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Suppose that $\phi_{X}(t)$ and $\phi_{Y}(t)$ are characteristic functions of $X, Y$, respectively. Moreover, $X$ and $Y$ are NOT independent random variables. I want to know if $\phi_{X}(t)\cdot\phi_{Y}(t)$ also a characteristic function?

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    $\begingroup$ Here is a hint: characteristic functions are exactly those functions that only take the values 0 and 1. If you multiply two characteristic functions together, what values can they take? Where will the product equal 1? $\endgroup$
    – Aaron
    Mar 11, 2020 at 7:20
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    $\begingroup$ @Aaron The term characteristic function has different meaning in Probability Theory. So your comment is not relevant to this question. Please see the tags in this question. $\endgroup$ Mar 11, 2020 at 7:25

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Let $X',Y'$ be independent random variables such that $X$ and $X'$ have the same distribution and $Y$ and $Y'$ have the same distribution. (Such random variables always exist). Then the characteristic function of $X'+Y'$ is $\phi_X \phi_Y$. So the answer is YES.

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  • $\begingroup$ Thanks for your answer! I have another question: why do such random variables always exist? $\endgroup$
    – MHMH
    Mar 11, 2020 at 9:16
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    $\begingroup$ @MinguQIU We can always construct independent random variables with given distrbution functions. This is based on the construction of product measures. For the most general result in this direction you can read Kolmogorov's Consistency Theorem (aka Kolmogorov's Existence Theorem ) in Wikipedia. $\endgroup$ Mar 11, 2020 at 9:25
  • $\begingroup$ Kolmogorov's Existence Theorem ensures the existence of $X'$ and $Y'$, however, can it provide the independence between $X'$ and $Y'$? $\endgroup$
    – MHMH
    Mar 11, 2020 at 12:54
  • $\begingroup$ Yes, we can instruct the random variables so that that distribution is the product measure. $\endgroup$ Mar 11, 2020 at 13:17
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$\phi_{X+Y}(t) = \mathbb{E}(e^{it(X+Y)}) = \mathbb{E}(e^{itX} \cdot e^{itY}) = \{ \text{independence} \} = \mathbb{E}(e^{itX}) \cdot \mathbb{E}(e^{itY}) = \phi_{X}(t) \cdot \phi_{Y}(t)$.

So if $X \perp Y \Rightarrow \phi_{X+Y}(t) = \phi_{X}(t) \cdot \phi_{Y}(t)$

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  • $\begingroup$ It is not given that $X$ and $Y$ are independent. The point of this question is $\phi_X \phi_Y$ is a characteristic function of some random variable even if $X$ and $Y$ are not independent. $\endgroup$ Mar 11, 2020 at 8:26
  • $\begingroup$ And my answer proves, that it is true in case when $X \perp Y$. So I proved the special case. $\endgroup$
    – Joitandr
    Mar 11, 2020 at 11:29

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