If the expectation and variance of x are both not affected by y, and vice versa, then must x and y be...
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I know that if $mathbb{E}[X]=mathbb{E}[X|Y] , mathbb{E}[Y]=mathbb{E}[Y|X]$, $X$ and $Y$ can be dependent, for example a ‘uniform’ distribution in a unit circle.
Now we add the variance, if
$$mathbb{E}[X]=mathbb{E}[X|Y], mathbb{E}[Y]=mathbb{E}[Y|X], $$$$Var(X)=Var(X|Y), Var(Y)=Var(Y|X).$$
Say the expectation and variance of $X$ are both not affected by $Y$, and vice versa, then must $X$ and $Y$ be independent?
In this case I can not find a counterexample just like the uniform circle.
If they are independent, how to prove it? If not, is there a counterexample?
Thanks!
probability conditional-expectation independence
New contributor
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$begingroup$
I know that if $mathbb{E}[X]=mathbb{E}[X|Y] , mathbb{E}[Y]=mathbb{E}[Y|X]$, $X$ and $Y$ can be dependent, for example a ‘uniform’ distribution in a unit circle.
Now we add the variance, if
$$mathbb{E}[X]=mathbb{E}[X|Y], mathbb{E}[Y]=mathbb{E}[Y|X], $$$$Var(X)=Var(X|Y), Var(Y)=Var(Y|X).$$
Say the expectation and variance of $X$ are both not affected by $Y$, and vice versa, then must $X$ and $Y$ be independent?
In this case I can not find a counterexample just like the uniform circle.
If they are independent, how to prove it? If not, is there a counterexample?
Thanks!
probability conditional-expectation independence
New contributor
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add a comment |
$begingroup$
I know that if $mathbb{E}[X]=mathbb{E}[X|Y] , mathbb{E}[Y]=mathbb{E}[Y|X]$, $X$ and $Y$ can be dependent, for example a ‘uniform’ distribution in a unit circle.
Now we add the variance, if
$$mathbb{E}[X]=mathbb{E}[X|Y], mathbb{E}[Y]=mathbb{E}[Y|X], $$$$Var(X)=Var(X|Y), Var(Y)=Var(Y|X).$$
Say the expectation and variance of $X$ are both not affected by $Y$, and vice versa, then must $X$ and $Y$ be independent?
In this case I can not find a counterexample just like the uniform circle.
If they are independent, how to prove it? If not, is there a counterexample?
Thanks!
probability conditional-expectation independence
New contributor
$endgroup$
I know that if $mathbb{E}[X]=mathbb{E}[X|Y] , mathbb{E}[Y]=mathbb{E}[Y|X]$, $X$ and $Y$ can be dependent, for example a ‘uniform’ distribution in a unit circle.
Now we add the variance, if
$$mathbb{E}[X]=mathbb{E}[X|Y], mathbb{E}[Y]=mathbb{E}[Y|X], $$$$Var(X)=Var(X|Y), Var(Y)=Var(Y|X).$$
Say the expectation and variance of $X$ are both not affected by $Y$, and vice versa, then must $X$ and $Y$ be independent?
In this case I can not find a counterexample just like the uniform circle.
If they are independent, how to prove it? If not, is there a counterexample?
Thanks!
probability conditional-expectation independence
probability conditional-expectation independence
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edited 2 hours ago
Brent Wagner
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asked 4 hours ago
Brent WagnerBrent Wagner
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Consider $Xsim N(0,1)$ and $Ysim N(0,2)$ if $X>0$ and $Ysim t_4$ otherwise. Then $X, Y$ are dependent, but your conditions hold (the $t_4$ distribution has mean 0 and variance 2).
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Here's a discrete example:
$$100P(x,y)=begin{array}{c|ccccc}xbackslash y&-2&-1&0&1&2\ hline -2&1&0&6&0&1\-1&0&9&0&9&0\0&6&0&36&0&6\1&0&9&0&9&0\2&1&0&6&0&1end{array}$$
Condition on any particular $x$, and $y$ has mean zero and variance $1$. Condition on any particular $y$, and $x$ has mean zero and variance $1$. But, of course, they're not independent, as the distribution's support isn't a Cartesian product.
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Thank you very much! But I can only adapt one answer so I adapted the earlier one... Your discrete example is wonderful!
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– Brent Wagner
2 hours ago
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2 Answers
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$begingroup$
Consider $Xsim N(0,1)$ and $Ysim N(0,2)$ if $X>0$ and $Ysim t_4$ otherwise. Then $X, Y$ are dependent, but your conditions hold (the $t_4$ distribution has mean 0 and variance 2).
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add a comment |
$begingroup$
Consider $Xsim N(0,1)$ and $Ysim N(0,2)$ if $X>0$ and $Ysim t_4$ otherwise. Then $X, Y$ are dependent, but your conditions hold (the $t_4$ distribution has mean 0 and variance 2).
$endgroup$
add a comment |
$begingroup$
Consider $Xsim N(0,1)$ and $Ysim N(0,2)$ if $X>0$ and $Ysim t_4$ otherwise. Then $X, Y$ are dependent, but your conditions hold (the $t_4$ distribution has mean 0 and variance 2).
$endgroup$
Consider $Xsim N(0,1)$ and $Ysim N(0,2)$ if $X>0$ and $Ysim t_4$ otherwise. Then $X, Y$ are dependent, but your conditions hold (the $t_4$ distribution has mean 0 and variance 2).
answered 3 hours ago
DashermanDasherman
1,065817
1,065817
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$begingroup$
Here's a discrete example:
$$100P(x,y)=begin{array}{c|ccccc}xbackslash y&-2&-1&0&1&2\ hline -2&1&0&6&0&1\-1&0&9&0&9&0\0&6&0&36&0&6\1&0&9&0&9&0\2&1&0&6&0&1end{array}$$
Condition on any particular $x$, and $y$ has mean zero and variance $1$. Condition on any particular $y$, and $x$ has mean zero and variance $1$. But, of course, they're not independent, as the distribution's support isn't a Cartesian product.
$endgroup$
$begingroup$
Thank you very much! But I can only adapt one answer so I adapted the earlier one... Your discrete example is wonderful!
$endgroup$
– Brent Wagner
2 hours ago
add a comment |
$begingroup$
Here's a discrete example:
$$100P(x,y)=begin{array}{c|ccccc}xbackslash y&-2&-1&0&1&2\ hline -2&1&0&6&0&1\-1&0&9&0&9&0\0&6&0&36&0&6\1&0&9&0&9&0\2&1&0&6&0&1end{array}$$
Condition on any particular $x$, and $y$ has mean zero and variance $1$. Condition on any particular $y$, and $x$ has mean zero and variance $1$. But, of course, they're not independent, as the distribution's support isn't a Cartesian product.
$endgroup$
$begingroup$
Thank you very much! But I can only adapt one answer so I adapted the earlier one... Your discrete example is wonderful!
$endgroup$
– Brent Wagner
2 hours ago
add a comment |
$begingroup$
Here's a discrete example:
$$100P(x,y)=begin{array}{c|ccccc}xbackslash y&-2&-1&0&1&2\ hline -2&1&0&6&0&1\-1&0&9&0&9&0\0&6&0&36&0&6\1&0&9&0&9&0\2&1&0&6&0&1end{array}$$
Condition on any particular $x$, and $y$ has mean zero and variance $1$. Condition on any particular $y$, and $x$ has mean zero and variance $1$. But, of course, they're not independent, as the distribution's support isn't a Cartesian product.
$endgroup$
Here's a discrete example:
$$100P(x,y)=begin{array}{c|ccccc}xbackslash y&-2&-1&0&1&2\ hline -2&1&0&6&0&1\-1&0&9&0&9&0\0&6&0&36&0&6\1&0&9&0&9&0\2&1&0&6&0&1end{array}$$
Condition on any particular $x$, and $y$ has mean zero and variance $1$. Condition on any particular $y$, and $x$ has mean zero and variance $1$. But, of course, they're not independent, as the distribution's support isn't a Cartesian product.
answered 3 hours ago
jmerryjmerry
4,982514
4,982514
$begingroup$
Thank you very much! But I can only adapt one answer so I adapted the earlier one... Your discrete example is wonderful!
$endgroup$
– Brent Wagner
2 hours ago
add a comment |
$begingroup$
Thank you very much! But I can only adapt one answer so I adapted the earlier one... Your discrete example is wonderful!
$endgroup$
– Brent Wagner
2 hours ago
$begingroup$
Thank you very much! But I can only adapt one answer so I adapted the earlier one... Your discrete example is wonderful!
$endgroup$
– Brent Wagner
2 hours ago
$begingroup$
Thank you very much! But I can only adapt one answer so I adapted the earlier one... Your discrete example is wonderful!
$endgroup$
– Brent Wagner
2 hours ago
add a comment |
Brent Wagner is a new contributor. Be nice, and check out our Code of Conduct.
Brent Wagner is a new contributor. Be nice, and check out our Code of Conduct.
Brent Wagner is a new contributor. Be nice, and check out our Code of Conduct.
Brent Wagner is a new contributor. Be nice, and check out our Code of Conduct.
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