the linearity of the Fourier transform, we get something like this (I'm not sure though):

$\hat{f}(\xi)=\Phi (-\frac{\mu}{\sigma})+{\displaystyle e^{\mu (i\xi)+\sigma ^{2}(i\xi)^{2}/2}\left[{\Phi (+\infty -\sigma (i\xi))-\Phi (0 -\sigma (i\xi))}\right]}$

how should I deal with the infinity?

  • 1
    $\begingroup$ This is correctly known as a censored Normal random variable (here censored below at 0). Not only is your nomenclature non-standard, but your pdf also makes no sense unless you define what U is and what $\delta$ is: in any event, requiring two extra functions to describe a censored normal is frankly clumsy, when all you need is a simple If $\endgroup$
    – wolfies
    Apr 18, 2020 at 11:26
  • $\begingroup$ It's not the truncated Normal if that is what you are implying. I have my reasons for prefering this formulation. Do you know the charactristic function? @wolfies $\endgroup$ Apr 18, 2020 at 12:48
  • $\begingroup$ or if you know how the second term is related to the truncated Normal. I can see that the second term is a censored normal, but I'm not sure if I just can replace it with the characteristic function of a truncated Normal. @wolfies $\endgroup$ Apr 18, 2020 at 13:01
  • $\begingroup$ All of the math has been done in this 2018 paper: Beauchamp - "On numerical computation for the distribution of the convolution of N independent rectified Gaussian variables" $\endgroup$ Nov 10, 2020 at 13:58

1 Answer 1


As this is a mixed continuous/discrete distribution , a point mass at zero, and a density above, we use an integral for the density part: $$ \DeclareMathOperator{\E}{\mathbb{E}} \E e^{i t X} = \Phi(-\frac\mu\sigma) e^{i t 0} +\int_0^\infty \frac1{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} e^{i t x}\; dx $$ which (with some help from maple) can be evaluated as $$ 1/2\,{\rm erf} \left(1/2\,{\frac {\sqrt {2} \left( it{\sigma}^{2}+\mu \right) }{\sigma}}\right){{\rm e}^{-1/2\,{\sigma}^{2}{t}^{2}+it\mu}}+ 1/2\,{{\rm e}^{-1/2\,{\sigma}^{2}{t}^{2}+it\mu}} + \Phi(-\frac\mu\sigma). $$ For the record, the maple command is

int( (1/(sqrt(2*Pi*sigma^2)))*exp(-(x-mu)^2/(2*sigma^2))*exp(I*t*x),x=0..infinity ) assuming t,real,mu,real,sigma>0;

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