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Random variables $x_1, x_2,...,x_n$ are independent. Then I want to show whether these functions $$y_1=f_1(x) \\ y_2=f_2(x) \\ ... \\ y_n=f_n(x)$$ are independent or not .

How to prove this?

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closed as unclear what you're asking by whuber Nov 14 '14 at 17:17

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ This question makes no sense: after introducing $n$ variables, you then ask about $n$ otherwise undefined functions. So that others are not lured into answering a question you did not mean to ask, I will close this until you have edited the question to your satisfaction, it is unlikely to change further, and then the community--if it agrees the question can be answered--votes to reopen it. $\endgroup$ – whuber Nov 14 '14 at 17:17
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Write down the joint distribution of all variables, marginalize out the $x$'s, and show that the result factorizes over the $y$'s.

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  • $\begingroup$ Sorry I made a mistake while writing functions. These should be eqaul to $f(x)$. Please explain the proof more explicitly for this changed functions. $\endgroup$ – Bb11 Nov 14 '14 at 15:43
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    $\begingroup$ The modified question does not make sense. I think the original question was the right one. $\endgroup$ – Tom Minka Nov 14 '14 at 15:56
  • $\begingroup$ My instructor said that for example $y_1= x_1+x_3+x_5$ other $y_2, y_3,...$ are different combinations of $x_i$ thus, $y_i$ is equl to $f(x)$. @TomMinka $\endgroup$ – Bb11 Nov 14 '14 at 16:26
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    $\begingroup$ You cannot prove independence for this modified question since the $y$'s are not going to be independent in general. $\endgroup$ – Tom Minka Nov 14 '14 at 17:12
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    $\begingroup$ Because if $f_1 = f_2$ then obviously $y_1$ and $y_2$ are dependent. This modified question really doesn't make sense at all. $\endgroup$ – Tom Minka Nov 14 '14 at 17:17

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