Truncated inverse Wishart distribution

Paul

unread,

May 20, 2012, 9:14:45 PM5/20/12

to

I would like to calculate the mean of a truncated inverse Wishart distribution? I.e., if B is an inverse Wishart variable, then I'd like to calculate E(B|B<W), where B<W means that W-B is positive definite.

I have the answer for the degenerate case where B and W are scalars -- that is, where B is an inverse chi-square variable and W is a scalar constant. The answer then is a ratio of gamma functions. It seems to me I should be able to work up from the scalar answer to the matrix answer, and maybe I can even fill in the matrix diagonal with the scalar answer. But I'm not sure. Any hints most appreciated.

Ray Koopman

unread,

May 21, 2012, 3:58:26 AM5/21/12

to

On May 20, 6:14 pm, Paul <paulvonhip...@yahoo.com> wrote:
> I would like to calculate the mean of a truncated inverse Wishart distribution? I.e., if B is an inverse Wishart variable, then I'd like to calculate E(B|B<W), where B<W means that W-B is positive definite.
>
> I have the answer for the degenerate case where B and W are scalars -- that is, where B is an inverse chi-square variable and W is a scalar constant. The answer then is a ratio of gamma functions. It seems to me I should be able to work up from the scalar answer to the matrix answer, and maybe I can even fill in the matrix diagonal with the scalar answer. But I'm not sure. Any hints most appreciated.

If W is positive definite then write W = VV' and get E(C|C<I),
where C = V^-1 B V'^-1.

Paul

unread,

May 21, 2012, 3:39:46 PM5/21/12

to

On Monday, May 21, 2012 2:58:26 AM UTC-5, Ray Koopman wrote:
> On May 20, 6:14 pm, Paul

Thanks for the hint! Two follow-up questions:
1. What is the distribution of C?
2. If I get E(C|C<I), how to I translate the solution into E(B|B<W)?

Ray Koopman

unread,

May 21, 2012, 4:07:55 PM5/21/12

to

If W ~ IW(S,n) then C ~ IW(V^-1 S V'^-1, n).

> 2. If I get E(C|C<I), how to I translate the solution into E(B|B<W)?

E(B|B<W) = V E(C|C<I) V'

Here is a further simplification.
Write the eigendecomposition V^-1 S V'^-1 = U T U',
where U is square orthonormal and T is diagonal,
and let D = U'CU. Then D ~ IW(T,n), and
E(B|B<W) = UV E(D|D<I) V'U'.

Paul

unread,

May 21, 2012, 4:17:04 PM5/21/12

to

On Monday, May 21, 2012 3:07:55 PM UTC-5, Ray Koopman wrote:
> On May 21, 12:39 pm, Paul

Thanks so much. So instead of E(B|B<W) we must now calculate E(C|C<I) or E(D|D<I). Are these expectations easier to calculate? Why?

Ray Koopman

unread,

May 21, 2012, 7:28:55 PM5/21/12

to

I don't *know* that they're easier. It's just a hunch,
based on the fact that in both cases I is simpler than W,
and in the second case T is diagonal.

Paul

unread,

May 22, 2012, 11:29:21 AM5/22/12

to

On Monday, May 21, 2012 6:28:55 PM UTC-5, Ray Koopman wrote:
> On May 21, 1:17 pm, Paul

Thanks -- I may understand now. Since T is diagonal, D represents the joint distribution of several uncorrelated (and let's assume independent) inverse chi-square variables. Then E(D|D<I) is a diagonal matrix whose diagonal entries can be calculated, element by element, by using the result that I already have for the scalar case of a truncated inverse chi-square variables.

Is that right?

Ray Koopman

unread,

May 24, 2012, 4:08:21 AM5/24/12

to

E(D) is diagonal, but D itself is not.
I don't know if E(D|D<I) is diagonal.
FWIW, the condition D<I is equivalent to
all the eigenvalues of D being < 1.
Let r_k be the k'th eigenvalue of D, and let
q_ik be the i'th element of the corresponding eigenvector.
If E(q_ik * q_jk | r_k < 1) = 0 for all i,j,k, i <> j,
then you have what you need.