Are functions of dependent random variables also dependent?

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I am aware of the fact that functions of independent random variables are also independent, but does similar reasoning apply to dependent variables?







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    Trivial counterexample: take a constant function $f$. (For instance, $f(x)=1$). Then, for any two random variables $X,Y$, $f(X),Y$ are independent, and so are $f(X),f(Y)$.
    – Clement C.
    Aug 9 at 22:54














up vote
0
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favorite












I am aware of the fact that functions of independent random variables are also independent, but does similar reasoning apply to dependent variables?







share|cite|improve this question
















  • 1




    Trivial counterexample: take a constant function $f$. (For instance, $f(x)=1$). Then, for any two random variables $X,Y$, $f(X),Y$ are independent, and so are $f(X),f(Y)$.
    – Clement C.
    Aug 9 at 22:54












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am aware of the fact that functions of independent random variables are also independent, but does similar reasoning apply to dependent variables?







share|cite|improve this question












I am aware of the fact that functions of independent random variables are also independent, but does similar reasoning apply to dependent variables?









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 9 at 22:47









Probably

353




353







  • 1




    Trivial counterexample: take a constant function $f$. (For instance, $f(x)=1$). Then, for any two random variables $X,Y$, $f(X),Y$ are independent, and so are $f(X),f(Y)$.
    – Clement C.
    Aug 9 at 22:54












  • 1




    Trivial counterexample: take a constant function $f$. (For instance, $f(x)=1$). Then, for any two random variables $X,Y$, $f(X),Y$ are independent, and so are $f(X),f(Y)$.
    – Clement C.
    Aug 9 at 22:54







1




1




Trivial counterexample: take a constant function $f$. (For instance, $f(x)=1$). Then, for any two random variables $X,Y$, $f(X),Y$ are independent, and so are $f(X),f(Y)$.
– Clement C.
Aug 9 at 22:54




Trivial counterexample: take a constant function $f$. (For instance, $f(x)=1$). Then, for any two random variables $X,Y$, $f(X),Y$ are independent, and so are $f(X),f(Y)$.
– Clement C.
Aug 9 at 22:54










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The answer is no: Functions of dependent variables need not be dependent. For example, let $X$ and $Y$ be independent standard normal variables. Define $W:=X+Y$. Then $X$ and $W$ are dependent random variables (their covariance is $1$). But $X$ and $W-X$ are functions of $(X,W)$, and they are independent.






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    1 Answer
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    The answer is no: Functions of dependent variables need not be dependent. For example, let $X$ and $Y$ be independent standard normal variables. Define $W:=X+Y$. Then $X$ and $W$ are dependent random variables (their covariance is $1$). But $X$ and $W-X$ are functions of $(X,W)$, and they are independent.






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      up vote
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      accepted










      The answer is no: Functions of dependent variables need not be dependent. For example, let $X$ and $Y$ be independent standard normal variables. Define $W:=X+Y$. Then $X$ and $W$ are dependent random variables (their covariance is $1$). But $X$ and $W-X$ are functions of $(X,W)$, and they are independent.






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        up vote
        2
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        accepted







        up vote
        2
        down vote



        accepted






        The answer is no: Functions of dependent variables need not be dependent. For example, let $X$ and $Y$ be independent standard normal variables. Define $W:=X+Y$. Then $X$ and $W$ are dependent random variables (their covariance is $1$). But $X$ and $W-X$ are functions of $(X,W)$, and they are independent.






        share|cite|improve this answer












        The answer is no: Functions of dependent variables need not be dependent. For example, let $X$ and $Y$ be independent standard normal variables. Define $W:=X+Y$. Then $X$ and $W$ are dependent random variables (their covariance is $1$). But $X$ and $W-X$ are functions of $(X,W)$, and they are independent.







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        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 9 at 22:52









        grand_chat

        18k11121




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