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If the characteristic function of a random variable is a real-valued function, does this imply that the random variable must be symmetric about zero?

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Yes. When the characteristic function $\phi_X$ of a random variable $X$ is real-valued, that means

$$\overline{\phi_X}(t) = \phi_X(t)$$

for all $t$. But for any characteristic function

$$\phi_{-X}(t) = \phi_X(-t) = \overline{\phi_X}(t),$$

showing that $X$ and $-X$ have identical distributions (since the cf determines the distribution), implying $X$ is symmetric about zero. (The converse is equally clear.)

These facts are given in the Wikipedia article, which also mentions this result.

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