Statistics of Medical Imaging

Performance Evaluation of Image Analysis Methods 403

If we use ln w ≃ α

1

w − α

2

and e

−

1

2

w

≃ β

1

− β

2

w, then we can show that

Eqs. (12.12) and (12.13) will be

ln L(

ˆ

r) ≃

α

1

√

2πK ˆσ

[β

1

J −

β

2

2

Y

K

] − α

2

J

, (12.73)

and

P

e

= P (

1

K

1

ˆσ

1

Y

K

1

−

1

K

0

ˆσ

0

Y

K

0

)

<

1

α

1

β

2

(

3

√

2π ln J)(K

0

− K

1

) + 2J

β

1

β

2

(

1

K

1

ˆσ

1

−

1

K

0

ˆσ

0

)

), (12.74)

respectively. Using Eq. (12.14), Eq. (12.74) becomes

P

e

= P (Z <

1

α

1

β

2

∆

1

+

β

1

β

2

∆

2

)

∆

= P (Z

< ∆

′

1

+ ∆

′

2

), (12.75)

where ∆

′

1

=

1

α

1

β

2

∆

1

a

nd ∆

′

2

=

β

1

β

2

∆

2

.

G

raphically, it is easy to verify that α

1

, α

2

> 1 and 0 < β

1

, β

2

< 1.

More precisely, these coefficients can be determined by minimizing the erro rs

|(α

1

w−α

2

)−ln w| and |(β

1

−β

2

w)−e

−

1

2

w

|.

This minimiza tion can be numeri-

cally implemented by a least s quares error fitting approa ch: min

α

1

,α

2

P

i

(α

1

w

i

−α

2

− ln w

i

)

2

and min

β

1

,β

2

P

i

(β

1

− β

2

w

i

− e

−

1

2

w

i

)

2

,

over a set of w

i

.

For example, by uniformly ta king ten data points (sa mples) w

i

over (0, 1)

and using the above least squares error appro ach, the numerical study shows

that α

1

β

2

= 0.9814 and

β

1

β

2

=

2.5068. α

1

β

2

≃ 1 could be the res ult of the

fact that the opposite effects caused by ln w and e

−

1

2

w

c

ancel each other. This

result leads to ∆

′

1

≃ ∆

1

and ∆

′

2

< ∆

2

(because ∆

2

< 0), that is, probability

P

′

e

= P (Z < ∆

′

1

+ ∆

′

2

) is less than P

e

= P (Z < ∆

1

+ ∆

2

). This observation

has also been verified by other data samples w

i

. Therefore, the probability P

e

of Eq. (12.13) is consistent with P

′

e

(which are obtained using more accurate

linear approximations of ln w and e

−

1

2

w

)

and also can be used as an upper

limit of P

′

e

.

12.4.2 Appendix 12B

This appendix derives Eq. (12.2 2).

Proof.

From [46], we have

K

X

k=1

J

k

X

j=1

(x

j

− ˆµ)

2

=

K

X

k=1

J

k

(ˆµ

k

− ˆµ)

2

+

K

X

k=1

J

k

X

j=1

(x

j

− ˆµ

k

)

2

, (12.76)