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Transcript of A Revealing Introduction to Hidden Markov Models Mark Stamp Revealing Introduction to HMMs 1.
1
A Revealing Introduction to Hidden Markov Models
Mark Stamp
Revealing Introduction to HMMs
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Hidden Markov Models
What is a hidden Markov model (HMM)?o A machine learning technique and…o A discrete hill climb techniqueo Two for the price of one!
Where are HMMs used?o Speech recognition, malware detection,
IDS, and many, many more applications Why is it useful?
o Easy to apply and efficient algorithms
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Markov Chain
Markov chaino “Memoryless random process”o Transitions depend only on current
state (Markov chain of order 1)… o …and transition probability matrix
Example? o See next slide…
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Markov Chain Suppose we’re interested
in average annual temperatureo Only consider Hot and Cold
From recorded history, obtain these probabilitieso That is, from thermometer
readings for “recent” years
H
C
0.7
0.6
0.3 0.4
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Markov Chain Transition probability matrix
Matrix is denoted as A
Note, A is “row stochastic”
H
C
0.7
0.6
0.3 0.4
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Markov Chain Can also include
begin, end states Matrix for begin
state is πo In this example,
Note that π also row stochastic
H
C
0.7
0.6
0.3 0.4begin
end
0.6
0.4
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Hidden Markov Model HMM includes a Markov chain
o But the Markov process is “hidden”o So we can’t observe the Markov processo Instead, observe things that are
probabilistically related to hidden stateso It’s as if there is a “curtain” between
Markov chain and observations Example on next few slides…
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HMM Example Consider H/C annual temp example Suppose we want to know H or C
annual temperature in distant pasto Before thermometers were inventedo We only distinguish between H and C
We assume transition between Hot and Cold years is same as todayo Then the A matrix is known
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HMM Example Temp in past determined by Markov
processo But, we cannot observe temperature in past
We find evidence that tree ring size is related to temperatureo Look at historical data to find the connection
We only consider 3 tree ring sizeso Small, Medium, Large (S, M, L, respectively)
Measure tree ring sizes and recorded temperatures to determine relationship
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HMM Example We find that tree ring sizes and
temperature related by
This is known as the B matrix
The matrix B is also row stochastic
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HMM Example
Can we now find H/C temps in past?
We cannot measure (observe) temps
But we can measure tree ring sizes…
…and tree ring sizes related to tempso By probabilities in the B matrix
Can we say something intelligent about temperatures in distant past?
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HMM Notation A lot of notation is required
o Notation may be the most difficult part…
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HMM Notation
Note that for simplicity, observations taken from V = {0,1,…,M-1}
That is, The matrix A = {aij} is N x N, where
o The matrix B = {bj(k)} is N x M,
whereo
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HMM Example
Consider our temperature example… What are the observations?
o V = {0,1,2}, corresponding to S,M,L What are states of Markov process?
o Q = {H,C} What are A,B, π, and T?
o A,B, π on previous slideso T is number of tree rings measured
What are N and M?o N = 2 and M = 3
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Generic HMM Generic view of HMM
HMM defined by A,B, and π We denote HMM “model” as λ =
(A,B,π)
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HMM Example
Suppose that we observe tree ring sizeso For some 4 year period of interest: S,M,S,Lo Then = (0, 1, 0, 2)
Most likely (hidden) state sequence?o That is, most likely X = (x0, x1, x2, x3)
Let πx0 be prob. of starting in state x0
Note prob. of initial observation o And ax0,x1 is prob. of transition x0 to x1
And so on…
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HMM Example
Bottom line? We can compute P(X) for any X For X = (x0, x1, x2, x3) we have
Suppose we observe (0,1,0,2), then what is probability of, say, HHCC?
Plug into formula above to find
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HMM Example
Do same for all 4-state seq’s
We find that the winner is…o CCCH
Not so fast my friend!
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HMM Example
The path CCCH scores the highest In dynamic programming (DP), we
find highest scoring path But, in HMM we maximize
expected number of correct stateso Sometimes called “EM algorithm”o For “Expectation Maximization”
How does HMM work in this example?Revealing Introduction to HMMs
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HMM Example For first position…
o Sum probabilities for all paths that have H in 1st position, compare to sum of probs for paths with C in 1st position --- biggest wins
Repeat for each position and we find
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HMM Example
So, HMM solution gives us CHCH While DP solution is CCCH Which solution is better? Neither!
o Just using different definitions of “best”
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HMM Paradox?
HMM maximizes expected number of correct stateso Whereas DP chooses “best” overall path
Possible for HMM to choose a “path” that is impossibleo Could be a transition probability of 0
Cannot get impossible path with DP Is this a flaw with HMM?
o No, it’s a feature
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Probability of Observations
Table computed for
= (0,1,0,2) For this sequence,
P( ) = .000412 + .000035
+ .000706 + … + .000847
= left to the reader Similarly for other
observationsRevealing Introduction to HMMs
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HMM Model
An HMM is defined by the three matrices, A, B, and π
Note that M and N are implied, since they are the dimensions of matrices
So, we denote an HMM “model” as λ = (A,B,π)
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The Three Problems HMMs used to solve 3 problems Problem 1: Given a model λ = (A,B,π) and
observation sequence O, find P(O|λ)o That is, we can score an observation sequence to
see how well it fits a given model Problem 2: Given λ = (A,B,π) and O, find an
optimal state sequence (in HMM sense)o Uncover hidden part (like previous example)
Problem 3: Given O, N, and M, find the model λ that maximizes probability of Oo That is, train a model to fit observations
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HMMs in Practice
Typically, HMMs used as follows: Given an observation sequence…
o Assume a (hidden) Markov process exists Train a model based on observations
o This is Problem 3 (optimal N by trial and error) Then given a sequence of observations,
score it versus the modelo This is Problem 1: high score implies similar to
training data, low score implies it’s not
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HMMs in Practice
Previous slide gives sense in which HMM is a “machine learning” techniqueo To train model, we do not need to specify
anything except the parameter No “Best” N often found by trial and error
That is, we don’t think too mucho Just train HMM and then use ito Best of all, efficient algorithms for HMMs
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The Three Solutions We give detailed solutions to 3 problems
o Note: We must have efficient solutions The three problems:
o Problem 1: Score an observation sequence versus a given model
o Problem 2: Given a model, “uncover” hidden parto Problem 3: Given an observation sequence, train a
model Recall that we considered example for 2 and 1
o But direct solutions are very inefficient
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Solution 1 Score observations versus a given model
o Given model λ = (A,B,π) and observation sequence O=(O0,O1,…,OT-1), find P(O|λ)
Denote hidden states as X = (x0, x1, . . . , xT-1)
Then from definition of B,P(O|X,λ)=bx0(O0) bx1(O1) … bxT-1(OT-1)
And from definition of A and π,P(X|λ)=πx0 ax0,x1 ax1,x2 … axT-2,xT-1
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Solution 1 Elementary conditional probability fact:
P(O,X|λ) = P(O|X,λ) P(X|λ) Sum over all possible state sequences X,
P(O|λ) = Σ P(O,X|λ) = Σ P(O|X,λ) P(X|λ)= Σπx0bx0(O0)ax0,x1bx1(O1)…axT-2,xT-1bxT-1(OT-1)
This “works” but way too costly Requires about 2TNT multiplications
o Why? There better be a better way…
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Forward Algorithm
Instead, use forward algorithmo Or “alpha pass”
For t = 0,1,…,T-1 and i=0,1,…,N-1, letαt(i) = P(O0,O1,…,Ot,xt=qi|λ)
Probability of “partial sum” to t, and Markov process is in state qi at step t
Can be computed recursively, efficiently
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Forward Algorithm
Let α0(i) = πibi(O0) for i = 0,1,…,N-1 For t = 1,2,…,T-1 and i=0,1,…,N-1, let
αt(i) = (Σαt-1(j)aji)bi(Ot)o Where the sum is from j = 0 to N-1
From definition of αt(i) we see
P(O|λ) = ΣαT-1(i) o Where the sum is from i = 0 to N-1
This requires only N2T multiplications
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Solution 2
Given model, find hidden states o Given λ = (A,B,π) and O, find an optimal
state sequenceo Recall that optimal means “maximize
expected number of correct states”o In contrast, DP finds best scoring path
For temp/tree ring example, solved thiso But hopelessly inefficient approach
A better way: backward algorithmo Or “beta pass”
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Backward Algorithm
For t = 0,1,…,T-1 and i = 0,1,…,N-1, let βt(i) = P(Ot+1,Ot+2,…,OT-1|xt=qi,λ)
Probability of partial sum from t to end and Markov process in state qi at step t
Analogous to the forward algorithm As with forward algorithm, this can be
computed recursively and efficiently
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Backward Algorithm Let βT-1(i) = 1 for i = 0,1,…,N-1
For t = T-2,T-3, …,1 and i = 0,1,…,N-1, let
βt(i) = Σaijbj(Ot+1)βt+1(j)
Where the sum is from j = 0 to N-1
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Solution 2 For t = 1,2,…,T-1 and i=0,1,…,N-1 define
γt(i) = P(xt=qi|O,λ)o Most likely state at t is qi that maximizes γt(i)
Note that γt(i) = αt(i)βt(i)/P(O|λ)o And recall P(O|λ) = ΣαT-1(i)
The bottom line?o Forward algorithm solves Problem 1o Forward/backward algorithms solve Problem 2
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Solution 3 Train a model: Given O, N, and M,
find λ that maximizes probability of O
Here, iteratively adjust λ = (A,B,π) to better fit the given observations Oo The size of matrices are fixed (N and
M)o But elements of matrices can change
It is amazing that this works!o And even more amazing that it’s
efficient
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Solution 3 For t=0,1,…,T-2 and i,j in {0,1,…,N-1},
define “di-gammas” asγt(i,j) = P(xt=qi, xt+1=qj|O,λ)
Note γt(i,j) is prob of being in state qi at time t and transiting to state qj at t+1
Then γt(i,j) = αt(i)aijbj(Ot+1)βt+1(j)/P(O|λ) And γt(i) = Σγt(i,j)
o Where sum is from j = 0 to N – 1
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Model Re-estimation Given di-gammas and gammas… For i = 0,1,…,N-1 let πi = γ0(i) For i = 0,1,…,N-1 and j = 0,1,…,N-1
aij = Σγt(i,j)/Σγt(i)o Where both sums are from t = 0 to T-2
For j = 0,1,…,N-1 and k = 0,1,…,M-1 bj(k) = Σγt(j)/Σγt(j)o Both sums from from t = 0 to T-2 but only t for
which Ot = k are counted in numerator
Why does this work?
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Solution 3
To summarize…1. Initialize λ = (A,B,π) 2. Compute αt(i), βt(i), γt(i,j), γt(i)
3. Re-estimate the model λ = (A,B,π) 4. If P(O|λ) increases, goto 2
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Solution 3 Some fine points… Model initialization
o If we have a good guess for λ = (A,B,π) then we can use it for initialization
o If not, let πi ≈ 1/N, ai,j ≈ 1/N, bj(k) ≈ 1/Mo Subject to row stochastic conditionso But, do not initialize to uniform values
Stopping conditionso Stop after some number of iterations and/or…o Stop if increase in P(O|λ) is “small”
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HMM as Discrete Hill Climb
Algorithm on previous slides shows that HMM is a “discrete hill climb”
HMM consists of discrete parameterso Specifically, the elements of the matrices
And re-estimation process improves model by modifying parameterso So, process “climbs” toward improved
modelo This happens in a high-dimensional space
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Dynamic Programming
Brief detour… For λ = (A,B,π) as above, it’s easy
to define a dynamic program (DP) Executive summary:
o DP is forward algorithm, with “sum” replaced by “max”
Precise details on next few slides
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Dynamic Programming
Let δ0(i) = πi bi(O0) for i=0,1,…,N-1 For t=1,2,…,T-1 and i=0,1,…,N-1 compute
δt(i) = max (δt-1(j)aji)bi(Ot) o 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 that point
So, probability of best path is max δT-1(j) This max gives the best probability
o Not the best path, for that, see next slide
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Dynamic Programming
To determine optimal patho While computing deltas, keep track of
pointers to previous stateo When finished, construct optimal path by
tracing back points For example, consider temp example:
recall that we observe (0,1,0,2) Probabilities for path of length 1:
These are the only “paths” of length 1
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Dynamic Programming
Probabilities for each path of length 2
Best path of length 2 ending with H is CH
Best path of length 2 ending with C is CC
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Dynamic Program
Continuing, we compute best path ending at H and C at each step
And save pointers --- why?
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Dynamic Program
Best final score is .002822o And, thanks to pointers, best path is CCCH
But what about underflow?o A serious problem in bigger cases
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Underflow Resistant DP
Common trick to prevent underflow:o Instead of multiplying probabilities…o …add logarithms of probabilities
Why does this work?o Because log(xy) = log x + log yo Adding logs does not tend to 0
Note that these logs are negative… …and we must avoid 0 probabilities
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Underflow Resistant DP
Underflow resistant DP algorithm: Let δ0(i) = log(πi bi(O0))
for i=0,1,…,N-1 For t=1,2,…,T-1 and i=0,1,…,N-1 compute
δt(i) = max (δt-1(j) + log(aji) + log(bi(Ot)))o Where the max is over j in {0,1,…,N-1}
And score of best path is max δT-1(j) As before, must also keep track of paths
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HMM Scaling
Trickier to prevent underflow in HMM We consider solution 3
o Since it includes solutions 1 and 2 Recall for t = 1,2,…,T-1, i=0,1,…,N-1,
αt(i) = (Σαt-1(j)aj,i)bi(Ot) The idea is to normalize alphas so
that they sum to 1o Algorithm on next slide
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HMM Scaling
Given αt(i) = (Σαt-1(j)aj,i)bi(Ot) Let a0(i) = α0(i) for i=0,1,…,N-1 Let c0 = 1/Σa0(j) For i = 0,1,…,N-1, let a0(i) = c0a0(i) This takes care of t = 0 case Algorithm continued on next slide…
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HMM Scaling
For t = 1,2,…,T-1 do the following: For i = 0,1,…,N-1,
at(i) = (Σat-1(j)aj,i)bi(Ot) Let ct = 1/Σat(j) For i = 0,1,…,N-1 let at(i) = ctat(i)
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HMM Scaling
Easy to show at(i) = c0c1…ct αt(i) (♯)o Simple proof by induction
So, c0c1…ct is scaling factor at step t Also, easy to show that
at(i) = αt(i)/Σαt(j) Which implies ΣaT-1(i) = 1 (♯♯)
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HMM Scaling By combining (♯) and (♯♯), we have
1 = ΣaT-1(i) = c0c1…cT-1 ΣαT-1(i)
= c0c1…cT-1 P(O|λ) Therefore, P(O|λ) = 1 / c0c1…cT-1
To avoid underflow, we computelog P(O|λ) = -Σ log(cj)o Where sum is from j = 0 to T-1
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HMM Scaling Similarly, scale betas as ctβt(i) For re-estimation,
o Compute γt(i,j) and γt(i) using original formulas, but with scaled alphas, betas
This gives us new values for λ = (A,B,π) “Easy exercise” to show re-estimate is
exact when scaled alphas and betas used Also, P(O|λ) cancels from formula
o Use log P(O|λ) = -Σ log(cj) to decide if iterate improves
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All Together Now Complete pseudo code for Solution 3 Given: (O0,O1,…,OT-1) and N and M Initialize: λ = (A,B,π)
o A is NxN, B is NxM and π is 1xNo πi ≈ 1/N, aij ≈ 1/N, bj(k) ≈ 1/M, each matrix row
stochastic, but not uniform Initialize:
o maxIters = max number of re-estimation stepso iters = 0o oldLogProb = -∞
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Forward Algorithm
Forward algorithmo With scaling
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Backward Algorithm Backward
algorithm or “beta pass”o With scaling
Note: same scaling factor as alphas
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Gammas Using scaled
alphas and betas
So formulas unchanged
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Re-Estimation Again, using
scaled gammas
So formulas unchanged
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Stopping Criteria Check that
probability increaseso In practice, wantlogProb >
oldLogProb + ε And don’t
exceed max iterations
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English Text Example
Suppose Martian arrives on eartho Sees written English texto Wants to learn something about ito Martians know about HMMs
So, strip out all non-letters, make all letters lower-caseo 27 symbols (26 letters, word-space)o Train HMM on long sequence of
symbolsRevealing Introduction to HMMs
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English Text
For first training case, initialize: o N = 2 and M = 27o Elements of A and π are all approx 1/2o Elements of B are each about 1/27
We use 50,000 symbols for training After 1st iter: log P(O|λ) ≈ -165097 After 100th iter: log P(O|λ) ≈ -137305
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English Text Matrices A and π converge to:
What does this tells us?o Started in hidden state 1 (not state 0)o And we know transition probabilities
between hidden states Nothing too interesting here
o We don’t (yet) know about hidden states
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English Text
What about B matrix?
This is very interestingo Why???
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A Security Application Suppose we want to detect metamorphic
computer viruseso Such viruses vary their internal structureo But function of malware stays sameo If sufficiently variable, standard signature
detection will fail Can we use HMM for detection?
o What to use as observation sequence?o Is there really a “hidden” Markov process?o What about N, M, and T?o How many Os needed for training, scoring?
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HMM for Metamorphic Detection
Set of “family” viruses into 2 subsets Extract opcodes from each virus Append opcodes from subset 1 to make
one long sequenceo Train HMM on opcode sequence (problem 3)o Obtain a model λ = (A,B,π)
Set threshold: score opcodes from files in subset 2 and “normal” files (problem 1)o Can you sets a threshold that separates sets?o If so, may have a viable detection method
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HMM for Metamorphic Detection
Virus detection results from recent papero Note the
separation This is
good!
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HMM Generalizations
Here, assumed Markov process of order 1o Current state depends only on previous state
and transition matrix A Can use higher order Markov process
o Current state depends on n previous stateso Higher order vs size of N ? “Depth” vs “width”
Can have A and B matrices depend on t HMM often combined with other
techniques (e.g., neural nets)
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Generalizations
In some cases, limitation of HMM is that position information is not usedo In many applications this is OK/desirableo In some apps, this is serious limitation
Bioinformatics applicationso DNA sequencing, protein alignment, etc.o Sequence alignment is crucialo Can use “profile HMMs” instead of
HMMs
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References
M. Stamp, A revealing introduction to hidden Markov models
L.R. Rabiner, A tutorial on hidden Markov models and selected applications in speech recognition
R.L. Cave & L.P. Neuwirth, Hidden Markov models for English
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