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Let $X(k)$ be i.i.d. random variables with arbitrary density $f_X$ and mean $\mathbb{E}[X(k)] =\mu$ for $k=0,1,...,N-1$. Let $a \in \mathbb{R}^1$, I would like to compute $$ \mathbb{E}\left[ {\prod\limits_{k = 0}^{N - 1} {\frac{{1 - aX\left( k \right)}}{{1 + aX\left( k \right)}}} } \right]. $$ Below is my thinking: \begin{align*} \mathbb{E}\left[ {\prod\limits_{k = 0}^{N - 1} {\frac{{1 - aX\left( k \right)}}{{1 + aX\left( k \right)}}} } \right] &= \prod\limits_{k = 0}^{N - 1} {\mathbb{E}\left[ {\frac{{1 - aX\left( k \right)}}{{1 + aX\left( k \right)}}} \right]} \hfill \\ &?= \prod\limits_{k = 0}^{N - 1} {\frac{{\mathbb{E}\left[ {1 - aX\left( k \right)} \right]}}{{\mathbb{E}\left[ {1 + aX\left( k \right)} \right]}}} \hfill \\ &= \prod\limits_{k = 0}^{N - 1} {\frac{{1 - a\mathbb{E}\left[ {X\left( k \right)} \right]}}{{1 + a\mathbb{E}\left[ {X\left( k \right)} \right]}}} \hfill \\ &= {\left( {\frac{{1 - a\mu }}{{1 + a\mu }}} \right)^N} \hfill \\ \end{align*} But I am somewhat bothers with my second equality above since I am not clear if it is allowed to take expectation on the denominator and numerator separately under i.i.d. condition. (I was thinking perhaps using iterated conditional expectation + i.i.d. might help me out). Any suggestion is appreciated.

Did
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Fianra
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    I'm pretty sure you're right that that step isn't allowed in general -- I can't think of an explicit counterexample at the moment though. – Chill2Macht Dec 17 '16 at 09:19
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    Two reasons why $$E\left(\frac{1-aX}{1+aX}\right)\ne\frac{E(1-aX)}{E(1+aX)}$$ First, $1-aX$ and $1+aX$ are not independent, second $$E\left(\frac{1}{1+aX}\right)\ne\frac{1}{E(1+aX)}$$ All you can do here is to write $$\mathbb{E}\left[ {\prod\limits_{k = 0}^{N - 1} {\frac{{1 - aX\left( k \right)}}{{1 + aX\left( k \right)}}} } \right] =g(a)^N\qquad g(a)=\mathbb{E}\left[ {\frac{{1 - aX\left( 0 \right)}}{{1 + aX\left( 0 \right)}}} \right]$$ and to compute $$g(a)=\int\frac{1-ax}{1+ax}f_X(x)dx$$ – Did Dec 17 '16 at 11:35
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    @William This is odd that you cannot, any (not almost surely constant) random variable works to disprove the OP's assertion. – Did Dec 17 '16 at 11:36

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