Joint work with Prof. Matthias Poloczek, System Engineering Department.

1. Stopping Problems
Mohamed Seif, PhD student
ECE Department
October 2017The University of Arizona

2. Outline
October 2017The University of Arizona
• Motivation
• Preliminaries on Hypothesis Testing
• Basic Classic of Stopping Problems
• Sequential Probability Ratio Test
• Proposed Solution
•Application
2

3. Motivation
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• Stopping problems are a simple but important class of learning problems.
•In this problem class, information arrives over time, and we have to
choose whether to view the information or stop and make a decision.
3
Event Detector
t=0 t=1 t=2 t=3
. . .
t=10
Input Signal
Sudden Change in observation
t=0 t=1 t=2 t=3
. . .
t=10
Ideal case: Stop observing and detect the change accurately !
Output Signal

4. Preliminaries on Hypothesis Testing
October 2017The University of Arizona
• Consider the following example
depicted in the figure.
• We have two hypotheses:
Fig. Radar System
(
H0, if No air fighter
H1, if Air fighter exist
4

5. Preliminaries on Hypothesis Testing (Cont’d)
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• Two sources of error in the
process of detecting air fighter:
1. False alarm
2. Mis detection
Fig. Radar System
=) Prob{ ˆH1|H0}
=) Prob{ ˆH0|H1}
Pe = P(H0)P( ˆH1|H0) + P(H1)P( ˆH0|H1)
• Average probability of error:
5

6. Basic Classes of Stopping Problems
October 2017The University of Arizona
1. Sequential probability ratio test (Talk focus)
2. Secretary problem
6
We have N candidates, interview K out of N
candidates. Then, choose the one who has highest
score.
Decide as quickly as possible when something is
changing.

7. Wn
=
(
¯µ0
+ ✏n
, H0
¯µ1
+ ✏n
, H1
Sequential Probability Ratio Test
October 2017The University of Arizona 7
• For example, we may think we are observing data being generated
by the sequence
• The observed signal at time slot n
• Prior probabilities
Prob(H0) = ⇢0 = ⇢0
0
Prob(H1) = ⇢1 = ⇢0
1

8. Sequential Probability Ratio Test
October 2017The University of Arizona 8
• After observing W1, update the prior using Baye’s rule as follows
⇢1
0 = P(H0|W1
)
=
P(W1
|H0)P(H0)
P(W1)
=
P(W1
|H0)P(H0)
⇢0P(W1|H0) + ⇢0
1P(W1|H1)
Similarly, we can update ⇢1
1 ⇢1
1 = P(H1|W1
)

9. Example
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• Consider the observed signal is drawn from Gaussian distribution,
then
9
P(W1
= w|H0) =
1
p
2⇡
exp
✓
1
2
(w ¯µ0
)2
◆
• Let, P0(Wn
) = P(Wn
= wn
|H0)

10. Example
October 2017The University of Arizona 10
⇢n
0 =
⇢0
Qn
k=1 P0(Wk
)
⇢0
Qn
k=1 P0(Wk) + ⇢0
1
Qn
k=1 P1(Wk)
• After n observations,
=
⇢0
n
(W1
, . . . , Wn
⇢0 + ⇢0
1
n(W1, . . . , Wn)
n
(W1
, . . . , Wn
) =
nY
k=1
P0(Wk
)
P1(Wk)
• Where,
)

11. Solution of the problem
October 2017The University of Arizona 11
• Denote the set of all experiments as Sn
= (W1
, . . . , Wn
)
• In the solution / policy , we have two decisions (X⇡
, Y ⇡
)⇡
X⇡
(Sn
) =
(
1, stop and decide
0, continue observing
Y ⇡
(Sn
) =
(
1, Decide H1
0, Decide H0

12. Solution of the problem (Cont’d)
October 2017The University of Arizona 12
• Two sources of error can happen:
1. False alarm: stop and conclude H1
but the true is the null hypothesis H0
2. Mis detection: we did not pick up
any change, i.e., concludeH0 but
H1the alternative hypothesis is true
P⇡
F = E [Y ⇡
(Sn
|H0)]
P⇡
M = E [1 Y ⇡
(Sn
|H1)]
Pe = (1 ⇢0)P⇡
F + ⇢0P⇡
M
• Average probability of error:

13. Solution of the problem (Cont’d)
October 2017The University of Arizona 13
• Let,
N⇡
= min{n|X⇡
(Sn
) = 1}
a random variable that depends on our policy , decision function
and the observations
⇡ X⇡
(W1
, W2
, . . . , Wn
)
• The cost function is defined as follows:
U⇡
(c) = Pe + cE(N⇡
)
where is a scaling coefficient.c

14. Solution of the problem (Cont’d)
October 2017The University of Arizona 14
• The cost function is decomposed into two conditionals risks
r⇡
0 = P⇡
F + cE(N⇡
|H0)
r⇡
1 = P⇡
M + cE(N⇡
|H1)
r⇡
= ⇢0r⇡
0 + (1 ⇢0)r⇡
1
• The total cost is in terms of the conditional risks:
• Now our objective is
R0
(⇢0) = min
⇡
r⇡

15. Solution of the problem (Cont’d)
October 2017The University of Arizona 15
• Special cases:
Then the cost function is
⇢0 = 1 =) N⇡
= 0, P⇡
F = 0, P⇡
M = 0
Similarly, for
R0
(1) = 0
⇢0 = 0 =) R0
(0) = 0
1.
2. If N⇡
= 0, decide Y = 0(i.e., H0) =) R0
(⇢0|Y = 0) = ⇢0
Similarly when N⇡
= 0, decide Y = 1(i.e., H1) =) R0
(⇢0|Y = 1) = 1 ⇢0

16. Solution of the problem (Cont’d)
October 2017The University of Arizona 16
⇢0
0
Risk
0.5
1 ⇢0
0 ⇢0
0
Risk is a concave function
⇢L ⇢U
Stop and decide H1Stop and decide H0
Continue updating priors

17. Solution of the problem (Cont’d)
October 2017The University of Arizona 17
• Now we are interested to obtain ⇢L
, ⇢U
• The updated prior is
⇢n+1
0 =
Ln
(Sn
)
Ln(Sn) + ⇢0/(1 ⇢0)
Ln
(Sn
) =
nY
k=1
P1(Wk
)
P0(Wk)
• The likelihood ratio is defined as

18. Solution of the problem (Cont’d)
October 2017The University of Arizona 18
• Determining if
⇢n+1
0 =
Ln
(Sn
)
Ln(Sn) + ⇢0/(1 ⇢0)
⇢n+1
0 (Sn
)  ⇢L
or ⇢n+1
0 (Sn
) ⇢U
is the same as testing
Ln
(Sn
)  Aor Ln
(Sn
) B
• Why?
Quasi linear function
0
Increasing function for Ln
(Sn
) 0

19. Solution of the problem (Cont’d)
October 2017The University of Arizona 19
• The new bounds A and B are obtained as
A =
⇢n
0 ⇢L
(1 ⇢n
0 )(1 ⇢L)
B =
⇢n
0 ⇢U
(1 ⇢n
0 )(1 ⇢U )
• Now the decision rule is
Ln
(Sn
) =
8
><
>:
B stop and chooseY n
= 1
 A stop and choose Y n
= 0
otherwise continue observing

20. Solution of the problem (Cont’d)
October 2017The University of Arizona 20
The following figure plots the log of the likelihood for a set of sample
observations. After 16 observations, we conclude that is true.H1

21. Solution of the problem (Cont’d)
October 2017The University of Arizona 21
• Since getting A, B exactly is difficult
P⇡
F ⇡
1 A
B A
P⇡
M ⇡
A(B 1)
B A
Wald’s approximation
• Then for an acceptable P⇡
F , P⇡
M
A =
P⇡
M
1 P⇡
F
B =
1 P⇡
M
P⇡
F

22. Application: Compressive Spectrum Sensing
October 2017The University of Arizona 22
Source: mdpi.com
Fig. Network Model