Is there a possibility that $|Corr(X,Y)|=1$ and there's no linear relationship between them?
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0
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Let us have $|Corr(X,Y)|=1$.
When is it possible for the variables $X,Y$ not be linearly related?
I would say that they must have infinite moments... But I'm not seeing if that's enough to get the result I want.
probability
asked Nov 16 at 20:37
An old man in the sea.
1,60411031
add a comment |
up vote
0
down vote
favorite
Let us have $|Corr(X,Y)|=1$.
When is it possible for the variables $X,Y$ not be linearly related?
I would say that they must have infinite moments... But I'm not seeing if that's enough to get the result I want.
probability
asked Nov 16 at 20:37
An old man in the sea.
1,60411031
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let us have $|Corr(X,Y)|=1$.
When is it possible for the variables $X,Y$ not be linearly related?
I would say that they must have infinite moments... But I'm not seeing if that's enough to get the result I want.
probability
asked Nov 16 at 20:37
An old man in the sea.
1,60411031
Let us have $|Corr(X,Y)|=1$.
When is it possible for the variables $X,Y$ not be linearly related?
I would say that they must have infinite moments... But I'm not seeing if that's enough to get the result I want.
probability
probability
asked Nov 16 at 20:37
An old man in the sea.
1,60411031
asked Nov 16 at 20:37
An old man in the sea.
1,60411031
asked Nov 16 at 20:37
An old man in the sea.
1,60411031
asked Nov 16 at 20:37
An old man in the sea.
1,60411031
asked Nov 16 at 20:37
An old man in the sea.
1,60411031
1,60411031
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
By the Cauchy-Schwarz inequality, $X$ is a constant multiple of $Y$ in this case.
answered Nov 16 at 20:43
Mike Hawk
1,312110
We can use Cauchy-Schwarz only if we have $E(XY)$ finite.
– An old man in the sea.
Nov 16 at 20:52
We can have $operatorname{Corr}(XY)$ only if $E(XY)$, $E(X^2)$, and $E(Y^2)$ are all finite.
– kimchi lover
Nov 16 at 21:03
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
By the Cauchy-Schwarz inequality, $X$ is a constant multiple of $Y$ in this case.
answered Nov 16 at 20:43
Mike Hawk
1,312110
We can use Cauchy-Schwarz only if we have $E(XY)$ finite.
– An old man in the sea.
Nov 16 at 20:52
We can have $operatorname{Corr}(XY)$ only if $E(XY)$, $E(X^2)$, and $E(Y^2)$ are all finite.
– kimchi lover
Nov 16 at 21:03
add a comment |
up vote
1
down vote
By the Cauchy-Schwarz inequality, $X$ is a constant multiple of $Y$ in this case.
answered Nov 16 at 20:43
Mike Hawk
1,312110
We can use Cauchy-Schwarz only if we have $E(XY)$ finite.
– An old man in the sea.
Nov 16 at 20:52
We can have $operatorname{Corr}(XY)$ only if $E(XY)$, $E(X^2)$, and $E(Y^2)$ are all finite.
– kimchi lover
Nov 16 at 21:03
add a comment |
up vote
1
down vote
up vote
1
down vote
By the Cauchy-Schwarz inequality, $X$ is a constant multiple of $Y$ in this case.
answered Nov 16 at 20:43
Mike Hawk
1,312110
By the Cauchy-Schwarz inequality, $X$ is a constant multiple of $Y$ in this case.
answered Nov 16 at 20:43
Mike Hawk
1,312110
answered Nov 16 at 20:43
Mike Hawk
1,312110
answered Nov 16 at 20:43
Mike Hawk
1,312110
answered Nov 16 at 20:43
Mike Hawk
1,312110
1,312110
We can use Cauchy-Schwarz only if we have $E(XY)$ finite.
– An old man in the sea.
Nov 16 at 20:52
We can have $operatorname{Corr}(XY)$ only if $E(XY)$, $E(X^2)$, and $E(Y^2)$ are all finite.
– kimchi lover
Nov 16 at 21:03
add a comment |
We can use Cauchy-Schwarz only if we have $E(XY)$ finite.
– An old man in the sea.
Nov 16 at 20:52
We can have $operatorname{Corr}(XY)$ only if $E(XY)$, $E(X^2)$, and $E(Y^2)$ are all finite.
– kimchi lover
Nov 16 at 21:03
We can use Cauchy-Schwarz only if we have $E(XY)$ finite.
– An old man in the sea.
Nov 16 at 20:52
We can use Cauchy-Schwarz only if we have $E(XY)$ finite.
– An old man in the sea.
Nov 16 at 20:52
We can have $operatorname{Corr}(XY)$ only if $E(XY)$, $E(X^2)$, and $E(Y^2)$ are all finite.
– kimchi lover
Nov 16 at 21:03
We can have $operatorname{Corr}(XY)$ only if $E(XY)$, $E(X^2)$, and $E(Y^2)$ are all finite.
– kimchi lover
Nov 16 at 21:03
add a comment |
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