What is the characteristic function of a rectified Normal distribution?

Question

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?

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;