University of Illinois Fall 2009
ECE 313: Problem Set 7: Solutions
Decision-Making Under Uncertainty
1. [A warm-up exercise]
(a) The likelihood matrix Lis as shown below and the maximum-likelihood decision rule is indicated
shading.
Hypothesis X=0 X=1 X=2 X=3
H10.4 0.3 0.2 0.1
H00.1 0.2 0.3 0.4
It is easy to get PFA = sum of unshaded entries on H0row = 0.1+0.2=0.3
and PMD = sum of unshaded entries on H1row = 0.1+0.2=0.3 also.
(b) By the law of total probability, P(E)=π0PFA +π1PMD =0.7×0.3+0.3×0.3=0.3.
(c) The joint probability matrix J=LD where D= diag[π1,π0] is as shown below, and the Bayesian
decision rule is as indicated by the shading.
Hypothesis X=0 X=1 X=2 X=3
H10.12 0.09 0.06 0.03
H00.07 0.14 0.21 0.28
P(E) = sum of all the unshaded entries in the Jmatrix = 0.25 is smaller than the average error
probability for the ML rule, as it should be.
(d) In general, the Jmatrix can be written as
Hypothesis X=0 X=1 X=2 X=3
H10.4π10.3π10.2π10.1π1
H00.1π00.2π00.3π00.4π0
Now, if 0.1π0>0.4π1=0.4(1 −π0), that is, π0>0.8, then the Bayesian decision is in favor of H0
whenever X= 0. But, if π0>0.8, then the same decision (favoring H0) holds whenever Xequals
1, 2, or 3 also. (Work it out!) Hence, if π0>0.8, the Bayesian decision always is in favor of H0.
From the symmetry of the problem, it should be obvious that if π1>0.8, the Bayesian decision
always is in favor of H1. The incredulous are invited to work out the details for themselves.
Alternatively, the likelihood ratio takes on values, 4, 3
2,2
3and 1
4, and hence always exceeds
π0
π1=π0
1−π0if π0<0.2, that is, if π1>0.8, and can never exceed π0
π1=π0
1−π0if π0>0.8. The
result to be proved follows from this.
2. [Which route did she fly?]
(a) If H0is the true hypothesis, then Xtakes on values 0, 1, 2 with probabilities (1 −p)2,2p(1 −p),
and p2respectively. If H1is the true hypothesis, then Xtakes on values 0, 2, 4 with probabilities
(1 −q)2,2q(1 −q), and q2respectively. The likelihood matrix is thus
Hypothesis X=0 X=1 X=2 X=4
H1(1 −q)20 2q(1 −q)q2
H0(1 −p)22p(1 −p)p20
(b) When X= 0, the ML decision is in favor of H1or H0according as (1 −q)2≷(1 −p)2, that is,
according as 1 −q≷1−p, or p≷q.
When X= 1, the ML decision is always in favor of H0regardless of the values of pand q.
When X= 2, the ML decision is in favor of H1or H0according as 2q(1 −q)≷p2. Note that the
maximum value of 2q(1 −q) is 1
2(prove this!) and so if p> 1
√2≈0.707 . . ., then the ML decision
will always be in favor of H0regardless of the value of q.
When X= 4, the ML decision is always in favor of H1regardless of the values of pand q.
