A Revealing Introduction to Hidden Markov Models.ppt

1、A Revealing Introduction to Hidden Markov Models,Mark Stamp,1,HMM,Hidden Markov Models,What is a hidden Markov model (HMM)? A machine learning technique A discrete hill climb technique Where are HMMs used? Speech recognition Malware detection, IDS, etc., etc. Why is it useful? Efficient algorithms,H

2、MM,2,Markov Chain,Markov chain is a “memoryless random process” Transitions depend only on current state and transition probabilities matrix Example on next slide,HMM,3,Markov Chain,We are interested in average annual temperature Only consider Hot and Cold From recorded history, we obtain probabilit

3、ies See diagram to the right,HMM,4,H,C,0.7,0.6,0.3,0.4,Markov Chain,Transition probability matrixMatrix is denoted as ANote, A is “row stochastic”,HMM,5,H,C,0.7,0.6,0.3,0.4,Markov Chain,Can also include begin, end states Begin state matrix is In this example,Note that is row stochastic,HMM,6,H,C,0.7

4、,0.6,0.3,0.4,begin,end,0.6,0.4,Hidden Markov Model,HMM includes a Markov chain But this Markov process is “hidden” Cannot observe the Markov process Instead, we observe something related to hidden states Its as if there is a “curtain” between Markov chain and observations Example on next slide,HMM,7

5、,HMM Example,Consider H/C temperature example Suppose we want to know H or C temperature in distant past Before humans (or thermometers) invented OK if we can just decide Hot versus Cold We assume transition between Hot and Cold years is same as today That is, the A matrix is same as today,HMM,8,HMM

6、 Example,Temp in past determined by Markov process But, we cannot observe temperature in past Instead, we note that tree ring size is related to temperature Look at historical data to see the connection We consider 3 tree ring sizes Small, Medium, Large (S, M, L, respectively) Measure tree ring size

7、s and recorded temperatures to determine relationship,HMM,9,HMM Example,We find that tree ring sizes and temperature related byThis is known as the B matrix:Note that B also row stochastic,HMM,10,HMM Example,Can we now find temps in distant past? We cannot measure (observe) temp But we can measure t

8、ree ring sizes and tree ring sizes related to temp By the B matrix So, we ought to be able to say something about temperature,HMM,11,HMM Notation,A lot of notation is required Notation may be the most difficult part,HMM,12,HMM Notation,To simplify notation, observations are taken from the set 0,1,M-

9、1 That is, The matrix A = aij is N x N, whereThe matrix B = bj(k) is N x M, where,HMM,13,HMM Example,Consider our temperature example What are the observations? V = 0,1,2, which corresponds to S,M,L What are states of Markov process? Q = H,C What are A,B, , and T? A,B, on previous slides T is number

10、 of tree rings measured What are N and M? N = 2 and M = 3,HMM,14,Generic HMM,Generic view of HMMHMM defined by A,B, and We denote HMM “model” as = (A,B,),HMM,15,HMM Example,Suppose that we observe tree ring sizes For 4 year period of interest: S,M,S,L Then = (0, 1, 0, 2) Most likely (hidden) state s

11、equence? We want most likely X = (x0, x1, x2, x3) Let x0 be prob. of starting in state x0 Note prob. of initial observation And ax0,x1 is prob. of transition x0 to x1 And so on,HMM,16,HMM Example,Bottom line? We can compute P(X) for any X For X = (x0, x1, x2, x3) we haveSuppose we observe (0,1,0,2),

12、 then what is probability of, say, HHCC? Plug into formula above to find,HMM,17,HMM Example,Do same for all 4-state sequences We find The winner is? CCCH Not so fast my friend,HMM,18,HMM Example,The path CCCH scores the highest In dynamic programming (DP), we find highest scoring path But, HMM maxim

13、izes expected number of correct states Sometimes called “EM algorithm” For “Expectation Maximization” How does HMM work in this example?,HMM,19,HMM Example,For first position Sum probabilities for all paths that have H in 1st position, compare to sum of probs for paths with C in 1st position - bigge

14、st wins Repeat for each position and we find:,HMM,20,HMM Example,So, HMM solution gives us CHCH While dynamic program solution is CCCH Which solution is better? Neither! Why is that? Different definitions of “best”,HMM,21,HMM Paradox?,HMM maximizes expected number of correct states Whereas DP choose

15、s “best” overall path Possible for HMM to choose “path” that is impossible Could be a transition probability of 0 Cannot get impossible path with DP Is this a flaw with HMM? No, its a feature,HMM,22,The Three Problems,HMMs used to solve 3 problems Problem 1: Given a model = (A,B,) and observation se

16、quence O, find P(O|) That is, we score an observation sequence to see how well it fits the given model Problem 2: Given = (A,B,) and O, find an optimal state sequence Uncover hidden part (as in previous example) Problem 3: Given O, N, and M, find the model that maximizes probability of O That is, tr

17、ain a model to fit the observations,HMM,23,HMMs in Practice,Typically, HMMs used as follows Given an observation sequence Assume a hidden Markov process exists Train a model based on observations Problem 3 (determine N by trial and error) Then given a sequence of observations, score it vs model from

18、 previous step Problem 1 (high score implies its similar to training data),HMM,24,HMMs in Practice,Previous slide gives sense in which HMM is a “machine learning” technique We do not need to specify anything except the parameter N And “best” N found by trial and error That is, we dont have to think

19、too much Just train HMM and then use it Best of all, efficient algorithms for HMMs,HMM,25,The Three Solutions,We give detailed solutions to the three problems Note: We must have efficient solutions Recall the three problems: Problem 1: Score an observation sequence versus a given model Problem 2: Gi

20、ven a model, “uncover” hidden part Problem 3: Given an observation sequence, train a model,HMM,26,Solution 1,Score observations versus a given model Given model = (A,B,) and observation sequence O=(O0,O1,OT-1), find P(O|) Denote hidden states as X = (x0, x1, . . . , xT-1) Then from definition of B,P

21、(O|X,)=bx0(O0) bx1(O1) bxT-1(OT-1) And from definition of A and ,P(X|)=x0 ax0,x1 ax1,x2 axT-2,xT-1,HMM,27,Solution 1,Elementary conditional probability fact:P(O,X|) = P(O|X,) P(X|) Sum over all possible state sequences X,P(O|) = P(O,X|) = P(O|X,) P(X|)= x0bx0(O0)ax0,x1bx1(O1)axT-2,xT-1bxT-1(OT-1) Th

22、is “works” but way too costly Requires about 2TNT multiplications Why? There better be a better way,HMM,28,Forward Algorithm,Instead of brute force: forward algorithm Or “alpha pass” For t = 0,1,T-1 and i=0,1,N-1, lett(i) = P(O0,O1,Ot,xt=qi|) Probability of “partial sum” to t, and Markov process is

23、in state qi at step t What the? Can be computed recursively, efficiently,HMM,29,Forward Algorithm,Let 0(i) = ibi(O0) for i = 0,1,N-1 For t = 1,2,T-1 and i=0,1,N-1, lett(i) = (t-1(j)aji)bi(Ot) Where the sum is from j = 0 to N-1 From definition of t(i) we seeP(O|) = T-1(i) Where the sum is from i = 0

24、to N-1 Note this requires only N2T multiplications,HMM,30,Solution 2,Given a model, find “most likely” hidden states: Given = (A,B,) and O, find an optimal state sequence Recall that optimal means “maximize expected number of correct states” In contrast, DP finds best scoring path For temp/tree ring

25、 example, solved this But hopelessly inefficient approach A better way: backward algorithm Or “beta pass”,HMM,31,Backward Algorithm,For t = 0,1,T-1 and i=0,1,N-1, lett(i) = P(Ot+1,Ot+2,OT-1|xt=qi,) Probability of partial sum from t to end and Markov process in state qi at step t Analogous to the for

26、ward algorithm As with forward algorithm, this can be computed recursively and efficiently,HMM,32,Backward Algorithm,Let T-1(i) = 1 for i = 0,1,N-1 For t = T-2,T-3, ,1 and i=0,1,N-1, lett(i) = ai,jbj(Ot+1)t+1(j) Where the sum is from j = 0 to N-1,HMM,33,Solution 2,For t = 1,2,T-1 and i=0,1,N-1 defin

27、et(i) = P(xt=qi|O,) Most likely state at t is qi that maximizes t(i) Note that t(i) = t(i)t(i)/P(O|) And recall P(O|) = T-1(i) The bottom line? Forward algorithm solves Problem 1 Forward/backward algorithms solve Problem 2,HMM,34,Solution 3,Train a model: Given O, N, and M, find that maximizes proba

28、bility of O Here, we iteratively adjust = (A,B,) to better fit the given observations O The size of matrices are fixed (N and M) But elements of matrices can change It is amazing that this works! And even more amazing that its efficient,HMM,35,Solution 3,For t=0,1,T-2 and i,j in 0,1,N-1, define “di-

29、gammas” ast(i,j) = P(xt=qi, xt+1=qj|O,) Note t(i,j) is prob of being in state qi at time t and transiting to state qj at t+1 Then t(i,j) = t(i)aijbj(Ot+1)t+1(j)/P(O|) And t(i) = t(i,j) Where sum is from j = 0 to N 1,HMM,36,Model Re-estimation,Given di-gammas and gammas For i = 0,1,N-1 let i = 0(i) F

30、or i = 0,1,N-1 and j = 0,1,N-1 aij = t(i,j)/t(i) Where both sums are from t = 0 to T-2 For j = 0,1,N-1 and k = 0,1,M-1 bj(k) = t(j)/t(j) Both sums from from t = 0 to T-2 but only t for which Ot = k are counted in numerator Why does this work?,HMM,37,Solution 3,To summarize Initialize = (A,B,) Comput

31、e t(i), t(i), t(i,j), t(i) Re-estimate the model = (A,B,) If P(O|) increases, goto 2,HMM,38,Solution 3,Some fine points Model initialization If we have a good guess for = (A,B,) then we can use it for initialization If not, let i 1/N, ai,j 1/N, bj(k) 1/M Subject to row stochastic conditions Note: Do

32、 not initialize to uniform values Stopping conditions Stop after some number of iterations Stop if increase in P(O|) is “small”,HMM,39,HMM as Discrete Hill Climb,Algorithm on previous slides shows that HMM is a “discrete hill climb” HMM consists of discrete parameters Specifically, the elements of t

33、he matrices And re-estimation process improves model by modifying parameters So, process “climbs” toward improved model This happens in a high-dimensional space,HMM,40,Dynamic Programming,Brief detour For = (A,B,) as above, its easy to define a dynamic program (DP) Executive summary: DP is forward a

34、lgorithm, with “sum” replaced by “max” Precise details on next slides,HMM,41,Dynamic Programming,Let 0(i) = i bi(O0) for i=0,1,N-1 For t=1,2,T-1 and i=0,1,N-1 computet(i) = max (t-1(j)aji)bi(Ot) Where the max is over j in 0,1,N-1 Note that at each t, the DP computes best path for each state, up to t

35、hat point So, probability of best path is max T-1(j) This max only gives best probability Not the best path, for that, see next slide,HMM,42,Dynamic Programming,To determine optimal path While computing optimal path, keep track of pointers to previous state When finished, construct optimal path by t

36、racing back points For example, consider temp example Probabilities for path of length 1:These are the only “paths” of length 1,HMM,43,Dynamic Programming,Probabilities for each path of length 2Best path of length 2 ending with H is CH Best path of length 2 ending with C is CC,HMM,44,Dynamic Program

37、,Continuing, we compute best path ending at H and C at each step And save pointers - why?,HMM,45,Dynamic Program,Best final score is .002822 And, thanks to pointers, best path is CCCH But what about underflow? A serious problem in bigger cases,HMM,46,Underflow Resistant DP,Common trick to prevent un

38、derflow Instead of multiplying probabilities we add logarithms of probabilities Why does this work? Because log(xy) = log x + log y And adding logs does not tend to 0 Note that we must avoid 0 probabilities,HMM,47,Underflow Resistant DP,Underflow resistant DP algorithm: Let 0(i) = log(i bi(O0) for i

39、=0,1,N-1 For t=1,2,T-1 and i=0,1,N-1 computet(i) = max (t-1(j) + log(aji) + log(bi(Ot) Where the max is over j in 0,1,N-1 And score of best path is max T-1(j) As before, must also keep track of paths,HMM,48,HMM Scaling,Trickier to prevent underflow in HMM We consider solution 3 Since it includes sol

40、utions 1 and 2 Recall for t = 1,2,T-1, i=0,1,N-1,t(i) = (t-1(j)aj,i)bi(Ot) The idea is to normalize alphas so that they sum to one Algorithm on next slide,HMM,49,HMM Scaling,Given t(i) = (t-1(j)aj,i)bi(Ot) Let a0(i) = 0(i) for i=0,1,N-1 Let c0 = 1/a0(j) For i = 0,1,N-1, let a0(i) = c0a0(i) This take

41、s care of t = 0 case Algorithm continued on next slide,HMM,50,HMM Scaling,For t = 1,2,T-1 do the following:For i = 0,1,N-1, at(i) = (at-1(j)aj,i)bi(Ot) Let ct = 1/at(j) For i = 0,1,N-1 let at(i) = ctat(i),HMM,51,HMM Scaling,Easy to show at(i) = c0c1ct t(i) () Simple proof by induction So, c0c1ct is

42、scaling factor at step t Also, easy to show thatat(i) = t(i)/t(j) Which implies aT-1(i) = 1 (),HMM,52,HMM Scaling,By combining () and (), we have1 = aT-1(i) = c0c1cT-1 T-1(i)= c0c1cT-1 P(O|) Therefore, P(O|) = 1 / c0c1cT-1 To avoid underflow, we computelog P(O|) = - log(cj) Where sum is from j = 0 t

43、o T-1,HMM,53,HMM Scaling,Similarly, scale betas as ctt(i) For re-estimation, Compute t(i,j) and t(i) using original formulas, but with scaled alphas and betas This gives us new values for = (A,B,) “Easy exercise” to show re-estimate is exact when scaled alphas and betas used Also, P(O|) cancels from

44、 formula Use log P(O|) = - log(cj) to decide if iterate improves,HMM,54,All Together Now,Complete pseudo code for Solution 3 Given: (O0,O1,OT-1) and N and M Initialize: = (A,B,) A is NxN, B is NxM and is 1xN i 1/N, aij 1/N, bj(k) 1/M, each matrix row stochastic, but not uniform Initialize: maxIters

45、= max number of re-estimation steps iters = 0 oldLogProb = -,HMM,55,Forward Algorithm,Forward algorithm With scaling,HMM,56,Backward Algorithm,Backward algorithm or “beta pass” With scaling Note: same scaling factor as alphas,HMM,57,Gammas,Here, use scaled alphas and betas So formulas unchanged,HMM,

46、58,Re-Estimation,Again, using scaled gammas So formulas unchanged,HMM,59,Stopping Criteria,Check that probability increases In practice, want logProb oldLogProb + And dont exceed max iterations,HMM,60,English Text Example,Suppose Martian arrives on earth Sees written English text Wants to learn some

47、thing about it Martians know about HMMs So, strip our all non-letters, make all letters lower-case 27 symbols (letters, plus word-space) Train HMM on long sequence of symbols,HMM,61,English Text,For first training case, initialize: N = 2 and M = 27 Elements of A and are about each Elements of B are

48、each about 1/27 We use 50,000 symbols for training After 1st iter: log P(O|) -165097 After 100th iter: log P(O|) -137305,HMM,62,English Text,Matrices A and converge:What does this tells us? Started in hidden state 1 (not state 0) And we know transition probabilities between hidden states Nothing too

49、 interesting here We dont care about hidden states,HMM,63,English Text,What about B matrix? This much more interesting Why?,HMM,64,A Security Application,Suppose we want to detect metamorphic computer viruses Such viruses vary their internal structure But function of malware stays same If sufficiently variable, standard signature detection will fail Can we use HMM for detection? What to use as observation sequence? Is there really a “hidden” Markov process? What about N, M, and T? How many Os needed for training, scoring?,