4
$\begingroup$

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?

$\endgroup$
4
  • 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

3
$\begingroup$

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;
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.