Degrees of Freedom in Covariance: Intuition?
$begingroup$
If we say $operatorname{Var}(x)$ has $n-1$ degrees of freedom which are lost after we estimate $operatorname{Var}(x)$, this matches how $n-1$ observations are now constrained to be sufficiently close to the remaining observation of $x$. In my class, $operatorname{Cov}(x,y)$ is also described as having $n-1$ degrees of freedom which are lost after we estimate $operatorname{Cov}(x,y)$. Reference for this use of "degrees of freedom" in covariance
My confusion is that the covariance does not actually seem to constrain $n-1$ $x$ and $y$ values, like the variance did for $x$.
Can you relate the $n-1$ degrees of freedom in covariance and the intuition for degrees of freedom as how many observations are not free to change after estimating $operatorname{Cov}(x,y)$? Is it possible to explain by counting newly restricted observations, like when explaining why the sample mean has degree of freedom $1,$ or variance degrees of freedom $n-1$?
The second related question is about the explanation in the reference linked above:
"Initially, we have $2n$ degrees of freedom in the bivariate data. We
lose two by computing the sample means $m(x)$ and $m(y)$. Of the
remaining $2n−2$ degrees of freedom, we lose $n−1$ by computing the
product deviations. Thus, we are left with $n−1$ degrees of freedom
total."
Do you see what are the "product deviations" and how does each one "lose" a degree of freedom?
statistics intuition estimation covariance
edited Oct 1 '18 at 22:05
Michael Hardy
1
asked Jan 22 '16 at 16:04
ikoiko
212
$endgroup$
add a comment |
$begingroup$
If we say $operatorname{Var}(x)$ has $n-1$ degrees of freedom which are lost after we estimate $operatorname{Var}(x)$, this matches how $n-1$ observations are now constrained to be sufficiently close to the remaining observation of $x$. In my class, $operatorname{Cov}(x,y)$ is also described as having $n-1$ degrees of freedom which are lost after we estimate $operatorname{Cov}(x,y)$. Reference for this use of "degrees of freedom" in covariance
My confusion is that the covariance does not actually seem to constrain $n-1$ $x$ and $y$ values, like the variance did for $x$.
Can you relate the $n-1$ degrees of freedom in covariance and the intuition for degrees of freedom as how many observations are not free to change after estimating $operatorname{Cov}(x,y)$? Is it possible to explain by counting newly restricted observations, like when explaining why the sample mean has degree of freedom $1,$ or variance degrees of freedom $n-1$?
The second related question is about the explanation in the reference linked above:
"Initially, we have $2n$ degrees of freedom in the bivariate data. We
lose two by computing the sample means $m(x)$ and $m(y)$. Of the
remaining $2n−2$ degrees of freedom, we lose $n−1$ by computing the
product deviations. Thus, we are left with $n−1$ degrees of freedom
total."
Do you see what are the "product deviations" and how does each one "lose" a degree of freedom?
statistics intuition estimation covariance
edited Oct 1 '18 at 22:05
Michael Hardy
1
asked Jan 22 '16 at 16:04
ikoiko
212
$endgroup$
add a comment |
$begingroup$
If we say $operatorname{Var}(x)$ has $n-1$ degrees of freedom which are lost after we estimate $operatorname{Var}(x)$, this matches how $n-1$ observations are now constrained to be sufficiently close to the remaining observation of $x$. In my class, $operatorname{Cov}(x,y)$ is also described as having $n-1$ degrees of freedom which are lost after we estimate $operatorname{Cov}(x,y)$. Reference for this use of "degrees of freedom" in covariance
My confusion is that the covariance does not actually seem to constrain $n-1$ $x$ and $y$ values, like the variance did for $x$.
Can you relate the $n-1$ degrees of freedom in covariance and the intuition for degrees of freedom as how many observations are not free to change after estimating $operatorname{Cov}(x,y)$? Is it possible to explain by counting newly restricted observations, like when explaining why the sample mean has degree of freedom $1,$ or variance degrees of freedom $n-1$?
The second related question is about the explanation in the reference linked above:
"Initially, we have $2n$ degrees of freedom in the bivariate data. We
lose two by computing the sample means $m(x)$ and $m(y)$. Of the
remaining $2n−2$ degrees of freedom, we lose $n−1$ by computing the
product deviations. Thus, we are left with $n−1$ degrees of freedom
total."
Do you see what are the "product deviations" and how does each one "lose" a degree of freedom?
statistics intuition estimation covariance
edited Oct 1 '18 at 22:05
Michael Hardy
1
asked Jan 22 '16 at 16:04
ikoiko
212
$endgroup$
If we say $operatorname{Var}(x)$ has $n-1$ degrees of freedom which are lost after we estimate $operatorname{Var}(x)$, this matches how $n-1$ observations are now constrained to be sufficiently close to the remaining observation of $x$. In my class, $operatorname{Cov}(x,y)$ is also described as having $n-1$ degrees of freedom which are lost after we estimate $operatorname{Cov}(x,y)$. Reference for this use of "degrees of freedom" in covariance
My confusion is that the covariance does not actually seem to constrain $n-1$ $x$ and $y$ values, like the variance did for $x$.
Can you relate the $n-1$ degrees of freedom in covariance and the intuition for degrees of freedom as how many observations are not free to change after estimating $operatorname{Cov}(x,y)$? Is it possible to explain by counting newly restricted observations, like when explaining why the sample mean has degree of freedom $1,$ or variance degrees of freedom $n-1$?
The second related question is about the explanation in the reference linked above:
"Initially, we have $2n$ degrees of freedom in the bivariate data. We
lose two by computing the sample means $m(x)$ and $m(y)$. Of the
remaining $2n−2$ degrees of freedom, we lose $n−1$ by computing the
product deviations. Thus, we are left with $n−1$ degrees of freedom
total."
Do you see what are the "product deviations" and how does each one "lose" a degree of freedom?
statistics intuition estimation covariance
statistics intuition estimation covariance
edited Oct 1 '18 at 22:05
Michael Hardy
1
asked Jan 22 '16 at 16:04
ikoiko
212
edited Oct 1 '18 at 22:05
Michael Hardy
1
asked Jan 22 '16 at 16:04
ikoiko
212
edited Oct 1 '18 at 22:05
Michael Hardy
1
edited Oct 1 '18 at 22:05
Michael Hardy
1
edited Oct 1 '18 at 22:05
Michael Hardy
1
1
asked Jan 22 '16 at 16:04
ikoiko
212
asked Jan 22 '16 at 16:04
ikoiko
212
asked Jan 22 '16 at 16:04
ikoiko
212
212
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Intuitively, the deduction of one degree of freedom is necessary to resolve a problem about the "biased"-ness of the estimator. An "unbiased" estimation for a (co)variance is one where it's "expected" to equal the population (co)variance I.E. if you take a SAMPLING distribution of (co)variance estimations and the average (or "expected value") of that distribution is the (co)variance of the population.
The "intuitive" explanation for the loss in degree of freedom in the variance and coveriance are EXACTLY the same issue in that it is concerning "biased"-ness of the estimator, which needs to be fixed by subtracting a degree of freedom.
Not sure if an informal proof of this fact would very helpful, but I'm going to cook one up myself (in a later edit of this post) to fine tune my econometrics knowledge so I can effectively tutor that next semester.
Hope this helps! :)
answered Jan 22 '16 at 17:38
Logician6Logician6
34314
$endgroup$
$begingroup$
Thanks for your reply. As you say, the estimator with $n-1$ has the true expectation. Can you say more explicitly how it is the same issue as degrees of freedom? In particular, I'm hoping to see how finding the covariance constrains $n-1$ observations of $x$ and $y$? Or, how every "product deviation" loses 1 degree of freedom?
$endgroup$
– iko
Jan 22 '16 at 19:06
$begingroup$
Well I think what you need are a few slightly technical definitions (but I promise to make them accessible). The problem with your question--and more generally with the "intuition" of statistics--is that sometimes the intuition that you want does not exist, since the answer you're looking for is buried beneath a very technical equation. But I think once you understand the basic notion of expected value for a "product distribution", and more specifically the covariance, I think a lot of confusion will be alleviated.
$endgroup$
– Logician6
Jan 22 '16 at 23:47
$begingroup$
I agree the expectation of the sample covariance (using $n-1$) equals the real covariance. Is that what you mean in: "expected value for a "product distribution", and more specifically the covariance"? Can this help me to see $n-1$ parameters (or observations) are no longer free to vary, after I find the covariance? It would help if you can add more explicit detail later :)
$endgroup$
– iko
Jan 23 '16 at 0:31
$begingroup$
Yes. That's what I mean to answer your first question. Not sure what you mean with your follow up question but hopefully my further elaboration will help. Expect it sometime tomorrow if not late tonight.
$endgroup$
– Logician6
Jan 23 '16 at 0:48
$begingroup$
Thanks for your continued input. To clarify, I can do the calculation to show the estimator is unbiased. It is the 2nd question in my comment above that I am asking about. If we say there are $n-1$ degrees of freedom in covariance, doesn't this mean that after we estimate the covariance, we lose the ability to freely choose $n-1$ observations/parameters?
$endgroup$
– iko
Jan 23 '16 at 0:54
|
show 2 more comments
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$begingroup$
Intuitively, the deduction of one degree of freedom is necessary to resolve a problem about the "biased"-ness of the estimator. An "unbiased" estimation for a (co)variance is one where it's "expected" to equal the population (co)variance I.E. if you take a SAMPLING distribution of (co)variance estimations and the average (or "expected value") of that distribution is the (co)variance of the population.
The "intuitive" explanation for the loss in degree of freedom in the variance and coveriance are EXACTLY the same issue in that it is concerning "biased"-ness of the estimator, which needs to be fixed by subtracting a degree of freedom.
Not sure if an informal proof of this fact would very helpful, but I'm going to cook one up myself (in a later edit of this post) to fine tune my econometrics knowledge so I can effectively tutor that next semester.
Hope this helps! :)
answered Jan 22 '16 at 17:38
Logician6Logician6
34314
$endgroup$
$begingroup$
Thanks for your reply. As you say, the estimator with $n-1$ has the true expectation. Can you say more explicitly how it is the same issue as degrees of freedom? In particular, I'm hoping to see how finding the covariance constrains $n-1$ observations of $x$ and $y$? Or, how every "product deviation" loses 1 degree of freedom?
$endgroup$
– iko
Jan 22 '16 at 19:06
$begingroup$
Well I think what you need are a few slightly technical definitions (but I promise to make them accessible). The problem with your question--and more generally with the "intuition" of statistics--is that sometimes the intuition that you want does not exist, since the answer you're looking for is buried beneath a very technical equation. But I think once you understand the basic notion of expected value for a "product distribution", and more specifically the covariance, I think a lot of confusion will be alleviated.
$endgroup$
– Logician6
Jan 22 '16 at 23:47
$begingroup$
I agree the expectation of the sample covariance (using $n-1$) equals the real covariance. Is that what you mean in: "expected value for a "product distribution", and more specifically the covariance"? Can this help me to see $n-1$ parameters (or observations) are no longer free to vary, after I find the covariance? It would help if you can add more explicit detail later :)
$endgroup$
– iko
Jan 23 '16 at 0:31
$begingroup$
Yes. That's what I mean to answer your first question. Not sure what you mean with your follow up question but hopefully my further elaboration will help. Expect it sometime tomorrow if not late tonight.
$endgroup$
– Logician6
Jan 23 '16 at 0:48
$begingroup$
Thanks for your continued input. To clarify, I can do the calculation to show the estimator is unbiased. It is the 2nd question in my comment above that I am asking about. If we say there are $n-1$ degrees of freedom in covariance, doesn't this mean that after we estimate the covariance, we lose the ability to freely choose $n-1$ observations/parameters?
$endgroup$
– iko
Jan 23 '16 at 0:54
|
show 2 more comments
$begingroup$
Intuitively, the deduction of one degree of freedom is necessary to resolve a problem about the "biased"-ness of the estimator. An "unbiased" estimation for a (co)variance is one where it's "expected" to equal the population (co)variance I.E. if you take a SAMPLING distribution of (co)variance estimations and the average (or "expected value") of that distribution is the (co)variance of the population.
The "intuitive" explanation for the loss in degree of freedom in the variance and coveriance are EXACTLY the same issue in that it is concerning "biased"-ness of the estimator, which needs to be fixed by subtracting a degree of freedom.
Not sure if an informal proof of this fact would very helpful, but I'm going to cook one up myself (in a later edit of this post) to fine tune my econometrics knowledge so I can effectively tutor that next semester.
Hope this helps! :)
answered Jan 22 '16 at 17:38
Logician6Logician6
34314
$endgroup$
$begingroup$
Thanks for your reply. As you say, the estimator with $n-1$ has the true expectation. Can you say more explicitly how it is the same issue as degrees of freedom? In particular, I'm hoping to see how finding the covariance constrains $n-1$ observations of $x$ and $y$? Or, how every "product deviation" loses 1 degree of freedom?
$endgroup$
– iko
Jan 22 '16 at 19:06
$begingroup$
Well I think what you need are a few slightly technical definitions (but I promise to make them accessible). The problem with your question--and more generally with the "intuition" of statistics--is that sometimes the intuition that you want does not exist, since the answer you're looking for is buried beneath a very technical equation. But I think once you understand the basic notion of expected value for a "product distribution", and more specifically the covariance, I think a lot of confusion will be alleviated.
$endgroup$
– Logician6
Jan 22 '16 at 23:47
$begingroup$
I agree the expectation of the sample covariance (using $n-1$) equals the real covariance. Is that what you mean in: "expected value for a "product distribution", and more specifically the covariance"? Can this help me to see $n-1$ parameters (or observations) are no longer free to vary, after I find the covariance? It would help if you can add more explicit detail later :)
$endgroup$
– iko
Jan 23 '16 at 0:31
$begingroup$
Yes. That's what I mean to answer your first question. Not sure what you mean with your follow up question but hopefully my further elaboration will help. Expect it sometime tomorrow if not late tonight.
$endgroup$
– Logician6
Jan 23 '16 at 0:48
$begingroup$
Thanks for your continued input. To clarify, I can do the calculation to show the estimator is unbiased. It is the 2nd question in my comment above that I am asking about. If we say there are $n-1$ degrees of freedom in covariance, doesn't this mean that after we estimate the covariance, we lose the ability to freely choose $n-1$ observations/parameters?
$endgroup$
– iko
Jan 23 '16 at 0:54
|
show 2 more comments
$begingroup$
Intuitively, the deduction of one degree of freedom is necessary to resolve a problem about the "biased"-ness of the estimator. An "unbiased" estimation for a (co)variance is one where it's "expected" to equal the population (co)variance I.E. if you take a SAMPLING distribution of (co)variance estimations and the average (or "expected value") of that distribution is the (co)variance of the population.
The "intuitive" explanation for the loss in degree of freedom in the variance and coveriance are EXACTLY the same issue in that it is concerning "biased"-ness of the estimator, which needs to be fixed by subtracting a degree of freedom.
Not sure if an informal proof of this fact would very helpful, but I'm going to cook one up myself (in a later edit of this post) to fine tune my econometrics knowledge so I can effectively tutor that next semester.
Hope this helps! :)
answered Jan 22 '16 at 17:38
Logician6Logician6
34314
$endgroup$
Intuitively, the deduction of one degree of freedom is necessary to resolve a problem about the "biased"-ness of the estimator. An "unbiased" estimation for a (co)variance is one where it's "expected" to equal the population (co)variance I.E. if you take a SAMPLING distribution of (co)variance estimations and the average (or "expected value") of that distribution is the (co)variance of the population.
The "intuitive" explanation for the loss in degree of freedom in the variance and coveriance are EXACTLY the same issue in that it is concerning "biased"-ness of the estimator, which needs to be fixed by subtracting a degree of freedom.
Not sure if an informal proof of this fact would very helpful, but I'm going to cook one up myself (in a later edit of this post) to fine tune my econometrics knowledge so I can effectively tutor that next semester.
Hope this helps! :)
answered Jan 22 '16 at 17:38
Logician6Logician6
34314
answered Jan 22 '16 at 17:38
Logician6Logician6
34314
answered Jan 22 '16 at 17:38
Logician6Logician6
34314
answered Jan 22 '16 at 17:38
Logician6Logician6
34314
34314
$begingroup$
Thanks for your reply. As you say, the estimator with $n-1$ has the true expectation. Can you say more explicitly how it is the same issue as degrees of freedom? In particular, I'm hoping to see how finding the covariance constrains $n-1$ observations of $x$ and $y$? Or, how every "product deviation" loses 1 degree of freedom?
$endgroup$
– iko
Jan 22 '16 at 19:06
$begingroup$
Well I think what you need are a few slightly technical definitions (but I promise to make them accessible). The problem with your question--and more generally with the "intuition" of statistics--is that sometimes the intuition that you want does not exist, since the answer you're looking for is buried beneath a very technical equation. But I think once you understand the basic notion of expected value for a "product distribution", and more specifically the covariance, I think a lot of confusion will be alleviated.
$endgroup$
– Logician6
Jan 22 '16 at 23:47
$begingroup$
I agree the expectation of the sample covariance (using $n-1$) equals the real covariance. Is that what you mean in: "expected value for a "product distribution", and more specifically the covariance"? Can this help me to see $n-1$ parameters (or observations) are no longer free to vary, after I find the covariance? It would help if you can add more explicit detail later :)
$endgroup$
– iko
Jan 23 '16 at 0:31
$begingroup$
Yes. That's what I mean to answer your first question. Not sure what you mean with your follow up question but hopefully my further elaboration will help. Expect it sometime tomorrow if not late tonight.
$endgroup$
– Logician6
Jan 23 '16 at 0:48
$begingroup$
Thanks for your continued input. To clarify, I can do the calculation to show the estimator is unbiased. It is the 2nd question in my comment above that I am asking about. If we say there are $n-1$ degrees of freedom in covariance, doesn't this mean that after we estimate the covariance, we lose the ability to freely choose $n-1$ observations/parameters?
$endgroup$
– iko
Jan 23 '16 at 0:54
|
show 2 more comments
$begingroup$
Thanks for your reply. As you say, the estimator with $n-1$ has the true expectation. Can you say more explicitly how it is the same issue as degrees of freedom? In particular, I'm hoping to see how finding the covariance constrains $n-1$ observations of $x$ and $y$? Or, how every "product deviation" loses 1 degree of freedom?
$endgroup$
– iko
Jan 22 '16 at 19:06
$begingroup$
Well I think what you need are a few slightly technical definitions (but I promise to make them accessible). The problem with your question--and more generally with the "intuition" of statistics--is that sometimes the intuition that you want does not exist, since the answer you're looking for is buried beneath a very technical equation. But I think once you understand the basic notion of expected value for a "product distribution", and more specifically the covariance, I think a lot of confusion will be alleviated.
$endgroup$
– Logician6
Jan 22 '16 at 23:47
$begingroup$
I agree the expectation of the sample covariance (using $n-1$) equals the real covariance. Is that what you mean in: "expected value for a "product distribution", and more specifically the covariance"? Can this help me to see $n-1$ parameters (or observations) are no longer free to vary, after I find the covariance? It would help if you can add more explicit detail later :)
$endgroup$
– iko
Jan 23 '16 at 0:31
$begingroup$
Yes. That's what I mean to answer your first question. Not sure what you mean with your follow up question but hopefully my further elaboration will help. Expect it sometime tomorrow if not late tonight.
$endgroup$
– Logician6
Jan 23 '16 at 0:48
$begingroup$
Thanks for your continued input. To clarify, I can do the calculation to show the estimator is unbiased. It is the 2nd question in my comment above that I am asking about. If we say there are $n-1$ degrees of freedom in covariance, doesn't this mean that after we estimate the covariance, we lose the ability to freely choose $n-1$ observations/parameters?
$endgroup$
– iko
Jan 23 '16 at 0:54
$begingroup$
Thanks for your reply. As you say, the estimator with $n-1$ has the true expectation. Can you say more explicitly how it is the same issue as degrees of freedom? In particular, I'm hoping to see how finding the covariance constrains $n-1$ observations of $x$ and $y$? Or, how every "product deviation" loses 1 degree of freedom?
$endgroup$
– iko
Jan 22 '16 at 19:06
$begingroup$
Thanks for your reply. As you say, the estimator with $n-1$ has the true expectation. Can you say more explicitly how it is the same issue as degrees of freedom? In particular, I'm hoping to see how finding the covariance constrains $n-1$ observations of $x$ and $y$? Or, how every "product deviation" loses 1 degree of freedom?
$endgroup$
– iko
Jan 22 '16 at 19:06
$begingroup$
Well I think what you need are a few slightly technical definitions (but I promise to make them accessible). The problem with your question--and more generally with the "intuition" of statistics--is that sometimes the intuition that you want does not exist, since the answer you're looking for is buried beneath a very technical equation. But I think once you understand the basic notion of expected value for a "product distribution", and more specifically the covariance, I think a lot of confusion will be alleviated.
$endgroup$
– Logician6
Jan 22 '16 at 23:47
$begingroup$
Well I think what you need are a few slightly technical definitions (but I promise to make them accessible). The problem with your question--and more generally with the "intuition" of statistics--is that sometimes the intuition that you want does not exist, since the answer you're looking for is buried beneath a very technical equation. But I think once you understand the basic notion of expected value for a "product distribution", and more specifically the covariance, I think a lot of confusion will be alleviated.
$endgroup$
– Logician6
Jan 22 '16 at 23:47
$begingroup$
I agree the expectation of the sample covariance (using $n-1$) equals the real covariance. Is that what you mean in: "expected value for a "product distribution", and more specifically the covariance"? Can this help me to see $n-1$ parameters (or observations) are no longer free to vary, after I find the covariance? It would help if you can add more explicit detail later :)
$endgroup$
– iko
Jan 23 '16 at 0:31
$begingroup$
I agree the expectation of the sample covariance (using $n-1$) equals the real covariance. Is that what you mean in: "expected value for a "product distribution", and more specifically the covariance"? Can this help me to see $n-1$ parameters (or observations) are no longer free to vary, after I find the covariance? It would help if you can add more explicit detail later :)
$endgroup$
– iko
Jan 23 '16 at 0:31
$begingroup$
Yes. That's what I mean to answer your first question. Not sure what you mean with your follow up question but hopefully my further elaboration will help. Expect it sometime tomorrow if not late tonight.
$endgroup$
– Logician6
Jan 23 '16 at 0:48
$begingroup$
Yes. That's what I mean to answer your first question. Not sure what you mean with your follow up question but hopefully my further elaboration will help. Expect it sometime tomorrow if not late tonight.
$endgroup$
– Logician6
Jan 23 '16 at 0:48
$begingroup$
Thanks for your continued input. To clarify, I can do the calculation to show the estimator is unbiased. It is the 2nd question in my comment above that I am asking about. If we say there are $n-1$ degrees of freedom in covariance, doesn't this mean that after we estimate the covariance, we lose the ability to freely choose $n-1$ observations/parameters?
$endgroup$
– iko
Jan 23 '16 at 0:54
$begingroup$
Thanks for your continued input. To clarify, I can do the calculation to show the estimator is unbiased. It is the 2nd question in my comment above that I am asking about. If we say there are $n-1$ degrees of freedom in covariance, doesn't this mean that after we estimate the covariance, we lose the ability to freely choose $n-1$ observations/parameters?
$endgroup$
– iko
Jan 23 '16 at 0:54
|
show 2 more comments
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