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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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Can you please describe what you have tried and where you are getting stuck? Is this a homework problem (that is okay), but please (tag) it as such? Regards –  Amzoti Jan 3 '13 at 4:41
    
This isn't a homework problem. I –  pedrosuavo Jan 3 '13 at 4:43
    
(Sorry, I pressed enter prematurely.) This isn't a homework problem - I'm trying to work out problems from old qualifying exams. I took a probability course last semester, and am still rusty. The characteristic function of X is exp(-|t|), and U is uniformly distributed on (0,1), independent of X. I tried computing E[exp{itUX}], and integrated out the marginal pdf of U in the hopes of getting something worthwhile, to no avail. I'm guessing there's something simple I don't know, and the two prob/stat books I've been referring to don't have much information about characteristic functions. –  pedrosuavo Jan 3 '13 at 4:53
    
PedroSuavo, no problem. It might help if you update you question with some of those additional tidbits. Regards –  Amzoti Jan 3 '13 at 4:59
    
Will do. Thanks for the suggestion. –  pedrosuavo Jan 3 '13 at 4:59
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1 Answer 1

up vote 4 down vote accepted

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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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? –  pedrosuavo Jan 3 '13 at 6:04
    
I used the law of total expectation. –  Sasha Jan 3 '13 at 6:16
    
Excellent. Thanks again! –  pedrosuavo Jan 3 '13 at 6:19
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