Correlation is zero but with non-zero correlation coefficient The Next CEO of Stack OverflowWhat is the correlation function in multivariable/vectoral case?Correlation of sums of correlated variablesMaximum and minimum Correlation CoefficientDistribution of the sum of normal random variablesDoes correlation have to be in the context of (Gaussian) normal distribution?Cross-correlation of identical sets: not getting expected resultCorrelation coefficient and orthogonalityIs there any difference between Correlation and Correlation coefficient?Correlation coefficient and regression line : Geometric intuitionCorrelation and Covariance on Standardized X
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Correlation is zero but with non-zero correlation coefficient
The Next CEO of Stack OverflowWhat is the correlation function in multivariable/vectoral case?Correlation of sums of correlated variablesMaximum and minimum Correlation CoefficientDistribution of the sum of normal random variablesDoes correlation have to be in the context of (Gaussian) normal distribution?Cross-correlation of identical sets: not getting expected resultCorrelation coefficient and orthogonalityIs there any difference between Correlation and Correlation coefficient?Correlation coefficient and regression line : Geometric intuitionCorrelation and Covariance on Standardized X
$begingroup$
The correlation coefficient is given by
$$rho_XY=fracR_XY-mu_X , mu_Ysigma_X , sigma_Y$$
If the product $mu_X , mu_Y neq 0$ and $rho_XYneq 0$, then we can have two cases:
$R_XY= 0$ when $X$ and $Y$ are orthogonal;
$R_XYneq 0$ when $X$ and $Y$ are not orthogonal to each other;
So I see from Case $1$ that $R_XY= 0$ is possible to have when $rho_XYneq 0$. Is my reasoning correct? Is this not counterintuitive?
correlation
asked Mar 28 at 9:03
macymacy
125
$endgroup$
add a comment |
$begingroup$
The correlation coefficient is given by
$$rho_XY=fracR_XY-mu_X , mu_Ysigma_X , sigma_Y$$
If the product $mu_X , mu_Y neq 0$ and $rho_XYneq 0$, then we can have two cases:
$R_XY= 0$ when $X$ and $Y$ are orthogonal;
$R_XYneq 0$ when $X$ and $Y$ are not orthogonal to each other;
So I see from Case $1$ that $R_XY= 0$ is possible to have when $rho_XYneq 0$. Is my reasoning correct? Is this not counterintuitive?
correlation
asked Mar 28 at 9:03
macymacy
125
$endgroup$
add a comment |
$begingroup$
The correlation coefficient is given by
$$rho_XY=fracR_XY-mu_X , mu_Ysigma_X , sigma_Y$$
If the product $mu_X , mu_Y neq 0$ and $rho_XYneq 0$, then we can have two cases:
$R_XY= 0$ when $X$ and $Y$ are orthogonal;
$R_XYneq 0$ when $X$ and $Y$ are not orthogonal to each other;
So I see from Case $1$ that $R_XY= 0$ is possible to have when $rho_XYneq 0$. Is my reasoning correct? Is this not counterintuitive?
correlation
asked Mar 28 at 9:03
macymacy
125
$endgroup$
The correlation coefficient is given by
$$rho_XY=fracR_XY-mu_X , mu_Ysigma_X , sigma_Y$$
If the product $mu_X , mu_Y neq 0$ and $rho_XYneq 0$, then we can have two cases:
$R_XY= 0$ when $X$ and $Y$ are orthogonal;
$R_XYneq 0$ when $X$ and $Y$ are not orthogonal to each other;
So I see from Case $1$ that $R_XY= 0$ is possible to have when $rho_XYneq 0$. Is my reasoning correct? Is this not counterintuitive?
correlation
correlation
asked Mar 28 at 9:03
macymacy
125
asked Mar 28 at 9:03
macymacy
125
edited Mar 28 at 12:36
macy
asked Mar 28 at 9:03
macymacy
125
asked Mar 28 at 9:03
macymacy
125
asked Mar 28 at 9:03
macymacy
125
125
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Yes, it is possible.
Following this very good paper on the topic, consider two random variables $X$ and $Y$ with the following realizations:
$$X = (1, -5, 3, -1), Y = (5, 1, 1, 3)$$
You have:
$$R_XY = E(XY') = E(1times5 - 5times1 + 3times1 - 1times3) = 0$$
but:
$$mu_X = -frac12, mu_Y = frac52 Rightarrow rho_X,Y neq 0$$
In general, recall that while both orthogonality and uncorrelation imply linear independence, there's no implication between orthogonality and uncorrelation themselves.
EDITING TO ANSWER COMMENT:
When it comes to stochastic processes, following your notation, we say that $(X_t)_t geq 1$ $(Y_t)_t geq 1$ are uncorrelated if:
$$forall t_1, t_2, COV_X,Y(t_1, t_2) = R_XY(t_1, t_2) - mu_X(t_1)mu_Y(t_2) = 0
$$
while we say that they're orthogonal if:
$$forall t_1, t_2, R_XY(t_1, t_2) = E[X(t_1)Y(t_2)'] = 0$$
so the same reasoning as before applies. Here for a broader analysis.
answered Mar 28 at 21:14
NicgNicg
514
New contributor
Nicg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Great! Also $sigma_x$ and $sigma_y$ are non-zero here. When you are calculating $R_XY$, you are doing it at lag zero i.e. no shift between X and Y. Is there a special reason. Perhaps a similar reasoning as evaluating an autocovariance which gives the variance?
$endgroup$
– macy
Mar 29 at 5:26
$begingroup$
See edited answer.
$endgroup$
– Nicg
Mar 29 at 10:32
$begingroup$
In your example, why is $R_XY$ computed at lag $0$?
$endgroup$
– macy
Mar 29 at 11:58
$begingroup$
Maybe I am missing what you mean by lag, but between $t_2$ and $t_1$ there's actually a one period lag. Perhaps it helps to see it using a single time series: Let $(X_t)_t geq 0$ be a time series. Then $$COV(X_t + h, X_t) = E[X_t + hX_t] - E(X_t + h)E(X_t)$$ is the autocovariance function, with $h$ being the lag, $1$ in our case. Hope this clarifies.
$endgroup$
– Nicg
Mar 29 at 13:44
$begingroup$
By lag, I mean the shift in one sequence relative to the other one when doing the correlation. I say lag or shift zero because you take the first sample of $X$ and multiply it by the first sample of $Y$ in your example etc. i.e. you are doing $X_1Y_1+X_2Y_2+X_3Y_3+ X_4Y_4$. So your $h$ in your explanation is $0$. For instance the first product is $X_1+0X_1$ with your $h$ as $0$. Your $h$ is what I called lag or shift. In effect you are computing $R_XX(0)$ with the argument $h=0$. We can have $R_XX(1)$, $R_XX(2)$ and $R_XX(3)$ i.e. by shifting one sequence by $1$, $2$ or $3$ unit.
$endgroup$
– macy
2 days ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes, it is possible.
Following this very good paper on the topic, consider two random variables $X$ and $Y$ with the following realizations:
$$X = (1, -5, 3, -1), Y = (5, 1, 1, 3)$$
You have:
$$R_XY = E(XY') = E(1times5 - 5times1 + 3times1 - 1times3) = 0$$
but:
$$mu_X = -frac12, mu_Y = frac52 Rightarrow rho_X,Y neq 0$$
In general, recall that while both orthogonality and uncorrelation imply linear independence, there's no implication between orthogonality and uncorrelation themselves.
EDITING TO ANSWER COMMENT:
When it comes to stochastic processes, following your notation, we say that $(X_t)_t geq 1$ $(Y_t)_t geq 1$ are uncorrelated if:
$$forall t_1, t_2, COV_X,Y(t_1, t_2) = R_XY(t_1, t_2) - mu_X(t_1)mu_Y(t_2) = 0
$$
while we say that they're orthogonal if:
$$forall t_1, t_2, R_XY(t_1, t_2) = E[X(t_1)Y(t_2)'] = 0$$
so the same reasoning as before applies. Here for a broader analysis.
answered Mar 28 at 21:14
NicgNicg
514
New contributor
Nicg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Great! Also $sigma_x$ and $sigma_y$ are non-zero here. When you are calculating $R_XY$, you are doing it at lag zero i.e. no shift between X and Y. Is there a special reason. Perhaps a similar reasoning as evaluating an autocovariance which gives the variance?
$endgroup$
– macy
Mar 29 at 5:26
$begingroup$
See edited answer.
$endgroup$
– Nicg
Mar 29 at 10:32
$begingroup$
In your example, why is $R_XY$ computed at lag $0$?
$endgroup$
– macy
Mar 29 at 11:58
$begingroup$
Maybe I am missing what you mean by lag, but between $t_2$ and $t_1$ there's actually a one period lag. Perhaps it helps to see it using a single time series: Let $(X_t)_t geq 0$ be a time series. Then $$COV(X_t + h, X_t) = E[X_t + hX_t] - E(X_t + h)E(X_t)$$ is the autocovariance function, with $h$ being the lag, $1$ in our case. Hope this clarifies.
$endgroup$
– Nicg
Mar 29 at 13:44
$begingroup$
By lag, I mean the shift in one sequence relative to the other one when doing the correlation. I say lag or shift zero because you take the first sample of $X$ and multiply it by the first sample of $Y$ in your example etc. i.e. you are doing $X_1Y_1+X_2Y_2+X_3Y_3+ X_4Y_4$. So your $h$ in your explanation is $0$. For instance the first product is $X_1+0X_1$ with your $h$ as $0$. Your $h$ is what I called lag or shift. In effect you are computing $R_XX(0)$ with the argument $h=0$. We can have $R_XX(1)$, $R_XX(2)$ and $R_XX(3)$ i.e. by shifting one sequence by $1$, $2$ or $3$ unit.
$endgroup$
– macy
2 days ago
add a comment |
$begingroup$
Yes, it is possible.
Following this very good paper on the topic, consider two random variables $X$ and $Y$ with the following realizations:
$$X = (1, -5, 3, -1), Y = (5, 1, 1, 3)$$
You have:
$$R_XY = E(XY') = E(1times5 - 5times1 + 3times1 - 1times3) = 0$$
but:
$$mu_X = -frac12, mu_Y = frac52 Rightarrow rho_X,Y neq 0$$
In general, recall that while both orthogonality and uncorrelation imply linear independence, there's no implication between orthogonality and uncorrelation themselves.
EDITING TO ANSWER COMMENT:
When it comes to stochastic processes, following your notation, we say that $(X_t)_t geq 1$ $(Y_t)_t geq 1$ are uncorrelated if:
$$forall t_1, t_2, COV_X,Y(t_1, t_2) = R_XY(t_1, t_2) - mu_X(t_1)mu_Y(t_2) = 0
$$
while we say that they're orthogonal if:
$$forall t_1, t_2, R_XY(t_1, t_2) = E[X(t_1)Y(t_2)'] = 0$$
so the same reasoning as before applies. Here for a broader analysis.
answered Mar 28 at 21:14
NicgNicg
514
New contributor
Nicg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Great! Also $sigma_x$ and $sigma_y$ are non-zero here. When you are calculating $R_XY$, you are doing it at lag zero i.e. no shift between X and Y. Is there a special reason. Perhaps a similar reasoning as evaluating an autocovariance which gives the variance?
$endgroup$
– macy
Mar 29 at 5:26
$begingroup$
See edited answer.
$endgroup$
– Nicg
Mar 29 at 10:32
$begingroup$
In your example, why is $R_XY$ computed at lag $0$?
$endgroup$
– macy
Mar 29 at 11:58
$begingroup$
Maybe I am missing what you mean by lag, but between $t_2$ and $t_1$ there's actually a one period lag. Perhaps it helps to see it using a single time series: Let $(X_t)_t geq 0$ be a time series. Then $$COV(X_t + h, X_t) = E[X_t + hX_t] - E(X_t + h)E(X_t)$$ is the autocovariance function, with $h$ being the lag, $1$ in our case. Hope this clarifies.
$endgroup$
– Nicg
Mar 29 at 13:44
$begingroup$
By lag, I mean the shift in one sequence relative to the other one when doing the correlation. I say lag or shift zero because you take the first sample of $X$ and multiply it by the first sample of $Y$ in your example etc. i.e. you are doing $X_1Y_1+X_2Y_2+X_3Y_3+ X_4Y_4$. So your $h$ in your explanation is $0$. For instance the first product is $X_1+0X_1$ with your $h$ as $0$. Your $h$ is what I called lag or shift. In effect you are computing $R_XX(0)$ with the argument $h=0$. We can have $R_XX(1)$, $R_XX(2)$ and $R_XX(3)$ i.e. by shifting one sequence by $1$, $2$ or $3$ unit.
$endgroup$
– macy
2 days ago
add a comment |
$begingroup$
Yes, it is possible.
Following this very good paper on the topic, consider two random variables $X$ and $Y$ with the following realizations:
$$X = (1, -5, 3, -1), Y = (5, 1, 1, 3)$$
You have:
$$R_XY = E(XY') = E(1times5 - 5times1 + 3times1 - 1times3) = 0$$
but:
$$mu_X = -frac12, mu_Y = frac52 Rightarrow rho_X,Y neq 0$$
In general, recall that while both orthogonality and uncorrelation imply linear independence, there's no implication between orthogonality and uncorrelation themselves.
EDITING TO ANSWER COMMENT:
When it comes to stochastic processes, following your notation, we say that $(X_t)_t geq 1$ $(Y_t)_t geq 1$ are uncorrelated if:
$$forall t_1, t_2, COV_X,Y(t_1, t_2) = R_XY(t_1, t_2) - mu_X(t_1)mu_Y(t_2) = 0
$$
while we say that they're orthogonal if:
$$forall t_1, t_2, R_XY(t_1, t_2) = E[X(t_1)Y(t_2)'] = 0$$
so the same reasoning as before applies. Here for a broader analysis.
answered Mar 28 at 21:14
NicgNicg
514
New contributor
Nicg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Yes, it is possible.
Following this very good paper on the topic, consider two random variables $X$ and $Y$ with the following realizations:
$$X = (1, -5, 3, -1), Y = (5, 1, 1, 3)$$
You have:
$$R_XY = E(XY') = E(1times5 - 5times1 + 3times1 - 1times3) = 0$$
but:
$$mu_X = -frac12, mu_Y = frac52 Rightarrow rho_X,Y neq 0$$
In general, recall that while both orthogonality and uncorrelation imply linear independence, there's no implication between orthogonality and uncorrelation themselves.
EDITING TO ANSWER COMMENT:
When it comes to stochastic processes, following your notation, we say that $(X_t)_t geq 1$ $(Y_t)_t geq 1$ are uncorrelated if:
$$forall t_1, t_2, COV_X,Y(t_1, t_2) = R_XY(t_1, t_2) - mu_X(t_1)mu_Y(t_2) = 0
$$
while we say that they're orthogonal if:
$$forall t_1, t_2, R_XY(t_1, t_2) = E[X(t_1)Y(t_2)'] = 0$$
so the same reasoning as before applies. Here for a broader analysis.
answered Mar 28 at 21:14
NicgNicg
514
New contributor
Nicg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 29 at 13:37
answered Mar 28 at 21:14
NicgNicg
514
New contributor
Nicg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 28 at 21:14
NicgNicg
514
answered Mar 28 at 21:14
NicgNicg
514
514
New contributor
Nicg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nicg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nicg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Great! Also $sigma_x$ and $sigma_y$ are non-zero here. When you are calculating $R_XY$, you are doing it at lag zero i.e. no shift between X and Y. Is there a special reason. Perhaps a similar reasoning as evaluating an autocovariance which gives the variance?
$endgroup$
– macy
Mar 29 at 5:26
$begingroup$
See edited answer.
$endgroup$
– Nicg
Mar 29 at 10:32
$begingroup$
In your example, why is $R_XY$ computed at lag $0$?
$endgroup$
– macy
Mar 29 at 11:58
$begingroup$
Maybe I am missing what you mean by lag, but between $t_2$ and $t_1$ there's actually a one period lag. Perhaps it helps to see it using a single time series: Let $(X_t)_t geq 0$ be a time series. Then $$COV(X_t + h, X_t) = E[X_t + hX_t] - E(X_t + h)E(X_t)$$ is the autocovariance function, with $h$ being the lag, $1$ in our case. Hope this clarifies.
$endgroup$
– Nicg
Mar 29 at 13:44
$begingroup$
By lag, I mean the shift in one sequence relative to the other one when doing the correlation. I say lag or shift zero because you take the first sample of $X$ and multiply it by the first sample of $Y$ in your example etc. i.e. you are doing $X_1Y_1+X_2Y_2+X_3Y_3+ X_4Y_4$. So your $h$ in your explanation is $0$. For instance the first product is $X_1+0X_1$ with your $h$ as $0$. Your $h$ is what I called lag or shift. In effect you are computing $R_XX(0)$ with the argument $h=0$. We can have $R_XX(1)$, $R_XX(2)$ and $R_XX(3)$ i.e. by shifting one sequence by $1$, $2$ or $3$ unit.
$endgroup$
– macy
2 days ago
add a comment |
$begingroup$
Great! Also $sigma_x$ and $sigma_y$ are non-zero here. When you are calculating $R_XY$, you are doing it at lag zero i.e. no shift between X and Y. Is there a special reason. Perhaps a similar reasoning as evaluating an autocovariance which gives the variance?
$endgroup$
– macy
Mar 29 at 5:26
$begingroup$
See edited answer.
$endgroup$
– Nicg
Mar 29 at 10:32
$begingroup$
In your example, why is $R_XY$ computed at lag $0$?
$endgroup$
– macy
Mar 29 at 11:58
$begingroup$
Maybe I am missing what you mean by lag, but between $t_2$ and $t_1$ there's actually a one period lag. Perhaps it helps to see it using a single time series: Let $(X_t)_t geq 0$ be a time series. Then $$COV(X_t + h, X_t) = E[X_t + hX_t] - E(X_t + h)E(X_t)$$ is the autocovariance function, with $h$ being the lag, $1$ in our case. Hope this clarifies.
$endgroup$
– Nicg
Mar 29 at 13:44
$begingroup$
By lag, I mean the shift in one sequence relative to the other one when doing the correlation. I say lag or shift zero because you take the first sample of $X$ and multiply it by the first sample of $Y$ in your example etc. i.e. you are doing $X_1Y_1+X_2Y_2+X_3Y_3+ X_4Y_4$. So your $h$ in your explanation is $0$. For instance the first product is $X_1+0X_1$ with your $h$ as $0$. Your $h$ is what I called lag or shift. In effect you are computing $R_XX(0)$ with the argument $h=0$. We can have $R_XX(1)$, $R_XX(2)$ and $R_XX(3)$ i.e. by shifting one sequence by $1$, $2$ or $3$ unit.
$endgroup$
– macy
2 days ago
$begingroup$
Great! Also $sigma_x$ and $sigma_y$ are non-zero here. When you are calculating $R_XY$, you are doing it at lag zero i.e. no shift between X and Y. Is there a special reason. Perhaps a similar reasoning as evaluating an autocovariance which gives the variance?
$endgroup$
– macy
Mar 29 at 5:26
$begingroup$
Great! Also $sigma_x$ and $sigma_y$ are non-zero here. When you are calculating $R_XY$, you are doing it at lag zero i.e. no shift between X and Y. Is there a special reason. Perhaps a similar reasoning as evaluating an autocovariance which gives the variance?
$endgroup$
– macy
Mar 29 at 5:26
$begingroup$
See edited answer.
$endgroup$
– Nicg
Mar 29 at 10:32
$begingroup$
See edited answer.
$endgroup$
– Nicg
Mar 29 at 10:32
$begingroup$
In your example, why is $R_XY$ computed at lag $0$?
$endgroup$
– macy
Mar 29 at 11:58
$begingroup$
In your example, why is $R_XY$ computed at lag $0$?
$endgroup$
– macy
Mar 29 at 11:58
$begingroup$
Maybe I am missing what you mean by lag, but between $t_2$ and $t_1$ there's actually a one period lag. Perhaps it helps to see it using a single time series: Let $(X_t)_t geq 0$ be a time series. Then $$COV(X_t + h, X_t) = E[X_t + hX_t] - E(X_t + h)E(X_t)$$ is the autocovariance function, with $h$ being the lag, $1$ in our case. Hope this clarifies.
$endgroup$
– Nicg
Mar 29 at 13:44
$begingroup$
Maybe I am missing what you mean by lag, but between $t_2$ and $t_1$ there's actually a one period lag. Perhaps it helps to see it using a single time series: Let $(X_t)_t geq 0$ be a time series. Then $$COV(X_t + h, X_t) = E[X_t + hX_t] - E(X_t + h)E(X_t)$$ is the autocovariance function, with $h$ being the lag, $1$ in our case. Hope this clarifies.
$endgroup$
– Nicg
Mar 29 at 13:44
$begingroup$
By lag, I mean the shift in one sequence relative to the other one when doing the correlation. I say lag or shift zero because you take the first sample of $X$ and multiply it by the first sample of $Y$ in your example etc. i.e. you are doing $X_1Y_1+X_2Y_2+X_3Y_3+ X_4Y_4$. So your $h$ in your explanation is $0$. For instance the first product is $X_1+0X_1$ with your $h$ as $0$. Your $h$ is what I called lag or shift. In effect you are computing $R_XX(0)$ with the argument $h=0$. We can have $R_XX(1)$, $R_XX(2)$ and $R_XX(3)$ i.e. by shifting one sequence by $1$, $2$ or $3$ unit.
$endgroup$
– macy
2 days ago
$begingroup$
By lag, I mean the shift in one sequence relative to the other one when doing the correlation. I say lag or shift zero because you take the first sample of $X$ and multiply it by the first sample of $Y$ in your example etc. i.e. you are doing $X_1Y_1+X_2Y_2+X_3Y_3+ X_4Y_4$. So your $h$ in your explanation is $0$. For instance the first product is $X_1+0X_1$ with your $h$ as $0$. Your $h$ is what I called lag or shift. In effect you are computing $R_XX(0)$ with the argument $h=0$. We can have $R_XX(1)$, $R_XX(2)$ and $R_XX(3)$ i.e. by shifting one sequence by $1$, $2$ or $3$ unit.
$endgroup$
– macy
2 days ago
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