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Why is this not a valid Variance Covariance matrix, and inherently not positive semi-definite?

Find the eigenvalues and their multiplicities of a special matrixUnder what circumstance will a covariance matrix be positive semi-definite rather than positive definite?“Fully correlated” definitioncell-by-cell constraints within a positive-definite matrix?Covariance matrix always positive semidefinite?How do I show the covariance matrix of a multivariate normal random vector is positive definite?Matrix transformation conserving the “positive semi-definite” aspectIf we have a full rank positive semi-definite symmetric matrix, will any square partition be full rank as well?How to create a positive definite covariance matrix from an adjacency matrix?Show a matrix is positive definiteOptimize $M$ such that $MM^T$ is the “smallest” (in a positive semi-definite sense)

2

2

$begingroup$

 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 1.0 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5
[2,] -0.5 1.0 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5
[3,] -0.5 -0.5 1.0 -0.5 -0.5 -0.5 -0.5 -0.5
[4,] -0.5 -0.5 -0.5 1.0 -0.5 -0.5 -0.5 -0.5
[5,] -0.5 -0.5 -0.5 -0.5 1.0 -0.5 -0.5 -0.5
[6,] -0.5 -0.5 -0.5 -0.5 -0.5 1.0 -0.5 -0.5
[7,] -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 1.0 -0.5
[8,] -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 1.0

The reason I know is because this results in a negative eigenvalue, and variance-covariance matrices are positive semi-definite.

My thinking here was to have the variances be one, so that the correlations were the covariances, and thus equal -0.5.

Is there something with the theory I am missing? I understand that it is not positive semi-definite and how to show as such, but I am more curious what assumptions this is violating in terms of probability/statistics.

I went to generate MVN data with this variance-covariance structure and realized this wasn't positive semi-definite, and then became curious what was inherently wrong with this matrix.

linear-algebra probability statistics normal-distribution

share|cite|improve this question

edited Mar 20 at 20:40

OGV

asked Mar 20 at 19:54

OGVOGV

1037

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I find it a very fascinating question. If you don't get more responses here you might try at stats.stackexchange.com
  $endgroup$
  – Vincent
  Mar 21 at 14:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have a bit of a resolution to this, I'll try to post it (and possibly an addendum to this question elsewhere like at stats) and be sure to leave a link if so!
  $endgroup$
  – OGV
  Mar 23 at 4:25
  

  
  
  
  

add a comment | 

2

2

$begingroup$

 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 1.0 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5
[2,] -0.5 1.0 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5
[3,] -0.5 -0.5 1.0 -0.5 -0.5 -0.5 -0.5 -0.5
[4,] -0.5 -0.5 -0.5 1.0 -0.5 -0.5 -0.5 -0.5
[5,] -0.5 -0.5 -0.5 -0.5 1.0 -0.5 -0.5 -0.5
[6,] -0.5 -0.5 -0.5 -0.5 -0.5 1.0 -0.5 -0.5
[7,] -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 1.0 -0.5
[8,] -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 1.0

The reason I know is because this results in a negative eigenvalue, and variance-covariance matrices are positive semi-definite.

My thinking here was to have the variances be one, so that the correlations were the covariances, and thus equal -0.5.

Is there something with the theory I am missing? I understand that it is not positive semi-definite and how to show as such, but I am more curious what assumptions this is violating in terms of probability/statistics.

I went to generate MVN data with this variance-covariance structure and realized this wasn't positive semi-definite, and then became curious what was inherently wrong with this matrix.

linear-algebra probability statistics normal-distribution

share|cite|improve this question

edited Mar 20 at 20:40

OGV

asked Mar 20 at 19:54

OGVOGV

1037

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I find it a very fascinating question. If you don't get more responses here you might try at stats.stackexchange.com
  $endgroup$
  – Vincent
  Mar 21 at 14:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have a bit of a resolution to this, I'll try to post it (and possibly an addendum to this question elsewhere like at stats) and be sure to leave a link if so!
  $endgroup$
  – OGV
  Mar 23 at 4:25
  

  
  
  
  

add a comment | 

$begingroup$

 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 1.0 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5
[2,] -0.5 1.0 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5
[3,] -0.5 -0.5 1.0 -0.5 -0.5 -0.5 -0.5 -0.5
[4,] -0.5 -0.5 -0.5 1.0 -0.5 -0.5 -0.5 -0.5
[5,] -0.5 -0.5 -0.5 -0.5 1.0 -0.5 -0.5 -0.5
[6,] -0.5 -0.5 -0.5 -0.5 -0.5 1.0 -0.5 -0.5
[7,] -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 1.0 -0.5
[8,] -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 1.0

The reason I know is because this results in a negative eigenvalue, and variance-covariance matrices are positive semi-definite.

My thinking here was to have the variances be one, so that the correlations were the covariances, and thus equal -0.5.

Is there something with the theory I am missing? I understand that it is not positive semi-definite and how to show as such, but I am more curious what assumptions this is violating in terms of probability/statistics.

I went to generate MVN data with this variance-covariance structure and realized this wasn't positive semi-definite, and then became curious what was inherently wrong with this matrix.

linear-algebra probability statistics normal-distribution

share|cite|improve this question

edited Mar 20 at 20:40

OGV

asked Mar 20 at 19:54

OGVOGV

1037

$endgroup$

 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 1.0 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5
[2,] -0.5 1.0 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5
[3,] -0.5 -0.5 1.0 -0.5 -0.5 -0.5 -0.5 -0.5
[4,] -0.5 -0.5 -0.5 1.0 -0.5 -0.5 -0.5 -0.5
[5,] -0.5 -0.5 -0.5 -0.5 1.0 -0.5 -0.5 -0.5
[6,] -0.5 -0.5 -0.5 -0.5 -0.5 1.0 -0.5 -0.5
[7,] -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 1.0 -0.5
[8,] -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 -0.5 1.0

share|cite|improve this question

edited Mar 20 at 20:40

OGV

asked Mar 20 at 19:54

OGVOGV

1037

share|cite|improve this question

edited Mar 20 at 20:40

OGV

asked Mar 20 at 19:54

OGVOGV

1037

asked Mar 20 at 19:54

OGVOGV

1037

asked Mar 20 at 19:54

OGVOGV

1037

1 Answer
1

active

oldest

votes

0

$begingroup$

As shown here, an $ntimes n$ matrix with $a$ on the diagonal and $b$ elsewhere, that is, the matrix
$$beginbmatrix a & b & ldots & b\ b & a & ldots & b\ vdots & vdots & ddots & vdots\ b & b & ldots & aendbmatrix$$
has $colorbluea +(n-1)b$ as one of its eigenvalues. For your particular matrix, we have $a = 1, b = -0.5, n =8$, so $$a + (n-1)b = 1 + 7times (-0.5) = -2.5< 0.$$ Hence your symmetric matrix has a negative eigenvalue ($-2.5$), so cannot be positive semi-definite. This implies that it is not a valid variance-covariance matrix.

EDIT: Just realised you said you already know this...

share|cite|improve this answer

edited Mar 20 at 20:36

answered Mar 20 at 20:31

Minus One-TwelfthMinus One-Twelfth

2,933413

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yup! Sorry, the title was misleading people I think. I already knew for a fact it wasn't positive semi-definite, but theoretically random variables should have this structure. In terms of probability something is specifically violated making it not positive semi-definite, and I was curious what it might be.
  $endgroup$
  – OGV
  Mar 20 at 20:43
  

  
  
  
  

add a comment | 

Your Answer

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0

$begingroup$

As shown here, an $ntimes n$ matrix with $a$ on the diagonal and $b$ elsewhere, that is, the matrix
$$beginbmatrix a & b & ldots & b\ b & a & ldots & b\ vdots & vdots & ddots & vdots\ b & b & ldots & aendbmatrix$$
has $colorbluea +(n-1)b$ as one of its eigenvalues. For your particular matrix, we have $a = 1, b = -0.5, n =8$, so $$a + (n-1)b = 1 + 7times (-0.5) = -2.5< 0.$$ Hence your symmetric matrix has a negative eigenvalue ($-2.5$), so cannot be positive semi-definite. This implies that it is not a valid variance-covariance matrix.

EDIT: Just realised you said you already know this...

share|cite|improve this answer

edited Mar 20 at 20:36

answered Mar 20 at 20:31

Minus One-TwelfthMinus One-Twelfth

2,933413

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yup! Sorry, the title was misleading people I think. I already knew for a fact it wasn't positive semi-definite, but theoretically random variables should have this structure. In terms of probability something is specifically violated making it not positive semi-definite, and I was curious what it might be.
  $endgroup$
  – OGV
  Mar 20 at 20:43
  

  
  
  
  

add a comment | 

0

$begingroup$

As shown here, an $ntimes n$ matrix with $a$ on the diagonal and $b$ elsewhere, that is, the matrix
$$beginbmatrix a & b & ldots & b\ b & a & ldots & b\ vdots & vdots & ddots & vdots\ b & b & ldots & aendbmatrix$$
has $colorbluea +(n-1)b$ as one of its eigenvalues. For your particular matrix, we have $a = 1, b = -0.5, n =8$, so $$a + (n-1)b = 1 + 7times (-0.5) = -2.5< 0.$$ Hence your symmetric matrix has a negative eigenvalue ($-2.5$), so cannot be positive semi-definite. This implies that it is not a valid variance-covariance matrix.

EDIT: Just realised you said you already know this...

share|cite|improve this answer

edited Mar 20 at 20:36

answered Mar 20 at 20:31

Minus One-TwelfthMinus One-Twelfth

2,933413

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yup! Sorry, the title was misleading people I think. I already knew for a fact it wasn't positive semi-definite, but theoretically random variables should have this structure. In terms of probability something is specifically violated making it not positive semi-definite, and I was curious what it might be.
  $endgroup$
  – OGV
  Mar 20 at 20:43
  

  
  
  
  

add a comment | 

$begingroup$

As shown here, an $ntimes n$ matrix with $a$ on the diagonal and $b$ elsewhere, that is, the matrix
$$beginbmatrix a & b & ldots & b\ b & a & ldots & b\ vdots & vdots & ddots & vdots\ b & b & ldots & aendbmatrix$$
has $colorbluea +(n-1)b$ as one of its eigenvalues. For your particular matrix, we have $a = 1, b = -0.5, n =8$, so $$a + (n-1)b = 1 + 7times (-0.5) = -2.5< 0.$$ Hence your symmetric matrix has a negative eigenvalue ($-2.5$), so cannot be positive semi-definite. This implies that it is not a valid variance-covariance matrix.

EDIT: Just realised you said you already know this...

share|cite|improve this answer

edited Mar 20 at 20:36

answered Mar 20 at 20:31

Minus One-TwelfthMinus One-Twelfth

2,933413

$endgroup$

share|cite|improve this answer

edited Mar 20 at 20:36

answered Mar 20 at 20:31

Minus One-TwelfthMinus One-Twelfth

2,933413

answered Mar 20 at 20:31

Minus One-TwelfthMinus One-Twelfth

2,933413

answered Mar 20 at 20:31

Minus One-TwelfthMinus One-Twelfth

2,933413