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?
probability probability-theory
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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?
probability probability-theory
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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 $f(X),f(Y)$.
â Clement C.
Aug 9 at 22:54
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up vote
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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?
probability probability-theory
I am aware of the fact that functions of independent random variables are also independent, but does similar reasoning apply to dependent variables?
probability probability-theory
asked Aug 9 at 22:47
Probably
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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 $f(X),f(Y)$.
â Clement C.
Aug 9 at 22:54
add a comment |Â
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 $f(X),f(Y)$.
â 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 $f(X),f(Y)$.
â 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 $f(X),f(Y)$.
â Clement C.
Aug 9 at 22:54
add a comment |Â
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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
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
add a comment |Â
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.
add a comment |Â
up vote
2
down vote
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.
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.
answered Aug 9 at 22:52
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18k11121
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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 $f(X),f(Y)$.
â Clement C.
Aug 9 at 22:54