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If you're given the characteristic function of a continuous random variable, say X, and the distribution of another discreet random variable, say U, which is dependent of X, how do you explicitly find the characteristic function of UX? Consider the case that X is normal random variable while U is a random variable defined as

$$ U=\begin{cases} v & \text{if }X<1 \\ r & \text{ }Otherwise \end{cases} $$

where $$ v=\begin{cases} 1 & \text{with probability }\frac{1}{2} \\ -1 & \text{with probability }\frac{1}{2}% \end{cases} $$

and $$ r=\begin{cases} 0.25 & \text{with probability }0.75 \\ 0.7& \text{with probability }0.25% \end{cases} $$

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  • $\begingroup$ Any reference book or some additional help is appreciated $\endgroup$
    – Anonymous
    May 5, 2015 at 14:16

1 Answer 1

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Can you determine $P(X < 1)$? If you can, let $p = P(X < 1)$, and then

$$ U = \begin{cases} \hfill 1 \hfill & \text{with probability } \frac{p}{2} \\ \hfill -1 \hfill & \text{with probability } \frac{p}{2} \\ \hfill \frac{1}{4} \hfill & \text{with probability } \frac{3(1-p)}{4} \\ \hfill \frac{7}{10} \hfill & \text{with probability } \frac{1-p}{4} \end{cases} $$

and you can continue as usual.

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  • $\begingroup$ Thanks Brian for your attempt to help me, but I do not think what you mentioned is feasible. The reason for this is because when you want to account for the Characteristic function the boundaries of the integral changes depending on X. $\endgroup$
    – Anonymous
    May 5, 2015 at 16:27
  • $\begingroup$ Ahh, you want the characteristic function of $UX$, not $U$. Sorry, I didn't read carefully. All right, I'll give that a bit of thought. What are the mean and variance of $X$? $\endgroup$
    – Brian Tung
    May 5, 2015 at 16:34
  • $\begingroup$ suppose X is the standard normal say 0 and 1 $\endgroup$
    – Anonymous
    May 5, 2015 at 16:59

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