Mixing
Does negative correlation survive monotone transformation?
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Let $X$ and $Y$ be two non-negative random variables and be negative correlated, i.e.,
$$mathbb{E}[XY] leq mathbb{E}[X]mathbb{E}[Y].$$
Let $h(cdot)$ and $g(cdot)$ be two non-negative, monotone increasing functions. Do we have
$h(X)$ and $g(Y)$ being negative correlated as well? Intuitively this quite makes sense to me..
probability probability-theory correlation expected-value
asked Nov 18 '18 at 5:23
Stupid_GuyStupid_Guy
567317
$endgroup$
|
show 1 more comment
$begingroup$
Let $X$ and $Y$ be two non-negative random variables and be negative correlated, i.e.,
$$mathbb{E}[XY] leq mathbb{E}[X]mathbb{E}[Y].$$
Let $h(cdot)$ and $g(cdot)$ be two non-negative, monotone increasing functions. Do we have
$h(X)$ and $g(Y)$ being negative correlated as well? Intuitively this quite makes sense to me..
probability probability-theory correlation expected-value
asked Nov 18 '18 at 5:23
Stupid_GuyStupid_Guy
567317
$endgroup$
1
$begingroup$
Counterexample: Consider the discrete uniform distribution on ${(0,1),(0.9,0),(1,0)}$, and let $h$ be the monotone function that maps ${0,0.9,1}$ to ${0,0.1,1}$.
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– Rahul
Nov 18 '18 at 6:56
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@Rahul Thanks. Isn't it still being negative correlated after the transformation, because the expectation of the product is still zero?
$endgroup$
– Stupid_Guy
Nov 18 '18 at 16:35
1
$begingroup$
Whoops, typo: the third point should be $(1,1)$.
$endgroup$
– Rahul
Nov 18 '18 at 16:40
$begingroup$
@Rahul Hi Rahul, your ideas make sense to me. Thanks so much. However, we may still need a little bit of work as after the transformation they are still negative correlated, because $h(Y) = Y$ thus $Cov(h(X),h(Y)) = -frac{2.3}{3}$. I'll think more about this. Many thanks.
$endgroup$
– Stupid_Guy
Nov 18 '18 at 16:48
$begingroup$
One of us must be making a mistake. I get covariances $-0.8/9$ and $0.8/9$ before and after transformation by $h$ respectively. $E[XY]=1/3$ and $E[Y]=2/3$ in both cases, but $E[X]$ changes from $1.9/3$ to $1.1/3$. It's also easy to visualize the change in correlation if you plot the data and consider the best fitting line.
$endgroup$
– Rahul
Nov 19 '18 at 5:29
|
show 1 more comment
$begingroup$
Let $X$ and $Y$ be two non-negative random variables and be negative correlated, i.e.,
$$mathbb{E}[XY] leq mathbb{E}[X]mathbb{E}[Y].$$
Let $h(cdot)$ and $g(cdot)$ be two non-negative, monotone increasing functions. Do we have
$h(X)$ and $g(Y)$ being negative correlated as well? Intuitively this quite makes sense to me..
probability probability-theory correlation expected-value
asked Nov 18 '18 at 5:23
Stupid_GuyStupid_Guy
567317
$endgroup$
Let $X$ and $Y$ be two non-negative random variables and be negative correlated, i.e.,
$$mathbb{E}[XY] leq mathbb{E}[X]mathbb{E}[Y].$$
Let $h(cdot)$ and $g(cdot)$ be two non-negative, monotone increasing functions. Do we have
$h(X)$ and $g(Y)$ being negative correlated as well? Intuitively this quite makes sense to me..
probability probability-theory correlation expected-value
probability probability-theory correlation expected-value
asked Nov 18 '18 at 5:23
Stupid_GuyStupid_Guy
567317
asked Nov 18 '18 at 5:23
Stupid_GuyStupid_Guy
567317
asked Nov 18 '18 at 5:23
Stupid_GuyStupid_Guy
567317
asked Nov 18 '18 at 5:23
Stupid_GuyStupid_Guy
567317
asked Nov 18 '18 at 5:23
Stupid_GuyStupid_Guy
567317
567317
1
$begingroup$
Counterexample: Consider the discrete uniform distribution on ${(0,1),(0.9,0),(1,0)}$, and let $h$ be the monotone function that maps ${0,0.9,1}$ to ${0,0.1,1}$.
$endgroup$
– Rahul
Nov 18 '18 at 6:56
$begingroup$
@Rahul Thanks. Isn't it still being negative correlated after the transformation, because the expectation of the product is still zero?
$endgroup$
– Stupid_Guy
Nov 18 '18 at 16:35
1
$begingroup$
Whoops, typo: the third point should be $(1,1)$.
$endgroup$
– Rahul
Nov 18 '18 at 16:40
$begingroup$
@Rahul Hi Rahul, your ideas make sense to me. Thanks so much. However, we may still need a little bit of work as after the transformation they are still negative correlated, because $h(Y) = Y$ thus $Cov(h(X),h(Y)) = -frac{2.3}{3}$. I'll think more about this. Many thanks.
$endgroup$
– Stupid_Guy
Nov 18 '18 at 16:48
$begingroup$
One of us must be making a mistake. I get covariances $-0.8/9$ and $0.8/9$ before and after transformation by $h$ respectively. $E[XY]=1/3$ and $E[Y]=2/3$ in both cases, but $E[X]$ changes from $1.9/3$ to $1.1/3$. It's also easy to visualize the change in correlation if you plot the data and consider the best fitting line.
$endgroup$
– Rahul
Nov 19 '18 at 5:29
|
show 1 more comment
1
$begingroup$
Counterexample: Consider the discrete uniform distribution on ${(0,1),(0.9,0),(1,0)}$, and let $h$ be the monotone function that maps ${0,0.9,1}$ to ${0,0.1,1}$.
$endgroup$
– Rahul
Nov 18 '18 at 6:56
$begingroup$
@Rahul Thanks. Isn't it still being negative correlated after the transformation, because the expectation of the product is still zero?
$endgroup$
– Stupid_Guy
Nov 18 '18 at 16:35
1
$begingroup$
Whoops, typo: the third point should be $(1,1)$.
$endgroup$
– Rahul
Nov 18 '18 at 16:40
$begingroup$
@Rahul Hi Rahul, your ideas make sense to me. Thanks so much. However, we may still need a little bit of work as after the transformation they are still negative correlated, because $h(Y) = Y$ thus $Cov(h(X),h(Y)) = -frac{2.3}{3}$. I'll think more about this. Many thanks.
$endgroup$
– Stupid_Guy
Nov 18 '18 at 16:48
$begingroup$
One of us must be making a mistake. I get covariances $-0.8/9$ and $0.8/9$ before and after transformation by $h$ respectively. $E[XY]=1/3$ and $E[Y]=2/3$ in both cases, but $E[X]$ changes from $1.9/3$ to $1.1/3$. It's also easy to visualize the change in correlation if you plot the data and consider the best fitting line.
$endgroup$
– Rahul
Nov 19 '18 at 5:29
1
1
$begingroup$
Counterexample: Consider the discrete uniform distribution on ${(0,1),(0.9,0),(1,0)}$, and let $h$ be the monotone function that maps ${0,0.9,1}$ to ${0,0.1,1}$.
$endgroup$
– Rahul
Nov 18 '18 at 6:56
$begingroup$
Counterexample: Consider the discrete uniform distribution on ${(0,1),(0.9,0),(1,0)}$, and let $h$ be the monotone function that maps ${0,0.9,1}$ to ${0,0.1,1}$.
$endgroup$
– Rahul
Nov 18 '18 at 6:56
$begingroup$
@Rahul Thanks. Isn't it still being negative correlated after the transformation, because the expectation of the product is still zero?
$endgroup$
– Stupid_Guy
Nov 18 '18 at 16:35
$begingroup$
@Rahul Thanks. Isn't it still being negative correlated after the transformation, because the expectation of the product is still zero?
$endgroup$
– Stupid_Guy
Nov 18 '18 at 16:35
1
1
$begingroup$
Whoops, typo: the third point should be $(1,1)$.
$endgroup$
– Rahul
Nov 18 '18 at 16:40
$begingroup$
Whoops, typo: the third point should be $(1,1)$.
$endgroup$
– Rahul
Nov 18 '18 at 16:40
$begingroup$
@Rahul Hi Rahul, your ideas make sense to me. Thanks so much. However, we may still need a little bit of work as after the transformation they are still negative correlated, because $h(Y) = Y$ thus $Cov(h(X),h(Y)) = -frac{2.3}{3}$. I'll think more about this. Many thanks.
$endgroup$
– Stupid_Guy
Nov 18 '18 at 16:48
$begingroup$
@Rahul Hi Rahul, your ideas make sense to me. Thanks so much. However, we may still need a little bit of work as after the transformation they are still negative correlated, because $h(Y) = Y$ thus $Cov(h(X),h(Y)) = -frac{2.3}{3}$. I'll think more about this. Many thanks.
$endgroup$
– Stupid_Guy
Nov 18 '18 at 16:48
$begingroup$
One of us must be making a mistake. I get covariances $-0.8/9$ and $0.8/9$ before and after transformation by $h$ respectively. $E[XY]=1/3$ and $E[Y]=2/3$ in both cases, but $E[X]$ changes from $1.9/3$ to $1.1/3$. It's also easy to visualize the change in correlation if you plot the data and consider the best fitting line.
$endgroup$
– Rahul
Nov 19 '18 at 5:29
$begingroup$
One of us must be making a mistake. I get covariances $-0.8/9$ and $0.8/9$ before and after transformation by $h$ respectively. $E[XY]=1/3$ and $E[Y]=2/3$ in both cases, but $E[X]$ changes from $1.9/3$ to $1.1/3$. It's also easy to visualize the change in correlation if you plot the data and consider the best fitting line.
$endgroup$
– Rahul
Nov 19 '18 at 5:29
|
show 1 more comment
1 Answer
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Copying from Rahul's comment since it answers the question:
No. Counterexample: Consider the discrete uniform distribution on {(0,1),(0.9,0),(1,1)}, and let h be the monotone function that maps {0,0.9,1} to {0,0.1,1}.
answered Jan 20 at 22:57
VeechVeech
1184
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$begingroup$
Copying from Rahul's comment since it answers the question:
No. Counterexample: Consider the discrete uniform distribution on {(0,1),(0.9,0),(1,1)}, and let h be the monotone function that maps {0,0.9,1} to {0,0.1,1}.
answered Jan 20 at 22:57
VeechVeech
1184
$endgroup$
add a comment |
$begingroup$
Copying from Rahul's comment since it answers the question:
No. Counterexample: Consider the discrete uniform distribution on {(0,1),(0.9,0),(1,1)}, and let h be the monotone function that maps {0,0.9,1} to {0,0.1,1}.
answered Jan 20 at 22:57
VeechVeech
1184
$endgroup$
add a comment |
$begingroup$
Copying from Rahul's comment since it answers the question:
No. Counterexample: Consider the discrete uniform distribution on {(0,1),(0.9,0),(1,1)}, and let h be the monotone function that maps {0,0.9,1} to {0,0.1,1}.
answered Jan 20 at 22:57
VeechVeech
1184
$endgroup$
Copying from Rahul's comment since it answers the question:
No. Counterexample: Consider the discrete uniform distribution on {(0,1),(0.9,0),(1,1)}, and let h be the monotone function that maps {0,0.9,1} to {0,0.1,1}.
answered Jan 20 at 22:57
VeechVeech
1184
answered Jan 20 at 22:57
VeechVeech
1184
answered Jan 20 at 22:57
VeechVeech
1184
answered Jan 20 at 22:57
VeechVeech
1184
1184
add a comment |
add a comment |
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$begingroup$
Counterexample: Consider the discrete uniform distribution on ${(0,1),(0.9,0),(1,0)}$, and let $h$ be the monotone function that maps ${0,0.9,1}$ to ${0,0.1,1}$.
$endgroup$
– Rahul
Nov 18 '18 at 6:56
$begingroup$
@Rahul Thanks. Isn't it still being negative correlated after the transformation, because the expectation of the product is still zero?
$endgroup$
– Stupid_Guy
Nov 18 '18 at 16:35
1
$begingroup$
Whoops, typo: the third point should be $(1,1)$.
$endgroup$
– Rahul
Nov 18 '18 at 16:40
$begingroup$
@Rahul Hi Rahul, your ideas make sense to me. Thanks so much. However, we may still need a little bit of work as after the transformation they are still negative correlated, because $h(Y) = Y$ thus $Cov(h(X),h(Y)) = -frac{2.3}{3}$. I'll think more about this. Many thanks.
$endgroup$
– Stupid_Guy
Nov 18 '18 at 16:48
$begingroup$
One of us must be making a mistake. I get covariances $-0.8/9$ and $0.8/9$ before and after transformation by $h$ respectively. $E[XY]=1/3$ and $E[Y]=2/3$ in both cases, but $E[X]$ changes from $1.9/3$ to $1.1/3$. It's also easy to visualize the change in correlation if you plot the data and consider the best fitting line.
$endgroup$
– Rahul
Nov 19 '18 at 5:29