Hidden Markov Model Continues …

Finite State Markov Chain A discrete time stochastic process, consisting of a domain D of m states {1,…,m} and 1.An m dimensional initial distribution vector ( p(1),.., p(m)). 2.An m×m transition probabilities matrix M= (a st ) For each integer L, a Markov Chain assigns probability to sequences (x 1 …x L ) over D (i.e, x i D) as follows: Similarly, (X 1,…, X i,…)is a sequence of probability distributions over D.

Use of Markov Chains: Sequences with CpG Islands In human genomes the pair CG often transforms to (methyl-C) G which often transforms to TG. Hence the pair CG appears less than expected from what is expected from the independent frequencies of C and G alone. Due to biological reasons, this process is sometimes suppressed in short stretches of genomes such as in the start regions of many genes. These areas are called CpG islands (p denotes “pair”).

Modeling sequences with CpG Island The “+” model: Use transition matrix A + = (a + st ), Where: a + st = (the probability that t follows s in a CpG island) The “-” model: Use transition matrix A - = (a - st ), Where: a - st = (the probability that t follows s in a non CpG island)

CpG Island: Question 1 We solve the following question: Question 1: Given a short stretch of genomic data, does it come from a CpG island ? By modeling strings with and without CpG islands as Markov Chains over the same states {A,C,G,T} but different transition probabilities:

(Stationary) Markov Chains X1X1 X2X2 X L-1 XLXL Every variable x i has a domain. For example, suppose the domain are the letters {a, c, t, g}. Every variable is associated with a local (transition) probability table p(X i = x i | X i-1 = x i-1 ) and p(X 1 = x 1 ). The joint distribution is given by In short: Stationary means that the transition probability tables do not depend on i.

Question 1: Using two Markov chains X1X1 X2X2 X L-1 XLXL For CpG islands: We need to specify p I (x i | x i-1 ) where I stands for CpG Island. X i-1 XiXi ACTG A C 0.4p(C | C)p(T| C)high T 0.1p(C | T) p(T | T)p(G | T) G 0.3p(C | G) p(T | G)p(G | G) =1 Lines must add up to one; columns need not.

Question 1: Using two Markov chains X1X1 X2X2 X L-1 XLXL For non-CpG islands: We need to specify p N (x i | x i-1 ) where N stands for Non CpG island. X i-1 XiXi ACTG A C 0.4p(C | C) p(T | C)low T 0.1p(C | T) p(T | T)high G 0.3p(C | G) p(T | G)p(G | G) Some entries may or may not change compared to p I (x i | x i-1 ).

Question 1: Log Odds-Ratio test Comparing the two options via odds-ratio test yields If logQ > 0, then CpG island is more likely. If logQ < 0, then non-CpG island is more likely.

CpG Island: Question 2 We now solve the 2 nd question: Question 2: Given a long piece of genomic data, does it contain CpG islands in it, and where? For this, we need to decide which parts of a given long sequence of letters is more likely to come from the “+” model, and which parts are more likely to come from the “–” model. This is done by using the Hidden Markov Model, to be defined.

Question 2: Finding CpG Islands Given a long genomic string with possible CpG Islands, we define a Markov Chain over 8 states, all interconnected (hence it is ergodic): C+C+ T+T+ G+G+ A+A+ C-C- T-T- G-G- A-A- The problem is that we don’t know the sequence of states which are traversed, but just the sequence of letters. Therefore we use here Hidden Markov Model

Hidden Markov Model A Markov chain (s 1,…,s L ): and for each state s and a symbol x we have p(X i =x|S i =s) Application in communication: message sent is (s 1,…,s m ) but we receive (x 1,…,x m ). Compute what is the most likely message sent ? Application in speech recognition: word said is (s 1,…,s m ) but we recorded (x 1,…,x m ). Compute what is the most likely word said ? S1S1 S2S2 S L-1 SLSL x1x1 x2x2 X L-1 xLxL M M M M TTTT

Hidden Markov Model Notations: Markov Chain transition probabilities: p(S i+1 = t|S i = s) = a st Emission probabilities: p(X i = b| S i = s) = e s (b) S1S1 S2S2 S L-1 SLSL x1x1 x2x2 X L-1 xLxL M M M M TTTT For Markov Chains we know: What is p( s,x ) = p(s 1,…,s L ;x 1,…,x L ) ?

Independence assumptions X1X1 X2X3Xi-1XiXi+1x1x1 x2x2 x3x3 x i-1 xixi x i+1 X1X1 X2X3Xi-1XiXi+1S1S1 S2S2 S3S3 S i-1 SiSi S i+1 We assume the following joint distribution for the full chain: This factorization encodes the following conditional independence assumptions: p(s i | s 1,…,s i-1,x 1,…,x i-1 ) = p(s i | s i-1 ) and p(x i | s 1,…,s i,x 1,…,x i-1 ) = p(x i | s i )

Hidden Markov Model p(X i = b| S i = s) = e s (b), means that the probability of x i depends only on the probability of s i. Formally, this is equivalent to the conditional independence assumption: p(X i =x i |x 1,..,x i-1,x i+1,..,x L,s 1,..,s i,..,s L ) = e s i (x i ) S1S1 S2S2 S L-1 SLSL x1x1 x2x2 X L-1 xLxL M M M M TTTT Thus

Hidden Markov Model for CpG Islands The states: S1S1 S2S2 S L-1 SLSL X1X1 X2X2 X L-1 XLXL Domain(S i )={+, -}  {A,C,T,G} (8 values) In this representation P(x i | s i ) = 0 or 1 depending on whether x i is consistent with s i. E.g. x i = G is consistent with s i =(+,G) and with s i =(-,G) but not with any other state of s i. The query of interest: ),,|,...,(argmax),...,( 11 ) (s ** 1 1 LL s L xxsspss L  

Use of HMM: A posteriori belief The conditional probability of a variable X given the evidence e: This is the a posteriori belief in X, given evidence e This query is also called Belief update. We use HMM to compute our posteriori belief on a sequence, given some information on it (usually (x 1,…,x L ))

Hidden Markov Model Questions: Given the “visible” sequence x = (x 1,…,x L ), find: 1.A most probable (hidden) path. 2.The probability of x. 3.For each i = 1,.., L, and for each state k, p(s i =k| x ) S1S1 S2S2 S L-1 SLSL x1x1 x2x2 X L-1 xLxL M M M M TTTT

1. Most Probable state path S1S1 S2S2 S L-1 SLSL x1x1 x2x2 X L-1 xLxL M M M M TTTT First Question: Given an output sequence x = (x 1,…,x L ), A most probable path s*= (s * 1,…,s * L ) is one which maximizes p( s | x ).

Viterbi’s Algorithm for Most Probable Path s1s1 s2s2 X1X1 X2X2 sisi XiXi The task: compute v l (i) = the probability p(s 1,..,s i ;x 1,..,x i |s i =l ) of a most probable path up to i, which ends in state l. Let the states be {1, …, m} Idea: for i = 1, …, L and for each state l, compute:

Viterbi’s algorithm for most probable path v l (i) = the probability p(s 1,..,s i ;x 1,..,x i |s i =l ) of a most probable path up to i, which ends in state l. Exercise: For i = 1,…,L and for each state l: s1s1 S i-1 X1X1 X i-1 l XiXi...

Viterbi’s algorithm s1s1 s2s2 s L-1 sLsL X1X1 X2X2 X L-1 XLXL sisi XiXi For i=1 to L do for each state l : v l (i) = e l (x i ) MAX k {v k (i-1)a kl } ptr i (l)=arg max k {v k (i-1)a kl } [storing previous state for reconstructing the path] Termination: Initialization: v 0 (0) = 1, v k (0) = 0 for k > 0 0 We add the special initial state 0. Result : p(s 1 *,…,s L * ;x 1,…,x l ) =

2. Computing p(x) S1S1 S2S2 S L-1 SLSL x1x1 x2x2 X L-1 xLxL M M M M TTTT Given an output sequence x = (x 1,…,x L ), Compute the probability that this sequence was generated: The summation taken over all state-paths s generating x.

Forward algorithm for computing p(x) ? ? X1X1 X2X2 sisi XiXi The task: compute Idea: for i=1,…,L and for each state l, compute: f l (i) = p(x 1,…,x i ;s i =l ), the probability of all the paths which emit (x 1,.., x i ) and end in state s i =l. Use the recursive formula:

Forward algorithm for computing p(x) s1s1 s2s2 s L-1 sLsL X1X1 X2X2 X L-1 XLXL sisi XiXi For i=1 to L do for each state l : f l (i) = e l (x i ) ∑ k f k (i-1)a kl Initialization: f 0 (0) := 1, f k (0) := 0 for k>0 0 Similar to Viterbi’s algorithm: Result : p(x 1,…,x L ) =

3. The distribution of S i, given x S1S1 S2S2 S L-1 SLSL x1x1 x2x2 X L-1 xLxL M M M M TTTT Given an output sequence x = (x 1,…,x L ), Compute for each i=1,…,l and for each state k the probability that s i = k ; i.e., p(s i =k| x ). This helps to reply queries like: what is the probability that s i is in a CpG island, etc.

Solution in two stages 1. For each i and each state k, compute p(s i =k | x 1,…,x L ). s1s1 s2s2 s L-1 sLsL X1X1 X2X2 X L-1 XLXL sisi XiXi 2. Do the same computation for every i = 1,.., L but without repeating the first task L times.

Computing for a single i: s1s1 s2s2 s L-1 sLsL X1X1 X2X2 X L-1 XLXL sisi XiXi

Decomposing the computation P(x 1,…,x L,s i ) = P(x 1,…,x i,s i ) P(x i+1,…,x L | x 1,…,x i,s i ) (by the equality p(A,B) = p(A) p(B|A ). s1s1 s2s2 s L-1 sLsL X1X1 X2X2 X L-1 XLXL sisi XiXi P(x 1,…,x i,s i )= f s i (i) ≡ F(s i ), so we are left with the task to compute P(x i+1,…,x L | x 1,…,x i,s i ) ≡ B(s i )

Decomposing the computation s1s1 s2s2 S i+1 sLsL X1X1 X2X2 X i+1 XLXL sisi XiXi Exercise: Show from the definitions of Markov Chain and Hidden Markov Chain that: P(x i+1,…,x L | x 1,…,x i,s i ) = P(x i+1,…,x L | s i ) Denote P(x i+1,…,x L | s i ) ≡ B(s i ).

Decomposing the computation Summary: P(x 1,…,x L,s i ) = P(x 1,…,x i,s i ) P(x i+1,…,x L | x 1,…,x i, s i ) s1s1 s2s2 s L-1 sLsL X1X1 X2X2 X L-1 XLXL sisi XiXi Equality due to independence of ( {x i+1,…,x L }, and {x 1,…,x i } | s i ) – by the Exercise. = P(x 1,…,x i,s i ) P(x i+1,…,x L | s i ) ≡ F(s i )·B(s i )

F(s i ): The Forward algorithm: s1s1 s2s2 s L-1 sLsL X1X1 X2X2 X L-1 XLXL sisi XiXi For i=1 to L do for each state l : F(s i ) = e s i (x i )·∑ s i-1 F (s i-1 )a s i-1 s i Initialization: F (0) = 1 0 The algorithm computes F(s i ) = P(x 1,…,x i,s i ) for i=1,…,L (namely, considering evidence up to time slot i).

B(s i ): The backward algorithm The task: Compute B(s i ) = P(x i+1,…,x L |s i ) for i=L-1,…,1 (namely, considering evidence after time slot i). S L-1 SLSL X L-1 XLXL SiSi S i+1 X i+1 {first step, step L-1: Compute B(s L-1 ).} {step i: compute B(s i ) from B(s i+1 )} P(x i+1,…,x L |s i ) =  P(s i+1 | s i ) P(x i+1 | s i+1 ) P(x i+2,…,x L | s i+1 ) s i+1 B(si) B(si) B(s i+1 ) P(x L | s L-1 ) =  s L P(x L,s L |s L-1 ) =  s L P(s L |s L-1 ) P(x L |s L )

The combined answer 1. To compute the probability that S i =s i given that {x 1,…,x L } run the forward algorithm and compute F(s i ) = P(x 1,…,x i,s i ), run the backward algorithm to compute B(s i ) = P(x i+1,…,x L |s i ), the product F(s i )B(s i ) is the answer (for every possible value s i ). 2. To compute these probabilities for every s i simply run the forward and backward algorithms once, storing F(s i ) and B(s i ) for every i (and every value of s i ). Compute F(s i )B(s i ) for every i. s1s1 s2s2 s L-1 sLsL X1X1 X2X2 X L-1 XLXL sisi XiXi

Time and Space Complexity of the forward/backward algorithms Time complexity is O(m 2 L) where m is the number of states. It is linear in the length of the chain, provided the number of states is a constant. s1s1 s2s2 s L-1 sLsL X1X1 X2X2 X L-1 XLXL sisi XiXi Space complexity is also O(m 2 L).