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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 $\phi_{X+Y}(u)=\phi_{X}(u)\phi_{Y\mid X}(u)$?

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  • $\begingroup$ I'd have to think through the formalities of notation, but I think you are in the right ballpark. When you add two independent RVs, you really have a convolution integral. The characteristic function is the Fourier transform of the density functions, and you can do convolution by taking a Fourier transform, adding pointwise, and doing the reverse Fourier transform. That's why your first equation is correct if they are uncorrelated. If they are correlated, it means that the distribution of Y depends on the specific value of X, and vice-versa. $\endgroup$ – eSurfsnake Dec 14 '19 at 6:11
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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).

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  • $\begingroup$ Thank you so much! That was helpful. $\endgroup$ – Carl Dec 14 '19 at 17:25

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