Characteristic function for positive part of random variables I need your help in solving the following problem:
I have to calculate the characteristic function for the positive side of a random variable - simplest case: let $Y$ be standard normal and $Y^+ =\max (0,Y)$.
What is then the characteristic function of $Y^+$? I need to calculate both the direct formula (used as a check), but also the relation between $\phi_t(Y)$ and $\phi_t(Y^+)$ (to be used for more complex examples).
I have tried to get a straightforward answer by computing $\int_{-\infty}^{\infty}e^{ity^+}f(y)\mathrm dy$, by splitting in two integrals, completing the square and getting in the end:
$\phi_t(Y) = \Phi(0) [ 1 - e^{\frac {-t^2} 2 } ]$
but I have no confidence in this result.
Any ideas on how this can be calculated? 
Thank you for any feedback you have,
Mihai
 A: There is no reason why $\varphi_Z(t)=\mathbb E(\mathrm e^{\mathrm itZ})$ with $Z=\max(Y,0)$ should be a simple function of $\varphi_Y(t)=\mathbb E(\mathrm e^{\mathrm itY})$, in general. 
In the specific case when $Y$ is standard normal, the tedious residue computations one imagines yield
$\varphi_Y(t)=\mathrm e^{-t^2/2}$ and
$$
\varphi_Z(t)=\tfrac12+\mathrm e^{-t^2/2}\left(\tfrac12+\mathrm iG(t)\right),\qquad G(t)=\int_0^t\frac1{\sqrt{2\pi}}\mathrm e^{u^2/2}\mathrm du.
$$
Edit: To prove this, write $\varphi_Z(t)=\mathbb P(Y\leqslant0)+\mathbb E(\mathrm e^{\mathrm itY};Y\gt0)=\frac12+\mathbb E(\mathrm e^{\mathrm itY};Y\gt0)$ and
$$
\mathbb E(\mathrm e^{\mathrm itY};Y\gt0)=\int_0^{+\infty}\mathrm e^{\mathrm itx}\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2}\mathrm dx=\mathrm e^{-t^2/2}I(t),
$$
with
$$
I(t)=\int_0^{+\infty}\frac1{\sqrt{2\pi}}\mathrm e^{-(x-\mathrm it)^2/2}\mathrm dx.
$$
Consider the rectangle with vertices $-\mathrm it$, $-\mathrm it+R$, $+R$ and $0$. The integral of the function $z\mapsto\mathrm e^{-z^2/2}$ along the boundary of this rectangle, considered clockwise, is zero, hence
$$
I_R(t)-J_R(t)-I_R(0)+J_0(t)=0,
$$
with
$$
I_R(t)=\int_{-\mathrm it}^{R-\mathrm it}\frac1{\sqrt{2\pi}}\mathrm e^{-z^2/2}\mathrm dz,
\quad
J_R(t)=\int_R^{R-\mathrm it}\frac1{\sqrt{2\pi}}\mathrm e^{-z^2/2}\mathrm dz.
$$
When $R\to+\infty$, $I_R(t)\to I(t)$ and $J_R(t)\to0$, hence
$I(t)=I(0)-J_0(t)$ with
$$
I(0)=\int_0^{+\infty}\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2}\mathrm dx=\frac12,
$$
and, by the change of variable $z=-\mathrm iu$,
$$
J_0(t)=-\mathrm i\int_0^t\frac1{\sqrt{2\pi}}\mathrm e^{u^2/2}\mathrm du.
$$
This completes the proof.
A: An expression for the characteristic function of the random variable $\max(0,X)$ in terms of the Hilbert transform of the characteristic function of an arbitrary random variable $X$ is given in the note just posted at http://arxiv.org/abs/1309.5928 .
