A question about characteristic functions A probability 101 question. We know that if two variables $X$ and $Y$ are independent then the characteristic function $\phi_{X+Y}(u)$ can be written as
\begin{equation}
\phi_{X+Y}(u)=\phi_{X}(u)\phi_{Y}(u)
\end{equation}
I have read somewhere "If two variables are not independent the above proposition concerning the characteristic functions involve the characteristic function of the conditional probability distribution."
What does this exactly mean? Does it imply something like $\phi_{X+Y}(u)=\phi_{X}(u)\phi_{Y\mid X}(u)$?
 A: The equation you have given is not quite right, but you are on the right track.  This is fairly simple if you remember that the characteristic function is just an expected value of a complex exponential.  As such, it obeys all the normal rules for the expected value of a function of variable.  In general, the characteristic function of $X+Y$ can be written as:
$$\begin{equation} \begin{aligned}
\phi_{X+Y}(t) 
&\equiv \mathbb{E}(e^{it(X+Y)}) \\[6pt]
&= \mathbb{E}(e^{itX} e^{itY}) \\[6pt]
&= \mathbb{E}(e^{itX}) \cdot \mathbb{E}(e^{itY}) + \mathbb{Cov}(e^{itX}, e^{itY}) \\[6pt]
&= \phi_{X}(t) \cdot \phi_{Y}(t) + \mathbb{Cov}(e^{itX}, e^{itY}). \\[6pt]
\end{aligned} \end{equation}$$
In the  case where $X$ and $Y$ are independent, we have $e^{itX} \bot \ e^{itY}$ which gives $\mathbb{Cov}(e^{itX}, e^{itY})=0$, so we get the simpler rule $\phi_{X+Y}(t) = \phi_{X}(t) \cdot \phi_{Y}(t)$.  Alternatively, taking $\phi_{Y|X}(t|x) \equiv \mathbb{E}(e^{itY} |X=x)$ you can use the law of iterated expectation to get:
$$\begin{equation} \begin{aligned}
\phi_{X+Y}(t) 
&\equiv \mathbb{E}(e^{it(X+Y)}) \\[6pt]
&= \mathbb{E}(e^{itX} e^{itY}) \\[6pt]
&= \mathbb{E}(\mathbb{E}(e^{itX} e^{itY}|X)) \\[6pt]
&= \mathbb{E}(e^{itX} \cdot \mathbb{E}(e^{itY} |X)) \\[6pt]
&= \mathbb{E}(e^{itX} \cdot \phi_{Y|X}(t|X)) \\[6pt]
\end{aligned} \end{equation}$$
Note that the conditional characteristic function is not generally separable from the expectation over $X$ in this case (which is why your equation is wrong).
