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How do you prove that the real part of the characteristic function of the continuous probability distribution $f(x)$ is a characteristic function, but the imaginary part is not?

The second part is simple, and, if I am correct, the solution is below: ${\mathop{\rm Im}\nolimits} \left( {{f_\xi }(0)} \right) = 0$, and it is impossible for the correspondent candidate for characteristic function $h(t) = {\mathop{\rm Im}\nolimits} \left( {f(t)} \right)$.

But could you help, please, how to prove the first part? There is a solution through Bochner’s theorem, but it seems there is a better solution.

Thank you in advance!

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    I edited the post to make it more readable. I hope I did not change the intended meaning of your question.2011-12-08

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If $\varphi$ is the characteristic function of $X$, then $\Re(\varphi)$ is the characteristic function of a symmetrized version $X_{\rm sym}$ of $X$.

That is, $X_{\rm sym}=\varepsilon X$, where $\varepsilon$ is independent of $X$ and $\mathbb{P}(\varepsilon=+1)=\mathbb{P}(\varepsilon=-1)=1/2$.

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    Glad to help. $ $2011-12-08
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Suppose that $\psi(t)$ is the characteristic function corresponding to $f(x)$. Using the properties of Fourier transforms, we have that

  • since $f(x)$ is a real-valued function, $\psi(t)$ has conjugate symmetry: $\psi(-t) = [\psi(t)]^*.$

  • the characteristic function corresponding to $f(-x)$, the density of $-X$, is $\psi(-t)$.

Now, $\dfrac{1}{2}(f(x)+f(-x))$ is a density and the corresponding characteristic function is $\frac{1}{2}(\psi(t) + \psi(-t)) = \frac{1}{2}(\psi(t) + [\psi(t)]^*) = \text{Re}(\psi(t)).$

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    You are welcome. Only one answer can be accepted for each question, and Byron Schmuland's answer is much shorter, more to the point, and was posted earlier than mine anyway (it appeared as I was finishing up writing mine). So it is not too surprising that you accepted his answer.2011-12-08