# Joint Gaussian With Tensors

1. Aug 28, 2009

### John Creighto

For vectors we can define the Joint Guasian as follows:

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

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

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

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

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:

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

$$( A - \bar A)^T : Q^{-1} : (A - \bar A)$$

Or maybe just get rid of the transpose operator?

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

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

Other notation issues:

is

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

equivalent to:

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

Seems like it should be for the case that $$Q$$ is symmetric but not in general.

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

so $$[Q^{-1}]^{(i_2,j_1,m,n)}$$ would be $$[Q^{-1}]^{(i,j,m,n)}$$ 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.

Last edited: Aug 28, 2009