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 $$ \begin{eqnarray*} E\left( e^{itUX}\right) &=&\int_{-\infty }^{+\infty }e^{itU\left( x\right) x}f_{UX}\left( x\right) dx \\ &=&\int_{3.5}^{+\infty }e^{itx}f_{X}\left( x\right) dx+\int_{-0.4}^{3.5}e^{it% \frac{x}{2}}f_{\frac{X}{2}}\left( x\right) dx+\int_{-\infty }^{-0.4}e^{-itx}f_{-X}\left( x\right) dx \end{eqnarray*} $$

  • 2
    $\begingroup$ Conditional expectation is your friend here. $\endgroup$
    – Calculon
    May 6, 2015 at 14:19
  • $\begingroup$ yeah it really is, but I did not see it working. What I mentioned above is not correct I believe. I do not think it is correct, do you? if so may you provide me with a reference? $\endgroup$
    – Anonymous
    May 6, 2015 at 14:21
  • $\begingroup$ $E[f(X,U) \mid X > 3.5]$, $E[f(X,U) \mid X < -0.4]$, $E[f(X,U) \mid -0.4 \leq X \leq 3.5]$. First calculate these three. That gives you $E[f(X,U) \mid X]$ as a random variable. Next, make use of this: $E[E[f(X,U) \mid X ]] = E[f(X,U)]$. Is this clear enough? By $f$ I mean whatever function you want to compute the expectation of. In your case it is the characteristic function of $UX$. $\endgroup$
    – Calculon
    May 6, 2015 at 14:24
  • $\begingroup$ what is $f(X,U)$ ? is it $f(X,U)= e^{itUX}$ while the other is summing if I am not mistaken right? Thank you very much Calculon for your help really appreciated $\endgroup$
    – Anonymous
    May 6, 2015 at 14:26

1 Answer 1


More like a hint

$$E\left[ e^{itUX}\right]=E\left[ e^{itUX}\right|X>3.5]P(X>3.5)+E\left[ e^{itUX}|X<-0.4\right]P(X<-0.4)+$$ $$+E\left[ e^{itUX}|-0.4 \le X \le 3.5\right]P(-0.4 \le X \le 3.5).$$


  • $E\left[ e^{itUX}|X>3.5\right]=E\left[ e^{itX}|X>3.5\right],$
  • $E\left[ e^{itUX}|X<-0.4\right]=E\left[ e^{-itX}|X<-0.4\right],$
  • $E\left[ e^{itUX}|-0.4\le X \le3.5\right]=E\left[ e^{-itX}|X<-0.4\right]$.

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.


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