multivariate poisson hidden markov models for analysis of spatial ...

and y
3
as illustrated above. Again, to rewrite equation 5.1 in compact form, we can
define the following notations:
⎡L
⎤
1
(2) ⎢
L
⎥ (3)
L
2
, = [ L 4 ]
= ⎢ ⎥
⎢⎣L
⎥
3 ⎦
L and
R = ( R ∪R ∪ R ) where
(1) (2) (3)
2 2 2 2
(1)
R
2
= {12,13},
(2)
R
2
= {12,23},
(3)
R
2
= {13, 23}.
In fact, substituting the X ’s for the Y ’s in (5.1) result in:
L
(2) (3)
L
∑∑
PY [ = y, Y = y , Y = y ] = P[ X = x, X = x , X = x, X = x , X = x , X = x , X = x ]
1 1 2 2 3 3 1 1 2 2 3 3 12 12 13 13 23 23 123 123
(2) (3)
= 0 = 0
x
x
and since the X ’s are independent univariate Poisson variables, the joint distribution
reduces to the product of the univariate distributions:
1 1 2 2 3 3
L
(2) (3)
L
∑∑∏
∑ ∑
PY [ = y, Y = y , Y = y ] = P[ X = ( y − x − x)] P[ X = x]
.
∏
j j k l i i
(2) (3) ( j )
= 0x
= 0 j∈R1 k∈R
l∈R
2
3 i∈R2∪R3
x
Now, the following three conditions on X
1,
X
2,
and X
3
must be satisfied, since the
X ’s are univariate Poisson distributions, and since the Poisson distribution is only
defined for positive integer values and zero:
y −x −x −x
≥0
1 12 13 123
y −x −x −x
≥0
2 12 23 123
y −x −x −x
≥0.
3 13 23 123
(5.2)
These conditions imply that all x ’s cannot just be any integer value, but depend on the
values of y
1,
y
2
, and y
3
. Accordingly, the distribution for PY [
1
= y1, Y2 = y2, Y3 = y3]
by summing up all the x ’s is:
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