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ECE/CS 313: Probability with Engineering Applications Fall 2002
Problem Set 5 Solutions
- Sensor fusion (a) The likelihood matrix for observation XY given the hypothesis is the following.
XY → 00 01 02 10 11 12 20 21 22 H 0 0. 56 0. 16 0. 08 0. 07 0. 02 0. 01 0. 07 0. 02 0. 01 H 1 0. 01 0. 01 0. 08 0. 03 0. 03 0. 24 0. 06 0. 06 0. 48
The ML decisions are indicated by the underlined elements in the likelihood matrix. The larger number in each column is underlined, with the tie broken in case 02 broken in favor of H 1 , as specified in the problem statement. (b) For the ML rule, pf alse alarm is the sum of the entries in the row for H 0 in the likelihood matrix that are not underlined. So pf alse alarm = 0.08 + 0.02 + 0.01 + 0.02 + 0.01 = 0. 14. For the ML rule, pmiss is the sum of the entries in the row for H 1 in the likelihood matrix that are not underlined. So pmiss = 0.01 + 0.01 + 0.03 + 0.06 = 0. 11. (c) The joint probability matrix is given by
XY → 00 01 02 10 11 12 20 21 22 H 0 0. 448 0. 128 0. 064 0. 056 0. 016 0. 008 0. 056 0. 016 0. 008 H 1 0. 002 0. 002 0. 016 0. 006 0. 006 0. 048 0. 012 0. 012 0. 096
The MAP decisions are indicated by the underlined elements in the joint probability matrix. The larger number in each column is underlined. (d) For the MAP rule,
pf alse alarm = P [XY ∈ { 12 , 22 }|H 0 ] = 0.01 + 0.01 = 0. 02 ,
and pmiss = P [XY 6 ∈ { 12 , 22 }|H 1 ] = 1 − P [XY ∈ { 12 , 22 }|H 1 ] = 1 − 0. 24 − 0 .48 = 0. 28.
Thus, for the MAP rule, pe = (0.8)(0.02) + (0.2)(0.28) = 0.072. (Check that for the MAP rule, pe is the sum of the probabilities in the joint probability matrix that are not underlined.) (e) Using the conditional probabilities found in (a) and the given values of π 0 and π 1 yields that for the ML rule: pe = (0.8)(0.14) + (0.2)(0.11) = 0.134 (Note: While the MAP rule has a smaller probability of error than the ML rule, the pmiss for the MAP rule might be unacceptably high. For this problem, it may be appropriate to take the costs of wrong decisions into account. )
- Simple discrete random variables (a) j = 1 j = 2 j = 3 j = 4 j = 5 j = 6 i = 1 1 2 3 4 5 6 i = 2 2 2 3 4 5 6 i = 3 3 3 3 4 5 6 i = 4 4 4 4 4 5 6 i = 5 5 5 5 5 5 6 i = 6 6 6 6 6 6 6
(b) The range of Y is { 1 , 2 , 3 , 4 , 5 , 6 }. (c) By inspection of the table, we see that |Y = 1| = 1, |Y = 2| = 3, |Y = 3| = 5, |Y = 4| = 7, |Y = 5| = 9, and |Y = 6| = 11, which we can compactly write as |Y = i| = 2i − 1. Thus, the probability mass function of Y is given by p(i) = 2 i 36 −^1 for i ∈ { 1 , 2 , 3 , 4 , 5 , 6 }, and p(i) = 0 otherwise.
- Computing a simple probability mass function Define the events for i = 0 or i = 1:
Di = “Salesman sold deluxe model on ith^ appointment” Si = “Salesman sold standard model on ith^ appointment” Ni = “Salesman sold nothing on ith^ appointment”
Then
p(0) = P (N 1 N 2 ) = (1 − 0 .3)(1 − 0 .6) = 0. 28
p(500) = P ((S 1 N 2 ) ∪ (N 1 S 2 )) = (
p(1000) = P ((D 1 N 2 ) ∪ (S 1 S 2 ) ∪ (N 1 D 2 )) = (
p(1500) = P ((D 1 S 2 ) ∪ (S 1 D 2 )) = 2(
p(2000) = P (D 1 D 2 ) = (
- Maximum likelihood parameter estimation (a)
P ((2, 1 , 0 , 2 , 0 , 1 , 2 , 2 , 0 , 2)) = (
1 + a 3
1 − a 3
1 + a 3
1 − a 3
1 + a 3
1 + a 3
1 − a 3
1 + a 3
1 − a 3
)^3 ∗ (
)^2 (
1 + a 3
)^5.
(b) Finding a to maximize ( 1 − 3 a)^3 ∗ ( 13 )^2 ( 1+ 3 a)^5 yields ˆaM LE = 0.2 (See part (c) for general case.) (c) Same answer as (b). The order of the output values does not effect the probability of the total output vector. (d) In general, the probability of a particular output sequence that has n 0 0’s, n 1 1’s, and n 2 2’s, satisfies
1 − a 3
)n^0 + (
)n^1 + (
1 + a 3
)n^2 =
(1 − a)n^1 (1 + a)n^3 3 n
. Maximizing this with respect to a over the interval [− 1 , 1] is equivalent to maximizing (1 − a)n^0 (1 + a)n^2. Or, taking the log, the problem is to find a to maximize n 0 log(1 − a) + n 2 log(1 − a). The derivative of the function is d(n 0 log(1 − a) + n 2 log(1 + a)) da
−n 0 1 − a
n 2 1 + a
In the trivial special case that n 0 = n 2 = 0, the probability of the observation is the same for all a, in which case ˆaM L can be chosen arbitrarily from the interval [− 1 , 1]. (We won’t take off credit if this fine point is missed, but congratulations if you spotted it!) Otherwise the derivative is strictly decreasing in a and is zero only at a = n n^22 −+nn^00. The maximum of the function thus occurs at that value, so ˆaM L = n n^22 −+nn^00.
- Bayes decision making for given cost (a)
Λ(n) =
25 if 1 ≤ n ≤ 300
- 25 if 301 ≤ n ≤ 400 0 if 401 ≤ n ≤ 780
(b) The ML rule is given by the LRT (see previous problem set) with the threshold one. So the LRT decides H 1 is true if 1 ≤ X ≤ 400 and decides H 0 is true otherwise. For this rule,
pf alse alarm = P [1 ≤ X ≤ 400 |H 0 ] = (300)(0.0001) + (100)(0.002) = 0. 23 pmiss = P [401 ≤ X ≤ 780 |H 1 ] = 0
(c) The MAP rule is given by the LRT (see previous problem set) with the threshold π π^01 = 9, 999. Since this always larger than the likelihood ratio, the MAP rule always decides that H 0 is true. For this rule, pf alse alarm = 0 and pmiss = 1. Therefore, pe = π 0 pf alse alarm + π 1 pmiss = 10−^4. Note: while the unconditional error probability is small, the sensor is useless! (d) As mentioned in the hint, the Bayes minimum cost rule is given by the LRT with threshold τ given by
τ =
π 0 (C 10 − C 00 ) π 1 (C 01 − C 11 )
(1 − 10)−^4 (1 − 0)
10 −^4 (1000 − 0)
