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Material Type: Notes; Professor: Jayaweera; Class: Detection and Estimation Theory; Subject: Electrical & Computer Engineer; University: University of New Mexico; Term: Fall 2007;
Typology: Study notes
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ECE642: Detection and Estimation Theory
Dr. Sudharman K. Jayaweera
Assistant Professor
Department of Electrical and Computer Engineering
University of New Mexico
Lecture 07 - September
th
, Thursday
Fall 2007
ECE642: Detection and Estimation Theory
Minimax Hypothesis Testing Example 1:
Location Testing with Gaussian Error (Uniform Costs)
Recall the location testing problem defined in (3.6) and (3.7):
(^) σ
2 )
(^) σ
2 )
Recall, also from (3.9), that the minimum Bayes risk
π 0 )
for
uniform costs is,
π 0 ) = π 0 P 0 ( Γ 1
(^) π
0 ) P 1 ( Γ 0 )
= π 0 P 0 ( Γ 1
(^) π
0 )(
1 (^) −
1 ( Γ
1 ))
π 0 ) = ( 1
(^) π
0 ) +
(^) π
0 P 0 ( Γ 1 )
(^) π
0 ) P 1 ( Γ 1 )
ECE642: Detection and Estimation Theory
Location Testing with Gaussian Error: Conditional Risks
Since (4) is a differentiable function of
π 0 , randomization is not
necessary (since
(^) ′ ( π 0 )
exists everywhere on
Thus, we can find
π L
by setting
R 0 ( δ π L
= R 1 ( δ π L
or by setting
(^) ′ ( π L ) =
From (3.19) and definition of conditional risks (for uniform costs)
R 0 ( δ π 0
00
P 0 ( Γ
0 ) +
10
0 ( Γ 1 ) =
0 ( Γ 1 ) =
τ ′ −
(^) μ
0
σ
R 1 ( δ π 0
01
1 ( Γ 0 ) +
11
P 1 ( Γ 1 )
τ ′ −
(^) μ
1
σ
since
1 ( Γ 0 ) =
1 ( Γ
1 ))
ECE642: Detection and Estimation Theory
Location Testing with Gaussian Error: Minimax Condition
Thus, setting
R 0 ( δ π L
R 1 ( δ π L
τ ′ −
(^) μ
0
σ
τ ′ −
(^) μ
1
σ
Note that from (5), the priors enter into (8) only through
τ ′ .
So we need to solve (8) for
τ ′ =
τ L ′ .
ECE642: Detection and Estimation Theory
Example 1: Minimax Rule for Location Testing with Gaussian Error
with Uniform Costs
Hence, the minimax rule for the Gaussian location testing problem is:
δ π L (^) ( y )
if
y ≥ μ 0 + μ 1
2
if
y < μ 0 + μ 1
2
ECE642: Detection and Estimation Theory
Location Testing with Gaussian Error with Uniform Costs: Least
Favorable Prior
We can find the least favorable prior
π L
by substituting (9) into (5):
σ 2
μ 1 −
(^) μ
0
log
π L
(^) π
L )
i.e.,
π L
(^) π
L
π L
Thus, the
least favorable prior
corresponds to the
equal prior
case.
Hence from (3.25) we have the minimum risk,
V ( π L ) = V (
μ^ 1 (^) −
(^) μ
0
σ
(Note: the minimum risk of the minimax rule is independent of
π
0 )
ECE642: Detection and Estimation Theory
Minimax Example 2: The Function
for Binary Channel
From lecture #3, the Bayes risk for
uniform costs
is,
r ( δ B ) = π 0 P 0 ( Γ 1
(^) π
0 ) P 1 ( Γ 0 )
The
minimum Bayes risk function
for the binary channel can
V shown to be, (^) ( π 0 ) =
min
(^) π
0 ) λ 1 , (^) π
0 ( 1 (^) −
(^) λ
0 ) } (^) +
(^) min
(^) π
0 )(
1 (^) −
(^) λ
1 ) , (^) π
0 λ 0 }
(13)
10
ECE642: Detection and Estimation Theory
Function
π 0 )
for Binary Channel with Uniform Costs (ctd...)
We can rewrite (13) as (see Appendix C):
π 0 )
π 0
if 0
π 0
≤
π
π (^) +
π 0 (^) −
(^) π
)
if
π < π 0 < π
(^) π
0
if
π ≤ π 0 ≤ 1
where
π
min
λ 1
(^) λ
0 (^) +
(^) λ
1 ,
(^) λ
1
(^) λ
1 (^) +
(^) λ
0 }
π
max
λ 1
(^) λ
0
(^) λ
1 ,
(^) λ
1
(^) λ
1
(^) λ
0 }
and
is a constant independent of
π 0 ,
π¯ (^) −
(^) π
π ¯
(^) −
(^) π
ECE642: Detection and Estimation Theory
Minimax Example 2: Shape of
π 0 )
Note that
π 0 )
is a piecewise linear function of
π 0 , with changes in
slope occurring at
π 0
=
π
and
π 0
=
π¯
. Clearly,
(i.) if
Then
π 0 )
is maximum at
π L
=
π
and
π L )
π ) =
π
(from (14))
(ii.) if
Then
π 0 )
is maximum at
π L
=
π¯
and
π L )
π¯ ) =
π¯
(from (14))
(iii.) if
Then
π
L
can be taken as any prior in
π , ¯π ]
ECE642: Detection and Estimation Theory
Binary Channel with Uniform Costs: Minimax Risk
Thus, from (17) and (19), we have that the
minimax risk in general
is,
π L )
max
π , (^1) (^) −
π¯ }
forIn order to determine the minimax test we need to find a Bayes test
π 0
=
π L
prior.
Since
π 0 )
is not differentiable at either
π L
=
π
or
π L
=
¯π , we need
to consider a
randomized test
ECE642: Detection and Estimation Theory
Substituting for
from (17):
q
π¯ (^) −
(^) π
¯π (^) −
(^) π
(^) + (
(^) ¯π (^) −
(^) π
)
¯π (^) −
(^) π
π
ECE642: Detection and Estimation Theory
Binary Channel: Bayes Rule from Right-hand Side
if
π 0
π¯ :
Due to uniform costs
τ
=
π 0
(^) π
0 .
Since,
y )
λ 1
1 − λ 0
if
y
=
1 − λ 1
λ 0
if
y
=
and
δ B ( y ) =
if
L ( y ) ≥ τ
otherwise
ECE642: Detection and Estimation Theory
Binary Channel: Bayes Rule from Right-hand Side (ctd...)
But since,
¯π
max
λ 1
(^) λ
0
(^) λ
1 ,
(^) λ
1
(^) λ
1
(^) λ
0 }
and
π 0
π¯ ,
π 0
λ 1
(^) λ
0 (^) +
(^) λ
1
and
π 0
(^) λ
1
(^) λ
1
(^) λ
0
Hence, ( 1 (^) −
(^) λ
0 ) π 0 > λ 1 −
(^) λ
1 π
0
and
π 0 λ 0 > ( 1
(^) λ
1 ) (^) −
(^) π
0 ( 1 (^) −
(^) λ
1 )
ECE642: Detection and Estimation Theory
Binary Channel: Bayes Rule from Right-hand Side (ctd...)
(^) π
0 ) λ 1 < π 0 ( 1
(^) λ
0 )
and
(^) π
0 )(
1 (^) −
(^) λ
1 ) < π 0 λ 0
if Hence from (27) and (30):
y = 0 δ B ( y
δ B ( 0 ) =
if and from (31) and (34)
y = 1 δ B ( y
δ
B
( 1 ) =
