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If you're given the characteristic function of a random variable, say $X$, and the distribution of another, say $U$, which is independent of $X$, how do you explicitly find the characteristic function of $UX$?

(Edit:) This is a problem from an old qualifying exam I'm trying to work through. We're given that the characteristic function of $X$ is $e^{-|t|}$, and that $U$ is uniformly distributed on $(0,1)$, independent of $X$.

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Let $Z = X \, U$. Then $$ \varphi_Z(t) = \mathbb{E}\left( \exp(i t X U) \right) = \mathbb{E}\left( \mathbb{E}\left( \exp(i t X U) | U\right) \right) = \mathbb{E}\left( \varphi_X\left( t U \right) \right) $$ Since $\varphi_X(t) = \exp(-|t|)$, and since $U$ is almost surely positive: $$ \varphi_Z(t) = \mathbb{E}\left( \exp\left( - |t| U \right) \right) = \int_0^1 \exp\left(-|t| \, u \right) \mathrm{d}u $$ You should be able to finish it now.

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  • $\begingroup$ Thanks for the response! I don't see how you got $$\mathbb{E}\left( \exp(i t X U) \right) = \mathbb{E}\left( \mathbb{E}\left( \exp(i t X U) | U\right) \right)$$ but I follow everything else. Do you mind briefly explaining that equality? $\endgroup$ – eeeeeeeeee Jan 3 '13 at 6:04
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    $\begingroup$ I used the law of total expectation. $\endgroup$ – Sasha Jan 3 '13 at 6:16
  • $\begingroup$ Excellent. Thanks again! $\endgroup$ – eeeeeeeeee Jan 3 '13 at 6:19

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