Multivariate Gaussian Definition when Covariance matrix is singular, What's wrong?
$begingroup$
Given
$$mathbf{Sigma} in mathbb R^{k times k}$$
$$mathbf{u} in mathbb R^k$$
The multivariate Gaussian pdf can be determined By definition:
$$f(mathbf{x})=frac{1}{2pi^{frac{-k}{2}}|Sigma|^{frac{1}{2}}}e^{frac{1}{2}(mathbf{x-u})^Tmathbf{Sigma}^{-1}(mathbf{x-u})}$$
The Covariance matrix is only limited to be positive semidefinite.
So it could be singular (Non-invertible)
This will also lead to a zero in the denumerator, and also the$Sigma^{-1}$ doesn't exist.
What we do in that case to write the joint pdf?
probability
asked Dec 27 '15 at 7:56
Allen KuoAllen Kuo
1639
$endgroup$
add a comment |
$begingroup$
Given
$$mathbf{Sigma} in mathbb R^{k times k}$$
$$mathbf{u} in mathbb R^k$$
The multivariate Gaussian pdf can be determined By definition:
$$f(mathbf{x})=frac{1}{2pi^{frac{-k}{2}}|Sigma|^{frac{1}{2}}}e^{frac{1}{2}(mathbf{x-u})^Tmathbf{Sigma}^{-1}(mathbf{x-u})}$$
The Covariance matrix is only limited to be positive semidefinite.
So it could be singular (Non-invertible)
This will also lead to a zero in the denumerator, and also the$Sigma^{-1}$ doesn't exist.
What we do in that case to write the joint pdf?
probability
asked Dec 27 '15 at 7:56
Allen KuoAllen Kuo
1639
$endgroup$
2
$begingroup$
This MO question might help you.
$endgroup$
– Artem Mavrin
Dec 27 '15 at 8:03
add a comment |
$begingroup$
Given
$$mathbf{Sigma} in mathbb R^{k times k}$$
$$mathbf{u} in mathbb R^k$$
The multivariate Gaussian pdf can be determined By definition:
$$f(mathbf{x})=frac{1}{2pi^{frac{-k}{2}}|Sigma|^{frac{1}{2}}}e^{frac{1}{2}(mathbf{x-u})^Tmathbf{Sigma}^{-1}(mathbf{x-u})}$$
The Covariance matrix is only limited to be positive semidefinite.
So it could be singular (Non-invertible)
This will also lead to a zero in the denumerator, and also the$Sigma^{-1}$ doesn't exist.
What we do in that case to write the joint pdf?
probability
asked Dec 27 '15 at 7:56
Allen KuoAllen Kuo
1639
$endgroup$
Given
$$mathbf{Sigma} in mathbb R^{k times k}$$
$$mathbf{u} in mathbb R^k$$
The multivariate Gaussian pdf can be determined By definition:
$$f(mathbf{x})=frac{1}{2pi^{frac{-k}{2}}|Sigma|^{frac{1}{2}}}e^{frac{1}{2}(mathbf{x-u})^Tmathbf{Sigma}^{-1}(mathbf{x-u})}$$
The Covariance matrix is only limited to be positive semidefinite.
So it could be singular (Non-invertible)
This will also lead to a zero in the denumerator, and also the$Sigma^{-1}$ doesn't exist.
What we do in that case to write the joint pdf?
probability
probability
asked Dec 27 '15 at 7:56
Allen KuoAllen Kuo
1639
asked Dec 27 '15 at 7:56
Allen KuoAllen Kuo
1639
edited Dec 27 '15 at 8:04
Allen Kuo
asked Dec 27 '15 at 7:56
Allen KuoAllen Kuo
1639
asked Dec 27 '15 at 7:56
Allen KuoAllen Kuo
1639
asked Dec 27 '15 at 7:56
Allen KuoAllen Kuo
1639
1639
2
$begingroup$
This MO question might help you.
$endgroup$
– Artem Mavrin
Dec 27 '15 at 8:03
add a comment |
2
$begingroup$
This MO question might help you.
$endgroup$
– Artem Mavrin
Dec 27 '15 at 8:03
2
2
$begingroup$
This MO question might help you.
$endgroup$
– Artem Mavrin
Dec 27 '15 at 8:03
$begingroup$
This MO question might help you.
$endgroup$
– Artem Mavrin
Dec 27 '15 at 8:03
add a comment |
1 Answer
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$begingroup$
Covariance matrix of multivariant Gaussian is square, symmetric (both for all covariance matrices) and positive definite (semi-positive definite for all covariance matrices).
So why is the matrix positive definite? It we has one or more dimensions that have variance = 0 this means that they are deterministic (constant) and hence we can neglect these dimensions. If we neglect all dimensions that do not provide any information regarding the RV we will remain with positive definite matrix instead of semi positive definite.
A square positive definite matrix is always invertible (det not equal zero).
answered Dec 7 '18 at 19:01
Chen ShaniChen Shani
11
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Covariance matrix of multivariant Gaussian is square, symmetric (both for all covariance matrices) and positive definite (semi-positive definite for all covariance matrices).
So why is the matrix positive definite? It we has one or more dimensions that have variance = 0 this means that they are deterministic (constant) and hence we can neglect these dimensions. If we neglect all dimensions that do not provide any information regarding the RV we will remain with positive definite matrix instead of semi positive definite.
A square positive definite matrix is always invertible (det not equal zero).
answered Dec 7 '18 at 19:01
Chen ShaniChen Shani
11
$endgroup$
add a comment |
$begingroup$
Covariance matrix of multivariant Gaussian is square, symmetric (both for all covariance matrices) and positive definite (semi-positive definite for all covariance matrices).
So why is the matrix positive definite? It we has one or more dimensions that have variance = 0 this means that they are deterministic (constant) and hence we can neglect these dimensions. If we neglect all dimensions that do not provide any information regarding the RV we will remain with positive definite matrix instead of semi positive definite.
A square positive definite matrix is always invertible (det not equal zero).
answered Dec 7 '18 at 19:01
Chen ShaniChen Shani
11
$endgroup$
add a comment |
$begingroup$
Covariance matrix of multivariant Gaussian is square, symmetric (both for all covariance matrices) and positive definite (semi-positive definite for all covariance matrices).
So why is the matrix positive definite? It we has one or more dimensions that have variance = 0 this means that they are deterministic (constant) and hence we can neglect these dimensions. If we neglect all dimensions that do not provide any information regarding the RV we will remain with positive definite matrix instead of semi positive definite.
A square positive definite matrix is always invertible (det not equal zero).
answered Dec 7 '18 at 19:01
Chen ShaniChen Shani
11
$endgroup$
Covariance matrix of multivariant Gaussian is square, symmetric (both for all covariance matrices) and positive definite (semi-positive definite for all covariance matrices).
So why is the matrix positive definite? It we has one or more dimensions that have variance = 0 this means that they are deterministic (constant) and hence we can neglect these dimensions. If we neglect all dimensions that do not provide any information regarding the RV we will remain with positive definite matrix instead of semi positive definite.
A square positive definite matrix is always invertible (det not equal zero).
answered Dec 7 '18 at 19:01
Chen ShaniChen Shani
11
edited Dec 8 '18 at 9:53
answered Dec 7 '18 at 19:01
Chen ShaniChen Shani
11
answered Dec 7 '18 at 19:01
Chen ShaniChen Shani
11
answered Dec 7 '18 at 19:01
Chen ShaniChen Shani
11
11
add a comment |
add a comment |
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– Artem Mavrin
Dec 27 '15 at 8:03