LKJ Bijection

[Michael Betancourt]

The manual claims that the LKJ transformation is a bijection
from R^{choose(n, 2)} to the space of correlation matrices
(the original LKJ paper does make as explicit of a claim)
but there is a geometric argument that this can’t be an
exact bijection. This tension would be resolved if there is
some measure-zero set of correlation matrices that cannot
be modeled in this way (obstructing a bijection) but I cannot
figure out any potential singularities in the mapping. Thoughts?

[Ben Goodrich]

The singular correlation matrices are a bit dicey. If your correlation matrix has nullity one, then the last canonical partial correlation can be plus or minus 1. The bijection with positive definiteness correlations is presented more explicitly in K and Cooke

https://scholar.google.com/scholar?cluster=8075834827972186247&hl=en&as_sdt=1,33

Ben

[Michael Betancourt]

I’m thinking purely about positive-definite correlation matrices

so I’m happy to ignore singular edge cases — it still leaves

the tension between the parameterization and the geometric

intuition.  I’ll take a look through the paper — thanks!

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[Ben Goodrich]

On Wednesday, July 1, 2015 at 11:04:36 AM UTC-4, Michael Betancourt wrote:

I’m thinking purely about positive-definite correlation matrices

so I’m happy to ignore singular edge cases — it still leaves

the tension between the parameterization and the geometric

intuition.  I’ll take a look through the paper — thanks!

I'm not sure whether we have the same geometric intuition. To me, the easiest way to understand it algebraicly is to look at the relationship between canonical partial correlations (the omegas) and Pearson correlations (the sigmas)

This relationship can be inverted recursively to solve for $\Omega_{ij}$ iff all the omegas are on (-1,1). If any omega^2 = 1, then the correlation matrix is singular.

Ben

[Michael Betancourt]

I agree that the algebra implies a bijection.  But, geometrically

the space of correlation matrices is a cone (think in terms of

the eigenvalues — they’re constrained to lie on a circle for any

given value of the determinant, and as the the value of the

determinant shrinks that circle shrinks until it collapses to a point

when at least one of the eigenvalue vanishes.  

Now if we ignore singularities then we have a cone minus that

endpoint, which is a cylinder.  So the space of non-singular

correlation matrices should be a cylinder, which is not equivalent

to a patch of R^{choose(n , 2)} for the same reason the circle

isn’t (the identification of the endpoints can’t be matched in

a Euclidean space).  I’ve attached a figure demonstrating how

we need at least two parameterizations (charts) to cover the

cylinder.

So geometrically there should be at least one line of points

in correlation matrix space that can’t be identified using the

single algebraic transformation.  Hence the tension.

[Ben Goodrich]

On Wednesday, July 1, 2015 at 2:17:49 PM UTC-4, Michael Betancourt wrote:

I agree that the algebra implies a bijection.  But, geometrically

the space of correlation matrices is a cone (think in terms of

the eigenvalues — they’re constrained to lie on a circle for any

given value of the determinant, and as the the value of the

determinant shrinks that circle shrinks until it collapses to a point

when at least one of the eigenvalue vanishes.  

Right. There is the cone of positive semi-definite matrices and we are ignoring the edges where it the matrix is singular and specializing the diagonal elements to 1 to reduce the dimensionality.
 

Now if we ignore singularities then we have a cone minus that

endpoint, which is a cylinder.  So the space of non-singular

correlation matrices should be a cylinder, which is not equivalent

I think here is where our intuition goes in different directions. In the 3x3 case, the space of correlation matrix (generated from canonical partial correlations) looks like

the outskirts are the singular correlation matrices.

 

So geometrically there should be at least one line of points

in correlation matrix space that can’t be identified using the

single algebraic transformation.  Hence the tension.

I have a hard time seeing what line through that space would be problematic. If I had to guess I would say some line through the origin, but if the 3 correlations are zero, that is just the identity correlation matrix and the canonical partial correlations are also all zero.

Ben

[Michael Betancourt]

What does the color represent here?

Are you saying that each “corner” of that triangular cone is singular?

[Ben Goodrich]

On Thursday, July 2, 2015 at 4:11:42 AM UTC-4, Michael Betancourt wrote:

What does the color represent here?

The red just indicates that the point is in the front and the black indicates the point is in the back.
 

Are you saying that each “corner” of that triangular cone is singular?

All along the edges the correlation matrix is singular but at the corner points where the edges intersect, the rank is 1. Namely,

     [,1] [,2] [,3]
[1,]    1    1    1
[2,]    1    1    1
[3,]    1    1    1

     [,1] [,2] [,3]
[1,]    1    1   -1
[2,]    1    1   -1
[3,]   -1   -1    1

and

     [,1] [,2] [,3]
[1,]    1   -1   -1
[2,]   -1    1    1
[3,]   -1    1    1

For the edges that are not corners, the rank is 2.

Ben
 

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