Download Solutions to Homework 2 for Statistical Signal Processing | ECE 567 and more Assignments Electrical and Electronics Engineering in PDF only on Docsity!
ECE 567 STATISTICAL SIGNAL PROCESSING SPRING 2008
Homework Assignment # Solutions
- We choose H 1 if p(x|H 1 )p(H 1 ) > p(x|H 0 )p(H 0 ). If p(H 0 ) = 1/2 = p(H 1 ), we choose H 1 if
1 √ 4 π
exp
x^2
2 π
exp
x^2
exp
x^2
2 exp
x^2
x^2 > ln(
x^2 1 4 x^2 > ln(
x^2 > 4 ln(
(2)) = ln(4) |x| >
ln(4).
Therefore we use
|x|
d > 1 d< 0
If p(H 1 ) = 3/4 (and hence p(H 0 ) = 1/4), then we choose H 1 if
3 / 4 √ 4 π
exp
x^2
2 π
exp
x^2
and the same type of analysis results in
exp
x^2
Since the right hand side is less than one, this expression holds for all x, and
we always choose H 1.
Because p(x|H 1 ) > p(x|H 0 ) only for larger x, we have the decision rule in the first case. However, in the second case, H 1 is three times more likely to occur than H 0 , resulting in 3p(x|H 1 ) > p(x|H 0 ) holding for all x, including small x. This results in the decision rule for the second case.
- We solve this using Bayes’ Risk, with assigned risks
C 00 = 0 C 01 = 0 C 10 = $10M C 11 = $10M − $110M = −$10M. Note that C 0 i = 0 for both i = 0 and i = 1 because we incur no cost and make no profit if we don’t excavate. For C 1 i, we always incur the excavation cost (i = 0 and i = 1) but reap profits only if mineral deposits are present (i = 1). Since C 10 > C 00 and C 01 > C 11 , we have p(x|H 1 ) p(x|H 0 )
d 1
< d 0
(C 10 − C 00 )p(H 0 ) (C 01 − C 11 )p(H 1 ) √^1 4 π exp^
−^14 (x − 10)^2
√^1 2 π exp^
−^12 (x − 8)^2
d 1
< d 0
exp
x^2 − 3 x + 7
) (^) d 1
< d 0
Noting that 1 4 x^2 − 3 x + 7 =
(x − 6)^2 − 2 , we have that 1 4 (x − 6)^2 − 2
d 1
< d 0
ln
|x − 6 |
d 1
< d 0
8 + 4 ln(0. 4
yielding
|x − 6 |
d > 1 d< 0
If the mineral deposit value is $20M, then C 11 = −10M, and the decision rule becomes
√^1 2
exp
x^2 − 3 x + 7
) (^) d 1
< d 0
or |x − 6 |
d 1
< d 0
8 + 4 ln(
yielding
