
Change detection with polarimetric SAR imagery 399
and under the alternative hypothesis H
1
, x
1
6= x
2
, the likelihood for x
1
and
x
2
is
L(x
1
, x
2
) = p(g
1
| x
1
)p(g
2
| x
2
)
=
1
(x
1
/m)
m
(x
2
/m)
m
Γ(m)
2
g
m−1
1
g
m−1
2
e
−(g
1
m/x
1
+g
2
m/x
2
)
.
(9.34)
By taking derivatives of the log-likelihoods, it is easy to show (Exercise 9)
that the likelihood L(x) is maximized by
ˆx =
g
1
+ g
2
2
and that L(x
1
, x
2
) is maximized by
ˆx
1
= g
1
, ˆx
2
= g
2
.
Then, ac cording to Equation (2.73), the likelihood ratio test has the critical
region
Q =
L(ˆx)
L(ˆx
1
, ˆx
2
)
=
(g
1
/m)
m
(g
2
/m)
n
g
1
+g
2
2m
2m
= 2
2m
g
m
1
g
m
2
(g
1
+ g
2
)
2m
≤ k.
Equivalently,
g
1
g
2
(g
1
+ g
2
)
2
m
≤
k
2
2m
or
g
1
g
2
(g
1
+ g
2
)
2
≤
˜
k,
where
˜
k depends on k. Inverting both sides of the ...