Can We Extend Joint Gaussian Distributions to Higher Dimensions Using Tensors?

[John Creighto]

For vectors we can define the Joint Guasian as follows:

[tex]f_X(x_1, \dots, x_N) = \frac {1} {(2\pi)^{N/2}|\Sigma|^{1/2}} \exp \left( -\frac{1}{2} ( x - \mu)^\top \Sigma^{-1} (x - \mu) \right)[/tex]

Now what if [tex](x - \mu)[/tex] is a matrix [tex]A[/tex] and [tex]\Sigma[/tex] is an order four covariance matrix [tex]Q[/tex] between ellements of [tex]A[/tex]. Can we define a higher dimensional version of the joint gausian in terms of the double dot product as follows:

[tex]f_X(x_1, \dots, x_N) = \frac {1} {(2\pi)^{N/2}|Q|^{1/2}} \exp \left( -\frac{1}{2} ( A - \bar A)^T : Q^{-1} : (A - \bar A) \right)[/tex]

What I see as possible problems are perhaps [tex](2\pi)^{N/2}[/tex] should be [tex](2\pi)^{N^2/2}[/tex]

The transpose operator is ambiguous so maybe index notation is necessary, although the double dot notation seems much neater.

I understand in index notation repeated indices are summed so should I write:

[tex][ A - \bar A]^{(i,j)} [Q^{-1}]^{(i,j,m,n)}[A - \bar A]^{m,n}[/tex]

instead of:

[tex]( A - \bar A)^T : Q^{-1} : (A - \bar A) [/tex]

Or maybe just get rid of the transpose operator?

Finally how well is the inverse and determinant of Q defined?

Is [tex]Q^{-1}[/tex] defined so that [tex]Q:Q=I[/tex] where [tex]I[/tex] is rank four and is [tex]1[/tex] on the diagonal and [tex]0[/tex] is every where else?

Other notation issues:

is

[tex][ A - \bar A]^{(i,j)} [Q^{-1}]^{(i,j,m,n)}[A - \bar A]^{(m,n)}[/tex]

equivalent to:

[tex] [Q^{-1}]^{(i,j,m,n)}[A - \bar A]^{(m,n)}[ A - \bar A]^{(i,j)}[/tex]

Seems like it should be for the case that [tex]Q[/tex] is symmetric but not in general.

Maybe subscrips on indicies would be a good way to define transposes:

so [tex][Q^{-1}]^{(i_2,j_1,m,n)}[/tex] would be [tex][Q^{-1}]^{(i,j,m,n)}[/tex] with the first two indicies permuted (I'm sure this isn't the standard convention. Also note I haven't taken any courses that cover tensors so my knowledge is quite limited.

 

[pat_connell]

Lastly, can the determinant of Q be written as a product of its eigenvalues?Yes, the determinant of Q can be written as a product of its eigenvalues.