Characteristic function of a product of two dependent random variables such that one is continuous the other is discreet If you're given the characteristic function of a continuous random variable, say $X$, and the distribution of another discreet random variable, say $U$, which is dependent of $X$, how do you explicitly find the characteristic function of $UX$? 
Consider the case that $X\sim N(\mu,\sigma)$ is normal random variable while $$U=\begin{cases}1,&X>3.5\\-1,&X<-0.4\\0.5,&\text{ otherwise}\end{cases}$$
I do not think it is the following, I believe that there might be a misconception or I might have understand it incorrectly. In the 
$$
\begin{eqnarray*}
E\left( e^{itUX}\right)  &=&\int_{-\infty }^{+\infty }e^{itU\left( x\right)
x}f_{UX}\left( x\right) dx \\
&=&\int_{3.5}^{+\infty }e^{itx}f_{X}\left( x\right) dx+\int_{-0.4}^{3.5}e^{it%
\frac{x}{2}}f_{\frac{X}{2}}\left( x\right) dx+\int_{-\infty
}^{-0.4}e^{-itx}f_{-X}\left( x\right) dx
\end{eqnarray*}
$$
 A: More like a hint
$$E\left[ e^{itUX}\right]=E\left[ e^{itUX}\right|X>3.5]P(X>3.5)+E\left[ e^{itUX}|X<-0.4\right]P(X<-0.4)+$$
$$+E\left[ e^{itUX}|-0.4 \le X \le 3.5\right]P(-0.4 \le X \le 3.5).$$ 
Then


*

*$E\left[ e^{itUX}|X>3.5\right]=E\left[ e^{itX}|X>3.5\right],$

*$E\left[ e^{itUX}|X<-0.4\right]=E\left[ e^{-itX}|X<-0.4\right],$

*$E\left[ e^{itUX}|-0.4\le X \le3.5\right]=E\left[ e^{-itX}|X<-0.4\right]$.


In order to calculate the conditional expectations above, we need the conditional cdf's. In general
$$f_{\{X|a\le X \le b\}}=\frac{dF_{\{X|a\le X \le b\}}(x)}{dx}=\begin{cases}0& \text{ if } x<a \\
\frac{f_X(x)}{F_X(b)-F_X(a)}&\text{ if } a\le x \le b\\
1& \text{ if } b<x
\end{cases}$$
where $f_X$  and $F_X$ are the pdf and the cdf of  $N(\mu,\sigma).$
With this, for the first expectation we have 
$$E\left[ e^{itUX}|X>3.5\right]=\frac{1}{\sqrt{2\pi}\sigma}\frac{1}{1-F_X(3.5)}\int_{3.5}^{\infty}e^{itx}e^{\frac{-(x-\mu)^2}{2\sigma^2}}dx.$$
The antiderivative of $e^{itx}e^{\frac{-(x-\mu)^2}{2\sigma^2}}$, according to Alpha, is
$$-\frac{1}{2}\sqrt{\pi}e^{2\mu^2+i\mu t-\frac{t^2}{4}}\text{erf}(\mu+\frac{it}{2}-x),$$
where $\text{erf}(u)=\int_0^u e^{-s^2}ds.$
This result should be checked. For instance, one should check if the integral formula above gives the characteristic function of the normal distribution if the integration limits are $-\infty$ and $\infty$.
From this point on some serious work is still to be done.
