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A testing scenario for probabilistic automata

A testing scenario for probabilistic automata. Marielle Stoelinga UC Santa Cruz Frits Vaandrager University of Nijmegen. ??. ´. P. Q. Characterization of process equivalences. process equivalences bisimulation / ready trace / ... equivalence

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A testing scenario for probabilistic automata

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  1. A testing scenario for probabilistic automata Marielle Stoelinga UC Santa Cruz Frits Vaandrager University of Nijmegen

  2. ?? ´ P Q Characterization of process equivalences • process equivalences • bisimulation / ready trace / ... equivalence • algebraical / logical / denotional / ... definitions • are these reasonable ? • justify equivalence via testing scenarios / button pushing experiments • comparative concurrency theory • De Nicola & Hennesy [DH86], Milner [Mil80], van Glabbeek [vGl01] • testing scenarios: • define intuitive notion of observation, fundamental • processes that cannot be distinguished by observation ce ´:P ´ Q iff Obs(P) = Obs(Q) • ´ does not distinguish too much/too little

  3. ?? ´ P Q Characterization of process equivalences • process equivalences • bisimulation / ready trace / ... equivalence • algebraical / logical / denotional / ... definitions • are these reasonable ? • justify equivalence via testing scenarios / button pushing experiments • testing scenarios: • define intuitive notion of observation, fundamental • processes that cannot be distinguished by observation are deemed to be equivalent • justify process equivalence ´:P ´ Q iff Obs(P) = Obs(Q) • ´ does not distinguish too much/too little

  4. Main results • testing scenarios in non-probabilistic case • trace equivalence • bisimulation • ... • we define • observations of a probabilistic automaton (PA) • observe probabilities through statistical methods (hypothesis testing) • characterization result • Obs(P) = Obs(Q) iff trd(P) = trd(Q), P, Q finitely branching • trd(P) extension of traces for PAs. [Segala] • justifies trace distr equivalence in terms of observations

  5. Model for testing scenarios display a machine buttons • machine M • a black box • inside: process described by LTS P • display • showing current action • buttons • for user interaction

  6. Model for testing scenarios display a machine buttons • an observer • records what s/he sees (over time) + buttons • define ObsM(P): • observations of P = what observer records, if LTS P is inside M

  7. a Trace Machine (TM) • no buttons for interaction • display shows current action

  8. P a a b c Trace Machine (TM) • no buttons for interaction • display shows current action • with P inside M, an observer sees one of e, a, ab, ac • ObsTM(P) = traces of P • testing scenario justifies trace (language) equivalence

  9. a Trace Machine (TM) • no distinguishing observation between a a a b c b c

  10. d reset Trace Distribution Machine (TDM) P h t 1/2 1/2 d l • reset button: start over • repeat experiments • each experiment yields trace of same length k (wlog) • observefrequencies of traces

  11. d reset Trace Distribution Machine (TDM) P h t 1/2 1/2 d l • 100 exps, length 2frequencies - hd tl tl hd tl .... hd hd 48 tl52 other 0 (hh,dh,hl,...)

  12. d reset Trace Distribution Machine (TDM) P h t 1/2 1/2 d l not every seq of experiments is an observation • with many experiments: #hd¼ # tl • 100 exps, length 2frequencies - hd tl tl hd tl .... hd hd 48 tl52 other 0

  13. d reset Trace Distribution Machine (TDM) P h t 1/2 1/2 d l • with many experiments: #hd¼ # tl • 100 exps, length 2frequencies - hd tl tl hd tl ... hd hd 48 tl52 other 0 - hd hd hd hd .... hd hd 100 tl 0 other 0 2 Obs(P) freq likely 2 Obs(P) freq unlikely /

  14. d reset Trace Distribution Machine (TDM) P d b c b 3/4 1/3 2/3 1/4 • nondeterministic choice • choose one transition probabilistically • in large outcomes: 1/2 #c + 1/3 #d ¼ #b • use statistics: • b,b,b,....b2 Obs(P) freqs likely • b,d,c,b,b,b,c,...2 Obs(P) freqs unlikey /

  15. a reset Observations TDM P h t 1/2 1/2 d l • ObsTDM(P) = { s | s is likely to be produced by P}

  16. a reset Observations TDM P h t 1/2 1/2 d l • perform m experiments (m resets) • each experiment yields trace of length k(wlog) • sequence s2 (Actk)m • Obs(P) = {s2 (Actk)m | s is likely to be produced by P, k,m 2N} • what is likely? use hypothesis testing

  17. h t 1/2 1/2 Which frequencies are likely? • I have a sequence s = h,t,t,t,h,t,...2 {h,t}100 • I claim: generated s with automaton P. • do you believe me • if s contains 15 h’s? • if s contains 42 h’s?

  18. h t 1/2 1/2 0 40 50 60 100 reject H0 K: accept H0 reject H0 Which frequencies are likely? • I have a sequence s = h,t,t,t,h,t,...2 {h,t}100 • I claim: generated s with automaton P. • do you believe me • if s contains 15 h’s? • if s contains 42 h’s? • use hypothesis testing: • fix confidence level a 2 (0,1) • H0 null hypothesis = s is generated by P #h

  19. h t 1/2 1/2 0 40 50 60 100 reject H0 K: accept H0 reject H0 Which frequencies are likely? • I have a sequence s = h,t,t,t,h,t,...2 {h,t}100 • I claim: generated s with automaton P. • do you believe me • if s contains 15 h’s? • if s contains 42 h’s? • use hypothesis testing: • fix confidence level a 2 (0,1) • H0 null hypothesis = s is generated by P • PH0[K] > 1-a: prob on false rejection ·a P: H0[K] minimal : prob on false acceptance minimal #h

  20. h t 1/2 1/2 0 40 50 60 100 reject H0 K: accept H0 reject H0 Which frequencies are likely? • I have a sequence s = h,t,t,t,h,t,...2 {h,t}100 • I claim: generated s with automaton P. • do you believe me • if s contains 15 h’s? NO • if s contains 42 h’s? YES • use hypothesis testing: • fix confidence level a 2 (0,1) • H0 null hypothesis = s is generated by P • PH0[K] > 1-a: prob on false rejection ·a P: H0[K] minimal : prob on false acceptance minimal #h

  21. h t 1/2 1/2 Example: Observations for a =0.05 • Obs(P) = {s2 (Actk)m | s is likely to be produced by P, k,m2N } • for k = 1 and m = 100 s2 (Act)100 is an observation iff 40 · freqs (hd) · 60 60 50 40 0 100 K1,100

  22. h t 1/2 1/2 Example: Observations for a =0.05 • Obs(P) = {s2 (Actk)m | s is likely to be produced by P, k,m2N } • for k = 1 and m = 100 s2 (Act)100 is an observation iff 40 · freqs (hd) · 60 • for k = 1 and m = 200 s2 (Act)200 is an observation iff 88 · freqs (h) · 112 • etc .... 112 100 88 0 200 K1,200

  23. h t 1/2 1/2 Example: Observations for a =0.05 • Obs(P) = {s2 (Actk)m | s is likely to be produced by P, k,m2N } • for k = 1 and m = 100, s2 (Act)100 is an observation iff 40 · freqs (hd) · 60 • for k = 1 and m = 200 s2 (Act)200 is an observation iff 88 · freqs (h) · 112 • etc .... exp freq EP alloweddeviation e ( = 12) 112 100 88 0 0 200 K1,200

  24. With nondeterminism P a c b a 3/4 1/3 2/3 1/4 • s = b,c,c,d,b,d,....,c2 Obs(P) ??

  25. With nondeterminism P a c b a 3/4 1/3 2/3 1/4 • s = b,c,c,d,b,d,....,c2 Obs(P) ?? • fix schedulers: p1, p2, p3... p100 • pi = P[take left trans in experiment i] • 1 - pi = P[takerighttrans in experiment i] • H0: s is generated by P under p1, p2, p3... p100 • critical section Kp1,...,p100

  26. With nondeterminism P a c b a 3/4 1/3 2/3 1/4 • s = b,c,c,d,b,d,....,c2 Obs(P) ?? • fix schedulers: p1, p2, p3... p100 • pi = P[take left trans in experiment i] • 1 - pi = P[takerighttrans in experiment i] • H0: s is generated by P under p1, p2, p3... p100 • critical sectionKp1,...,p100 s2 Obs(P) iff s2Kp1,...,p100 for some p1, p2, p3... p100

  27. Example: Observations b d c b 3/4 1/3 2/3 1/4 Observations for k =1, m = 100. • s contains b,c only with 54 · freqs (c) · 78 • take pi = 1 for all i • s contains b,d only with 62 · freqs (d) · 88 • take pi = 0 for all i

  28. Example: Observations b d c b 3/4 1/3 2/3 1/4 Observations for k =1, m = 100. • s contains a,b only and 54 · freqs (c) · 78 • take pi = 1 for all i • s contains b,d only and 62 · freqs (d) · 88 • take pi = 0 for all i Observations for k=1, m = 200 • 61 · freqs (c) · 71 and 70 · freqs (d) · 80 • pi = ½ for all i (exp 66 c’s; 74 d’s; 60 b’s) • (these are not all observations; they form a sphere)

  29. Main result • TDM characterizes trace distr equivalence ObsTDM(P) = ObsTDM(Q)ifftrd(P) = trd(Q) if P, Q are finitely branching • justifies trace distribution equivalence in an observational way

  30. Main result TDM characterizes trace distr equiv ´TD ObsTDM(P) = ObsTDM(Q) iff trd(P) = trd(Q) • “only if” part is immediate, “if”-part is hard. • find a distinguishing observation if P, Q have different trace distributions. • IAP for P. Q fin branching • P, Q have the same infinite trace distrs iff P, Q have the same finite trace distrs • the set of trace distrs is a polyhedron • Law of large numbers • for random vars with different distributions

  31. h t 1/2 1/2 0 40 50 60 100 K Observations a =0.05 • Obs(P) = {s2 (Actk)m | s likely to be produced by P} • Obs(P) = {s2 (Actk)m | freq_s in K} • for k = 1 and m = 100, b2 (Act)100 is an observation iff 40 · freqb (hd) · 60

  32. h t 1/2 1/2 Nondeterministic case • \sigma = \beta_1 ,...\beta_m • fixed adversaries • take in • expect_freq • for \gamma\in Act^k, freq_\gamma(\beta) freq \in \ • we consider only frequency of traces in an outcome

  33. a! a b c b! a? b?

  34. a! b! a?

  35. d reset Trace Distribution Machine (TDM) h t 1/2 1/2 d l • reset button: start over • repeat experiments: yields sequenceof traces • in large outcomes: #hd¼ # tl • use statistics: • hd,hd,hd,...,hd2 Obs(P) too unlikely • hd,tl,tl,hd,...tl,hd2 Obs(P) likely /

  36. h t 1/2 1/2 0 40 50 60 100 exp freq

  37. Process equivalences P Q

  38. Testing scenario’s display a machine buttons • a black box with display and buttons • inside: process described by LTS P • display: current action • what do we see (over time)? ObsM(P) • P, Q are deemed equivalent iff ObsM(Q) = ObsM(Q) • desired characterization:

  39. h t 1/3 1/3 Observations a =0.05 • Obs(P) = {s2 (Actk)m | s is likely to be produced by P} • for k = 1 and m = 99, • expectation E = (33,33,33) • Obs(P) = {s2 (Act)99 | | s – E | < 15}

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