1. Analysis of uncertain data: Evaluation of Given Hypotheses Selection of probes for information gathering
Anatole Gershman, Eugene Fink,
Bin Fu, and Jaime G. Carbonell

2. Analysis of uncertain data: Evaluation of Given Hypotheses Selection of probes for information gathering
Anatole Gershman, Eugene Fink,
Bin Fu, and Jaime G. Carbonell

3. The analyst has to distinguish between two hypotheses:
Example
The analyst has to distinguish between two hypotheses:
0.4
0.6
Retires
Joins Vikings

4. Example Observations:
Without the tearful public ceremony that accompanied his retirement announcement from the Green Bay Packers just 11 months ago, quarterback Brett Favre has told the New York Jets he is retiring.
Minnesota coach Brad Childress, jilted at the altar Tuesday afternoon by Brett Farve telling him he wasn’t going to play for the Vikings in 2009.
According to many rumors, quarterback Brett Favre has closed on the purchase of a home in Eden Prairie, MN, where the Minnesota Vikings' team facility is located.

5. Example Observation distributions: Bayesian induction:
Without the tearful public ceremony that accompanied his retirement announcement from the Green Bay Packers just 11 months ago, quarterback Brett Favre has told the New York Jets he is retiring.
P(says retire | Retires) = 0.9
P(says retire | Joins Vikings) = 0.6
Bayesian induction:
P (Retire|says retire) = P (Retire) ∙ P(says retire|Retire) /
(P (Retire) ∙ P(says retire|Retire)
+ P (Joins Vikings) ∙ P(Joins Vikings|Retire))
= 0.5

6. General problem
We have to distinguish among n mutually exclusive hypotheses, denoted H1, H2,…, Hn.
0.4
0.6
For every hypothesis, we know its prior;
thus, we have an array of n of priors,
P(H1), P(H2), P(Hn)

7. General problem I will RETIRE! 0.9 0.4 I won’t RETIRE! 0.1 0.6
For every hypothesis, we know the related probability distribution of each observation. P(oa,j | Hi) represents the probabilities of possible values of OBSa.
For every observation, OBSa, we know the number of its possible values, num[a]. Thus, we have num[1..m] with the number of values for each observation.
We base the analysis on m observable features, denoted OBS1, OBS2, …, OBSm. Each observation is a variable that takes one of several discrete values.
I will RETIRE!
OBS1
0.9
0.4
I won’t RETIRE!
0.1
0.6
num[1] = 2
We know a specific value of each observation val [1..m].

8. General problem
We have to evaluate the posterior probabilities of the n given hypotheses, denoted Post(H1), Post(H2), Post(Hn)
0.5
0.5

9. Something else H0 (“surprise”)
Extension #1
Prior:
0.6
0.35
0.05
Something else H0 (“surprise”)

10. Extension #1 After discovering val, Posterior probability of H0:
Post(H0)
= P(H0) ∙ P(val | H0) / P(val)
= P(H0) ∙ P(val | H0)
/ (P(H0) ∙ P(val | H0) + likelihood(val)).
Bad news: We do not know P(val | H0).
Good news: Post(H0) monotonically depends on P(val | H0); thus, if we obtain lower and upper bounds for P(val | H0), we also get bounds for Post(H0).

11. Plausibility principle
Unlikely events normally do not happen; thus, if we have observed val, then its likelihood must not be too small.
Plausibility threshold: We use a global constant plaus, which must be between 0.0 and 1.0. If we have observed val, we assume that
P(val) ≥ plaus / num.
We use it to obtains bounds for P(val | H0), :
Lower: (plaus / num − likelihood(val)) / prior[0].
Upper: 1.0.

12. Plausibility principle
We use it to obtains bounds for P(val | H0):
Lower: (plaus / num − likelihood(val)) / P(H0) .
Upper: 1.0.
We substitute these bounds into the dependency of Post(H0) on P(val | H0), , thus obtaining the bounds for Post(H0):
Lower: 1.0 − likelihood(val) ∙ num / plaus.
Upper: P(H0) / (P(H0) + likelihood(val)).
We have derived bounds for the probability that none of the given hypotheses is correct.

13. Extension #2 Multiple observations: Which one(s) to Use?
Bayesian analysis: Use their joint distribution? Difficult to get.
Independence assumption: usually does not work.
We identify the highest-utility observation and do not use other observations to corroborate it.

14. Extension #2 0.5 0.5 0.4 0.6 0.35 0.65 Which one is “better”?
Utility Function
0.5
0.5
0.4
0.6
0.35
0.65
Which one is “better”?

15. Extension #2 0.5 0.5 0.4 0.6 0.35 0.65 Shannon’s Entropy (negation)
Utility Function
0.5
0.5
0.4
0.6
0.35
0.65
Shannon’s Entropy (negation)

16. Extension #2
Utility Function
0.5
0.5
0.4
0.6
0.35
0.65
KL-divergence

17. Extension #2 0.5 0.5 0.4 0.6 0.35 0.65 Self-defined function
Utility Function
0.5
0.5
0.4
0.6
0.35
0.65
Self-defined function

18. Analysis of uncertain data: Evaluation of Given Hypotheses Selection of probes for information gathering
Anatole Gershman, Eugene Fink,
Bin Fu, and Jaime G. Carbonell

19. The analyst has to distinguish between two hypotheses:
Example
The analyst has to distinguish between two hypotheses:
0.4
0.6
Retires
Joins Vikings

20. Example
I will RETIRE!
0.5
0.4
0.5
0.6
Ask
Probe: Execute external action and observe its response, to gather more information.

21. Example Probe: Probe Cost Observation Probability
Gain (utility function)
Probe
Cost
Observation
Probability

22. Probe Selection single-obs-gain(probej) = visible[i, a, j]
· (likelihood(1) · probe-gain(1)
+ … + likelihood(num[a]) · probe-gain(num[a]))
+ (1.0 − visible[i, a, j]) · cost[j]
Utility
Function
Observation
Probability
Probe
Cost
gain(probj)
= max (single-obs-gain(probj, obs1),…,
single-obs-gain(probj, obsm))

23. Experiment Task: Evaluating hypothesizes (H1, H2, H3, H4).
No Probe, Accuracy of distinguishing between H1 and other hypotheses

24. Experiment Task: Evaluating hypothesizes (H1, H2, H3, H4).
Probe Selection to distinguish H1 and other hypotheses

25. Experiment Task: Evaluating hypothesizes (H1, H2, H3, H4).
Probe Selection to distinguish four hypotheses

26. Summary
Use Bayesian inference to distinguish among mutually exclusive hypotheses.
H0 hypothesis
Multiple observations
Use Probe to gather more information for better analysis
Cost, Utility function, Observation Probability,...

27. Thank you