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.
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.
On Monday, May 21, 2012 2:58:26 AM UTC-5, Ray Koopman wrote:
> On May 20, 6:14 pm, Paul > 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.
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)?
> On Monday, May 21, 2012 2:58:26 AM UTC-5, Ray Koopman wrote:
>> On May 20, 6:14 pm, Paul 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.
> Thanks for the hint! Two follow-up questions:
> 1. What is the distribution of C?
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'.
On Monday, May 21, 2012 3:07:55 PM UTC-5, Ray Koopman wrote:
> On May 21, 12:39 pm, Paul > wrote:
> > On Monday, May 21, 2012 2:58:26 AM UTC-5, Ray Koopman wrote:
> >> On May 20, 6:14 pm, Paul 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.
> > Thanks for the hint! Two follow-up questions:
> > 1. What is the distribution of C?
> 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'.
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?
> On Monday, May 21, 2012 3:07:55 PM UTC-5, Ray Koopman wrote:
>> On May 21, 12:39 pm, Paul wrote:
>>> On Monday, May 21, 2012 2:58:26 AM UTC-5, Ray Koopman wrote:
>>>> On May 20, 6:14 pm, Paul 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.
>>> Thanks for the hint! Two follow-up questions:
>>> 1. What is the distribution of C?
>> 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'.
> 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?
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.
On Monday, May 21, 2012 6:28:55 PM UTC-5, Ray Koopman wrote:
> On May 21, 1:17 pm, Paul > wrote:
> > On Monday, May 21, 2012 3:07:55 PM UTC-5, Ray Koopman wrote:
> >> On May 21, 12:39 pm, Paul wrote:
> >>> On Monday, May 21, 2012 2:58:26 AM UTC-5, Ray Koopman wrote:
> >>>> On May 20, 6:14 pm, Paul 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.
> >>> Thanks for the hint! Two follow-up questions:
> >>> 1. What is the distribution of C?
> >> 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'.
> > 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?
> 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.
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.
> On Monday, May 21, 2012 6:28:55 PM UTC-5, Ray Koopman wrote:
>> On May 21, 1:17 pm, Paul
>> wrote:
>>> On Monday, May 21, 2012 3:07:55 PM UTC-5, Ray Koopman wrote:
>>>> On May 21, 12:39 pm, Paul wrote:
>>>>> On Monday, May 21, 2012 2:58:26 AM UTC-5, Ray Koopman wrote:
>>>>>> On May 20, 6:14 pm, Paul 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.
>>>>> Thanks for the hint! Two follow-up questions:
>>>>> 1. What is the distribution of C?
>>>> 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'.
>>> 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?
>> 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.
> 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?
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.
