Does negative correlation survive monotone transformation?
up vote
0
down vote
favorite
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 at 5:23
Stupid_Guy
547317
|
show 1 more comment
up vote
0
down vote
favorite
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 at 5:23
Stupid_Guy
547317
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 ${0,0.1,1}$.
– Rahul
Nov 18 at 6:56
@Rahul Thanks. Isn't it still being negative correlated after the transformation, because the expectation of the product is still zero?
– Stupid_Guy
Nov 18 at 16:35
Whoops, typo: the third point should be $(1,1)$.
– Rahul
Nov 18 at 16:40
@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.
– Stupid_Guy
Nov 18 at 16:48
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.
– Rahul
Nov 19 at 5:29
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 at 5:23
Stupid_Guy
547317
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 at 5:23
Stupid_Guy
547317
asked Nov 18 at 5:23
Stupid_Guy
547317
asked Nov 18 at 5:23
Stupid_Guy
547317
asked Nov 18 at 5:23
Stupid_Guy
547317
asked Nov 18 at 5:23
Stupid_Guy
547317
547317
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 ${0,0.1,1}$.
– Rahul
Nov 18 at 6:56
@Rahul Thanks. Isn't it still being negative correlated after the transformation, because the expectation of the product is still zero?
– Stupid_Guy
Nov 18 at 16:35
Whoops, typo: the third point should be $(1,1)$.
– Rahul
Nov 18 at 16:40
@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.
– Stupid_Guy
Nov 18 at 16:48
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.
– Rahul
Nov 19 at 5:29
|
show 1 more comment
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 ${0,0.1,1}$.
– Rahul
Nov 18 at 6:56
@Rahul Thanks. Isn't it still being negative correlated after the transformation, because the expectation of the product is still zero?
– Stupid_Guy
Nov 18 at 16:35
Whoops, typo: the third point should be $(1,1)$.
– Rahul
Nov 18 at 16:40
@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.
– Stupid_Guy
Nov 18 at 16:48
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.
– Rahul
Nov 19 at 5:29
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 ${0,0.1,1}$.
– Rahul
Nov 18 at 6:56
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 ${0,0.1,1}$.
– Rahul
Nov 18 at 6:56
@Rahul Thanks. Isn't it still being negative correlated after the transformation, because the expectation of the product is still zero?
– Stupid_Guy
Nov 18 at 16:35
@Rahul Thanks. Isn't it still being negative correlated after the transformation, because the expectation of the product is still zero?
– Stupid_Guy
Nov 18 at 16:35
Whoops, typo: the third point should be $(1,1)$.
– Rahul
Nov 18 at 16:40
Whoops, typo: the third point should be $(1,1)$.
– Rahul
Nov 18 at 16:40
@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.
– Stupid_Guy
Nov 18 at 16:48
@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.
– Stupid_Guy
Nov 18 at 16:48
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.
– Rahul
Nov 19 at 5:29
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.
– Rahul
Nov 19 at 5:29
|
show 1 more comment
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
draft saved
draft discarded
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3003157%2fdoes-negative-correlation-survive-monotone-transformation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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 ${0,0.1,1}$.
– Rahul
Nov 18 at 6:56
@Rahul Thanks. Isn't it still being negative correlated after the transformation, because the expectation of the product is still zero?
– Stupid_Guy
Nov 18 at 16:35
Whoops, typo: the third point should be $(1,1)$.
– Rahul
Nov 18 at 16:40
@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.
– Stupid_Guy
Nov 18 at 16:48
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.
– Rahul
Nov 19 at 5:29