What is the unbiased estimator of covariance matrix of N-dimensional random variable?
$begingroup$
Suppose $x$ is a random vector in $mathbb{R}^n$ which is distributed according to $D$.
What is the unbiased estimator of covariance matrix of an N-dimensional random variable?
When $y$ is a i.i.d. random variable and we have access to $(y_1,y_2,cdots,y_n)$, the sample mean is an unbiased estimator of $hat{mu}=frac{sum_{i=1}^N}{N}$ and $hat{sigma}^2=frac{1}{N-1}sum_{i=1}^N(y_i-hat{mu})^2$ is an unbiased estimator of variance.
By going to higher dimension in addition to variance we have covariance between each element of the random vector. My question is
$$
hat{C}=?
$$
where $hat{C}$ is an unbiased estimator of $C = mathbb{E}[(x-mu)(x-mu)^T]$.
statistics estimation covariance estimation-theory
asked Dec 26 '18 at 18:05
SaeedSaeed
1,036310
$endgroup$
add a comment |
$begingroup$
Suppose $x$ is a random vector in $mathbb{R}^n$ which is distributed according to $D$.
What is the unbiased estimator of covariance matrix of an N-dimensional random variable?
When $y$ is a i.i.d. random variable and we have access to $(y_1,y_2,cdots,y_n)$, the sample mean is an unbiased estimator of $hat{mu}=frac{sum_{i=1}^N}{N}$ and $hat{sigma}^2=frac{1}{N-1}sum_{i=1}^N(y_i-hat{mu})^2$ is an unbiased estimator of variance.
By going to higher dimension in addition to variance we have covariance between each element of the random vector. My question is
$$
hat{C}=?
$$
where $hat{C}$ is an unbiased estimator of $C = mathbb{E}[(x-mu)(x-mu)^T]$.
statistics estimation covariance estimation-theory
asked Dec 26 '18 at 18:05
SaeedSaeed
1,036310
$endgroup$
$begingroup$
Essentially what you might expect, with $hat{vec{mu}}= frac1n sum vec{x}_i$ for the estimator of the mean vector and $frac1{n-1} sum (vec{x}_i - hat{vec{mu}})(vec{x}_i - hat{vec{mu}})^T$ for the estimator of the covariance matrix. See math.stackexchange.com/questions/2019122/…
$endgroup$
– Henry
Dec 26 '18 at 18:12
$begingroup$
@Henry: That answer is for the $(X,Y)$ random vector, I need a proof in terms of vector.
$endgroup$
– Saeed
Dec 26 '18 at 19:33
add a comment |
$begingroup$
Suppose $x$ is a random vector in $mathbb{R}^n$ which is distributed according to $D$.
What is the unbiased estimator of covariance matrix of an N-dimensional random variable?
When $y$ is a i.i.d. random variable and we have access to $(y_1,y_2,cdots,y_n)$, the sample mean is an unbiased estimator of $hat{mu}=frac{sum_{i=1}^N}{N}$ and $hat{sigma}^2=frac{1}{N-1}sum_{i=1}^N(y_i-hat{mu})^2$ is an unbiased estimator of variance.
By going to higher dimension in addition to variance we have covariance between each element of the random vector. My question is
$$
hat{C}=?
$$
where $hat{C}$ is an unbiased estimator of $C = mathbb{E}[(x-mu)(x-mu)^T]$.
statistics estimation covariance estimation-theory
asked Dec 26 '18 at 18:05
SaeedSaeed
1,036310
$endgroup$
Suppose $x$ is a random vector in $mathbb{R}^n$ which is distributed according to $D$.
What is the unbiased estimator of covariance matrix of an N-dimensional random variable?
When $y$ is a i.i.d. random variable and we have access to $(y_1,y_2,cdots,y_n)$, the sample mean is an unbiased estimator of $hat{mu}=frac{sum_{i=1}^N}{N}$ and $hat{sigma}^2=frac{1}{N-1}sum_{i=1}^N(y_i-hat{mu})^2$ is an unbiased estimator of variance.
By going to higher dimension in addition to variance we have covariance between each element of the random vector. My question is
$$
hat{C}=?
$$
where $hat{C}$ is an unbiased estimator of $C = mathbb{E}[(x-mu)(x-mu)^T]$.
statistics estimation covariance estimation-theory
statistics estimation covariance estimation-theory
asked Dec 26 '18 at 18:05
SaeedSaeed
1,036310
asked Dec 26 '18 at 18:05
SaeedSaeed
1,036310
asked Dec 26 '18 at 18:05
SaeedSaeed
1,036310
asked Dec 26 '18 at 18:05
SaeedSaeed
1,036310
asked Dec 26 '18 at 18:05
SaeedSaeed
1,036310
1,036310
$begingroup$
Essentially what you might expect, with $hat{vec{mu}}= frac1n sum vec{x}_i$ for the estimator of the mean vector and $frac1{n-1} sum (vec{x}_i - hat{vec{mu}})(vec{x}_i - hat{vec{mu}})^T$ for the estimator of the covariance matrix. See math.stackexchange.com/questions/2019122/…
$endgroup$
– Henry
Dec 26 '18 at 18:12
$begingroup$
@Henry: That answer is for the $(X,Y)$ random vector, I need a proof in terms of vector.
$endgroup$
– Saeed
Dec 26 '18 at 19:33
add a comment |
$begingroup$
Essentially what you might expect, with $hat{vec{mu}}= frac1n sum vec{x}_i$ for the estimator of the mean vector and $frac1{n-1} sum (vec{x}_i - hat{vec{mu}})(vec{x}_i - hat{vec{mu}})^T$ for the estimator of the covariance matrix. See math.stackexchange.com/questions/2019122/…
$endgroup$
– Henry
Dec 26 '18 at 18:12
$begingroup$
@Henry: That answer is for the $(X,Y)$ random vector, I need a proof in terms of vector.
$endgroup$
– Saeed
Dec 26 '18 at 19:33
$begingroup$
Essentially what you might expect, with $hat{vec{mu}}= frac1n sum vec{x}_i$ for the estimator of the mean vector and $frac1{n-1} sum (vec{x}_i - hat{vec{mu}})(vec{x}_i - hat{vec{mu}})^T$ for the estimator of the covariance matrix. See math.stackexchange.com/questions/2019122/…
$endgroup$
– Henry
Dec 26 '18 at 18:12
$begingroup$
Essentially what you might expect, with $hat{vec{mu}}= frac1n sum vec{x}_i$ for the estimator of the mean vector and $frac1{n-1} sum (vec{x}_i - hat{vec{mu}})(vec{x}_i - hat{vec{mu}})^T$ for the estimator of the covariance matrix. See math.stackexchange.com/questions/2019122/…
$endgroup$
– Henry
Dec 26 '18 at 18:12
$begingroup$
@Henry: That answer is for the $(X,Y)$ random vector, I need a proof in terms of vector.
$endgroup$
– Saeed
Dec 26 '18 at 19:33
$begingroup$
@Henry: That answer is for the $(X,Y)$ random vector, I need a proof in terms of vector.
$endgroup$
– Saeed
Dec 26 '18 at 19:33
add a comment |
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$begingroup$
Essentially what you might expect, with $hat{vec{mu}}= frac1n sum vec{x}_i$ for the estimator of the mean vector and $frac1{n-1} sum (vec{x}_i - hat{vec{mu}})(vec{x}_i - hat{vec{mu}})^T$ for the estimator of the covariance matrix. See math.stackexchange.com/questions/2019122/…
$endgroup$
– Henry
Dec 26 '18 at 18:12
$begingroup$
@Henry: That answer is for the $(X,Y)$ random vector, I need a proof in terms of vector.
$endgroup$
– Saeed
Dec 26 '18 at 19:33