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The dimension of the parameter space of multivariate normal distribution?
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In multivariate normal distribution, $Xsim N(mu,Sigma)$, where $muin mathbbR^p,~Sigma inmathbbR^ptimes p,~ Sigma succ0text and Sigma$ is symmetric. My professor told me that the dimension of the parameters space is $p+p+1 choose 2$. I know the dimension of $mu$ is $p$, but I don't know where does the $p+1 choose 2$ come from.
matrices statistics normal-distribution
asked Jul 24 at 15:11
coolcat
1018
add a comment |Â
up vote
1
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In multivariate normal distribution, $Xsim N(mu,Sigma)$, where $muin mathbbR^p,~Sigma inmathbbR^ptimes p,~ Sigma succ0text and Sigma$ is symmetric. My professor told me that the dimension of the parameters space is $p+p+1 choose 2$. I know the dimension of $mu$ is $p$, but I don't know where does the $p+1 choose 2$ come from.
matrices statistics normal-distribution
asked Jul 24 at 15:11
coolcat
1018
Maybe there is a typo and it supposed to be $p+binomp-12$
– callculus
Jul 24 at 16:11
@callculus: No, it's correct. Try $p=1$.
– joriki
Jul 24 at 16:14
@joriki For $p=1$ we have an univariate distribution. Is that right? And the parameter space for the variance then is $1+1=2$. It doesn´t looks right to me.
– callculus
Jul 24 at 16:33
@callculus: No, for a univariate distribution the variance contributes only $1$ to the dimension of the parameter space. So the total comes out correctly as $1+binom1+12=1+1=2$.
– joriki
Jul 24 at 17:36
@joriki Thanks for your comment. But I´ve to admit I cannot comprehend what you said. But it doesn´t matter. At the moment I´m more interested enjoying the sunny weather outside.
– callculus
Jul 24 at 18:07
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In multivariate normal distribution, $Xsim N(mu,Sigma)$, where $muin mathbbR^p,~Sigma inmathbbR^ptimes p,~ Sigma succ0text and Sigma$ is symmetric. My professor told me that the dimension of the parameters space is $p+p+1 choose 2$. I know the dimension of $mu$ is $p$, but I don't know where does the $p+1 choose 2$ come from.
matrices statistics normal-distribution
asked Jul 24 at 15:11
coolcat
1018
In multivariate normal distribution, $Xsim N(mu,Sigma)$, where $muin mathbbR^p,~Sigma inmathbbR^ptimes p,~ Sigma succ0text and Sigma$ is symmetric. My professor told me that the dimension of the parameters space is $p+p+1 choose 2$. I know the dimension of $mu$ is $p$, but I don't know where does the $p+1 choose 2$ come from.
matrices statistics normal-distribution
asked Jul 24 at 15:11
coolcat
1018
edited Jul 24 at 15:41
asked Jul 24 at 15:11
coolcat
1018
asked Jul 24 at 15:11
coolcat
1018
asked Jul 24 at 15:11
coolcat
1018
1018
Maybe there is a typo and it supposed to be $p+binomp-12$
– callculus
Jul 24 at 16:11
@callculus: No, it's correct. Try $p=1$.
– joriki
Jul 24 at 16:14
@joriki For $p=1$ we have an univariate distribution. Is that right? And the parameter space for the variance then is $1+1=2$. It doesn´t looks right to me.
– callculus
Jul 24 at 16:33
@callculus: No, for a univariate distribution the variance contributes only $1$ to the dimension of the parameter space. So the total comes out correctly as $1+binom1+12=1+1=2$.
– joriki
Jul 24 at 17:36
@joriki Thanks for your comment. But I´ve to admit I cannot comprehend what you said. But it doesn´t matter. At the moment I´m more interested enjoying the sunny weather outside.
– callculus
Jul 24 at 18:07
add a comment |Â
Maybe there is a typo and it supposed to be $p+binomp-12$
– callculus
Jul 24 at 16:11
@callculus: No, it's correct. Try $p=1$.
– joriki
Jul 24 at 16:14
@joriki For $p=1$ we have an univariate distribution. Is that right? And the parameter space for the variance then is $1+1=2$. It doesn´t looks right to me.
– callculus
Jul 24 at 16:33
@callculus: No, for a univariate distribution the variance contributes only $1$ to the dimension of the parameter space. So the total comes out correctly as $1+binom1+12=1+1=2$.
– joriki
Jul 24 at 17:36
@joriki Thanks for your comment. But I´ve to admit I cannot comprehend what you said. But it doesn´t matter. At the moment I´m more interested enjoying the sunny weather outside.
– callculus
Jul 24 at 18:07
Maybe there is a typo and it supposed to be $p+binomp-12$
– callculus
Jul 24 at 16:11
Maybe there is a typo and it supposed to be $p+binomp-12$
– callculus
Jul 24 at 16:11
@callculus: No, it's correct. Try $p=1$.
– joriki
Jul 24 at 16:14
@callculus: No, it's correct. Try $p=1$.
– joriki
Jul 24 at 16:14
@joriki For $p=1$ we have an univariate distribution. Is that right? And the parameter space for the variance then is $1+1=2$. It doesn´t looks right to me.
– callculus
Jul 24 at 16:33
@joriki For $p=1$ we have an univariate distribution. Is that right? And the parameter space for the variance then is $1+1=2$. It doesn´t looks right to me.
– callculus
Jul 24 at 16:33
@callculus: No, for a univariate distribution the variance contributes only $1$ to the dimension of the parameter space. So the total comes out correctly as $1+binom1+12=1+1=2$.
– joriki
Jul 24 at 17:36
@callculus: No, for a univariate distribution the variance contributes only $1$ to the dimension of the parameter space. So the total comes out correctly as $1+binom1+12=1+1=2$.
– joriki
Jul 24 at 17:36
@joriki Thanks for your comment. But I´ve to admit I cannot comprehend what you said. But it doesn´t matter. At the moment I´m more interested enjoying the sunny weather outside.
– callculus
Jul 24 at 18:07
@joriki Thanks for your comment. But I´ve to admit I cannot comprehend what you said. But it doesn´t matter. At the moment I´m more interested enjoying the sunny weather outside.
– callculus
Jul 24 at 18:07
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
You have $muinmathbb R^p$ and $Sigmain mathbb R^ptimes p$ and $Sigma$ is symmetric.
If there were no other constraints on the parameter space that would give you the number of scalar components of $mu,$ which is $p,$ plus the number of scalar entries on or above the diagonal of $Sigma$ (since those below the diagonal are completely determined by those above. The number on the diagonal is $p.$ The number above the diagonal is $dbinom p 2,$ since you have to choose one of the $p$ rows and one of the $p$ columns, and they can't both have the same index, so you have to choose two of $1,ldots, p.$ That means you have
$$
p + p + binom p 2.
$$
And then
$$
p + binom p 2 = p + fracp(p-1) 2 = frac2p2 + fracp^2-p 2 = fracp(p+1) 2 = binom p+1 2.
$$
The subtler part is showing that the constraint that $Sigma$ must be nonnegative-definite does not further reduce the dimension.
That can be shown by induction on $p$ if you know a certain trick, described in this posting.
PS: I suppose it should also be mentioned that every nonnegative-definite matrix is the variance of some random vector. Thus the constraint that $Sigma$ must be such a variance does not diminish the parameter space beyond saying $Sigma$ must be nonnegative-definite. This "every" can be deduced from the finite-dimensional version of the spectral theorem.
answered Jul 24 at 19:51
Michael Hardy
204k23186461
That's what I thought too.
– callculus
Jul 24 at 20:30
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You have $muinmathbb R^p$ and $Sigmain mathbb R^ptimes p$ and $Sigma$ is symmetric.
If there were no other constraints on the parameter space that would give you the number of scalar components of $mu,$ which is $p,$ plus the number of scalar entries on or above the diagonal of $Sigma$ (since those below the diagonal are completely determined by those above. The number on the diagonal is $p.$ The number above the diagonal is $dbinom p 2,$ since you have to choose one of the $p$ rows and one of the $p$ columns, and they can't both have the same index, so you have to choose two of $1,ldots, p.$ That means you have
$$
p + p + binom p 2.
$$
And then
$$
p + binom p 2 = p + fracp(p-1) 2 = frac2p2 + fracp^2-p 2 = fracp(p+1) 2 = binom p+1 2.
$$
The subtler part is showing that the constraint that $Sigma$ must be nonnegative-definite does not further reduce the dimension.
That can be shown by induction on $p$ if you know a certain trick, described in this posting.
PS: I suppose it should also be mentioned that every nonnegative-definite matrix is the variance of some random vector. Thus the constraint that $Sigma$ must be such a variance does not diminish the parameter space beyond saying $Sigma$ must be nonnegative-definite. This "every" can be deduced from the finite-dimensional version of the spectral theorem.
answered Jul 24 at 19:51
Michael Hardy
204k23186461
That's what I thought too.
– callculus
Jul 24 at 20:30
add a comment |Â
up vote
2
down vote
accepted
You have $muinmathbb R^p$ and $Sigmain mathbb R^ptimes p$ and $Sigma$ is symmetric.
If there were no other constraints on the parameter space that would give you the number of scalar components of $mu,$ which is $p,$ plus the number of scalar entries on or above the diagonal of $Sigma$ (since those below the diagonal are completely determined by those above. The number on the diagonal is $p.$ The number above the diagonal is $dbinom p 2,$ since you have to choose one of the $p$ rows and one of the $p$ columns, and they can't both have the same index, so you have to choose two of $1,ldots, p.$ That means you have
$$
p + p + binom p 2.
$$
And then
$$
p + binom p 2 = p + fracp(p-1) 2 = frac2p2 + fracp^2-p 2 = fracp(p+1) 2 = binom p+1 2.
$$
The subtler part is showing that the constraint that $Sigma$ must be nonnegative-definite does not further reduce the dimension.
That can be shown by induction on $p$ if you know a certain trick, described in this posting.
PS: I suppose it should also be mentioned that every nonnegative-definite matrix is the variance of some random vector. Thus the constraint that $Sigma$ must be such a variance does not diminish the parameter space beyond saying $Sigma$ must be nonnegative-definite. This "every" can be deduced from the finite-dimensional version of the spectral theorem.
answered Jul 24 at 19:51
Michael Hardy
204k23186461
That's what I thought too.
– callculus
Jul 24 at 20:30
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You have $muinmathbb R^p$ and $Sigmain mathbb R^ptimes p$ and $Sigma$ is symmetric.
If there were no other constraints on the parameter space that would give you the number of scalar components of $mu,$ which is $p,$ plus the number of scalar entries on or above the diagonal of $Sigma$ (since those below the diagonal are completely determined by those above. The number on the diagonal is $p.$ The number above the diagonal is $dbinom p 2,$ since you have to choose one of the $p$ rows and one of the $p$ columns, and they can't both have the same index, so you have to choose two of $1,ldots, p.$ That means you have
$$
p + p + binom p 2.
$$
And then
$$
p + binom p 2 = p + fracp(p-1) 2 = frac2p2 + fracp^2-p 2 = fracp(p+1) 2 = binom p+1 2.
$$
The subtler part is showing that the constraint that $Sigma$ must be nonnegative-definite does not further reduce the dimension.
That can be shown by induction on $p$ if you know a certain trick, described in this posting.
PS: I suppose it should also be mentioned that every nonnegative-definite matrix is the variance of some random vector. Thus the constraint that $Sigma$ must be such a variance does not diminish the parameter space beyond saying $Sigma$ must be nonnegative-definite. This "every" can be deduced from the finite-dimensional version of the spectral theorem.
answered Jul 24 at 19:51
Michael Hardy
204k23186461
You have $muinmathbb R^p$ and $Sigmain mathbb R^ptimes p$ and $Sigma$ is symmetric.
If there were no other constraints on the parameter space that would give you the number of scalar components of $mu,$ which is $p,$ plus the number of scalar entries on or above the diagonal of $Sigma$ (since those below the diagonal are completely determined by those above. The number on the diagonal is $p.$ The number above the diagonal is $dbinom p 2,$ since you have to choose one of the $p$ rows and one of the $p$ columns, and they can't both have the same index, so you have to choose two of $1,ldots, p.$ That means you have
$$
p + p + binom p 2.
$$
And then
$$
p + binom p 2 = p + fracp(p-1) 2 = frac2p2 + fracp^2-p 2 = fracp(p+1) 2 = binom p+1 2.
$$
The subtler part is showing that the constraint that $Sigma$ must be nonnegative-definite does not further reduce the dimension.
That can be shown by induction on $p$ if you know a certain trick, described in this posting.
PS: I suppose it should also be mentioned that every nonnegative-definite matrix is the variance of some random vector. Thus the constraint that $Sigma$ must be such a variance does not diminish the parameter space beyond saying $Sigma$ must be nonnegative-definite. This "every" can be deduced from the finite-dimensional version of the spectral theorem.
answered Jul 24 at 19:51
Michael Hardy
204k23186461
edited Jul 24 at 23:01
answered Jul 24 at 19:51
Michael Hardy
204k23186461
answered Jul 24 at 19:51
Michael Hardy
204k23186461
answered Jul 24 at 19:51
Michael Hardy
204k23186461
204k23186461
That's what I thought too.
– callculus
Jul 24 at 20:30
add a comment |Â
That's what I thought too.
– callculus
Jul 24 at 20:30
That's what I thought too.
– callculus
Jul 24 at 20:30
That's what I thought too.
– callculus
Jul 24 at 20:30
add a comment |Â
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Maybe there is a typo and it supposed to be $p+binomp-12$
– callculus
Jul 24 at 16:11
@callculus: No, it's correct. Try $p=1$.
– joriki
Jul 24 at 16:14
@joriki For $p=1$ we have an univariate distribution. Is that right? And the parameter space for the variance then is $1+1=2$. It doesn´t looks right to me.
– callculus
Jul 24 at 16:33
@callculus: No, for a univariate distribution the variance contributes only $1$ to the dimension of the parameter space. So the total comes out correctly as $1+binom1+12=1+1=2$.
– joriki
Jul 24 at 17:36
@joriki Thanks for your comment. But I´ve to admit I cannot comprehend what you said. But it doesn´t matter. At the moment I´m more interested enjoying the sunny weather outside.
– callculus
Jul 24 at 18:07