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Subject:

dirichlet used as the data sampling distribution (summary)

From:

Claire Chabanet #3354 <[log in to unmask]>

Reply-To:

Claire Chabanet #3354 <[log in to unmask]>

Date:

Wed, 24 May 2000 13:35:33 +0200 (MET DST)

Content-Type:

text/plain

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Thank  you  very  much  to  Jim Garrett,  Patrick  Bélisle  and  David
Spiegelhalter for their answers. I'll resume their suggestions.

First, it is not possible to have uncertainty on the parameters of the
Dirichlet distribution and learn about them.  It's just not a facility
within BUGS at the moment and it is not a priority.

1/ first suggestion : 
The  Dirichlet  distribution   is  the  distribution  of  independent,
equally-scaled gamma variates divided  by their sum, perhaps you could
work with gamma variables instead.  For example, if

z_1 ~ gamma(alpha_1, 1),
z_2 ~ gamma(alpha_2, 1), and
z_3 ~ gamma(alpha_3, 1),

and they are independent, and

K = z_1 + z_2 + z_3,

then z_1/K, z_2/K, and z_3/K are jointly Dirichlet. 

2/ second suggestion : logistic-normal  models.

Having  said this,  I read  an  excellent book  on compositional  data
"Analysis of  Compositional Data" by Aitchison, from  Chapman and Hall
publishers.

The author  points out several  limitations to the  Dirichlet stemming
from  the independence  assumption, and  recommends an  alternative in
which P  variables that  sum to one  are transformed to  P-1 variables
that  are unconstrained, so  that the  (P-1)-vector may  reasonably be
modeled by a multivariate  normal.  This allows for correlations among
the  proportions  in  addition  to  the correlations  implied  by  the
unit-sum constraint.  If I recall  correctly, the author refers to the
generalized  logistic transformation,  which would  be  (following the
example above)

s_1 = log(z_1 / z_3)
s_2 = log(z_2 / z_3)
s_3 = 0 (i.e., = log(z_3 / z_3) )

The choice of which component will be in the denominator is completely
arbitrary.

This transformation has the inverse

z_1 = exp(s_1) / (exp(s_1) + exp(s_2) + exp(s_3) ),
z_2 = exp(s_2) / (exp(s_1) + exp(s_2) + exp(s_3) ), and
z_3 = exp(s_3) / (exp(s_1) + exp(s_2) + exp(s_3) ).

You could then suppose that (s_1, s_2) ~ bivariate normal and estimate
both  the mean  and variance  matrix.   It seems  to me  likely to  be
straightforward.  It is a  richer model  for the  data. (but  it cannot
handle zero compositions).

Claire Chabanet                                 
INRA - Laboratoire de Recherches sur les Aromes 
17 rue SULLY, BV1540                            
21034 DIJON cedex  FRANCE

e-mail : [log in to unmask]


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