Re: Help me : Transformation of multivariate distribution

duvin103time wrote:
Thanks a lot to everyone of you!

David,
I followed your suggestion and got the following result.
F(q)=f(inv(P3)*q3)/|det(P3)|*delta(qn-Pn*inv(P3)*q3)
where
q3 is the first three elements of q
qn is the rest of q
P3 is the upper 3 by 3 part of P
Pn is the rest of P
Is that right?
In addtion, K have a spherical distribution not Multivariate Normal.
So I have to interest in the general case.

Looks OK, except:

(i) the usual notation has "f" for a density amd "F" for a cumulative
distribution function, while yours are both densities. But what
matters is that you know what you mean.
(ii) for a complete solution you strictly need to cover cases where
the upper 3 by 3 matrix of P is non-singular. Part of this may be
covered by re-arranging the elements. However, you would also need to
cover the cases where the column-rank of P is 0, 1 or 2 where there
would be no 3 by 3 submatrix which is invertible (but you could then
reduce the number of "special elements" from 3 to 2 or 1, and treat
the case where P=0 separately).

You may be able to carry over some of what would be most natural from
the Multivariate Normal to the spherical case. For example, for the
MVN you could work by describing the distribution of Q by its
covariance matrix (even though this covariance matrix is singular),
which is easily evaluated. In your case the matrix characterising the
distribution of Q is (P times transpose(P)). It may be that if you are
looking for properties other than the density that you don't
particularly need to worry about the fact that P isn't invertible.
Examples might be in finding the moments of Q. If you have not done so
already, you might want to look into deriving the (joint)
characteristic function of Q in terms of the (joint) characteristic
function of K.

David Jones



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