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A modeling Example

A+. C+. G+. T+. A-. C-. G-. T-. A modeling Example. CpG islands in DNA sequences. Methylation & Silencing. One way cells differentiate is methylation Addition of CH 3 in C-nucleotides Silences genes in region CG (denoted CpG) often mutates to TG, when methylated

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A modeling Example

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  1. A+ C+ G+ T+ A- C- G- T- A modeling Example CpG islands in DNA sequences

  2. Methylation & Silencing • One way cells differentiate is methylation • Addition of CH3 in C-nucleotides • Silences genes in region • CG (denoted CpG) often mutates to TG, when methylated • In each cell, one copy of X is silenced, methylation plays role • Methylation is inherited during cell division

  3. Example: CpG Islands CpG nucleotides in the genome are frequently methylated (Write CpG not to confuse with CG base pair) C  methyl-C  T Methylation often suppressed around genes, promoters  CpG islands

  4. Example: CpG Islands • In CpG islands, • CG is more frequent • Other pairs (AA, AG, AT…) have different frequencies Question: Detect CpG islands computationally

  5. A model of CpG Islands – (1) Architecture A+ C+ G+ T+ CpG Island A- C- G- T- Not CpG Island

  6. A model of CpG Islands – (2) Transitions How do we estimate parameters of the model? Emission probabilities: 1/0 • Transition probabilities within CpG islands Established from known CpG islands (Training Set) • Transition probabilities within other regions Established from known non-CpG islands (Training Set) Note: these transitions out of each state add up to one—no room for transitions between (+) and (-) states = 1 = 1 = 1 = 1 = 1 = 1 = 1 = 1

  7. Log Likehoods—Telling “CpG Island” from “Non-CpG Island” Another way to see effects of transitions: Log likelihoods L(u, v) = log[ P(uv | + ) / P(uv | -) ] Given a region x = x1…xN A quick-&-dirty way to decide whether entire x is CpG P(x is CpG) > P(x is not CpG)  i L(xi, xi+1) > 0

  8. A model of CpG Islands – (2) Transitions • What about transitions between (+) and (-) states? • They affect • Avg. length of CpG island • Avg. separation between two CpG islands 1-p Length distribution of region X: P[lX = 1] = 1-p P[lX = 2] = p(1-p) … P[lX= k] = pk-1(1-p) E[lX] = 1/(1-p) Geometric distribution, with mean 1/(1-p) X Y p q 1-q

  9. What if a new genome comes? • We just sequenced the porcupine genome • We know CpG islands play the same role in this genome • However, we have no known CpG islands for porcupines • We suspect the frequency and characteristics of CpG islands are quite different in porcupines How do we adjust the parameters in our model? LEARNING

  10. Problem 3: Learning Re-estimate the parameters of the model based on training data

  11. Two learning scenarios • Estimation when the “right answer” is known Examples: GIVEN: a genomic region x = x1…x1,000,000 where we have good (experimental) annotations of the CpG islands GIVEN: the casino player allows us to observe him one evening, as he changes dice and produces 10,000 rolls • Estimation when the “right answer” is unknown Examples: GIVEN: the porcupine genome; we don’t know how frequent are the CpG islands there, neither do we know their composition GIVEN: 10,000 rolls of the casino player, but we don’t see when he changes dice QUESTION: Update the parameters  of the model to maximize P(x|)

  12. 1. When the right answer is known Given x = x1…xN for which the true  = 1…N is known, Define: Akl = # times kl transition occurs in  Ek(b) = # times state k in  emits b in x We can show that the maximum likelihood parameters  (maximize P(x|)) are: Akl Ek(b) akl = ––––– ek(b) = ––––––– i AkicEk(c)

  13. 1. When the right answer is known Intuition: When we know the underlying states, Best estimate is the normalized frequency of transitions & emissions that occur in the training data Drawback: Given little data, there may be overfitting: P(x|) is maximized, but  is unreasonable 0 probabilities – BAD Example: Given 10 casino rolls, we observe x = 2, 1, 5, 6, 1, 2, 3, 6, 2, 3  = F, F, F, F, F, F, F, F, F, F Then: aFF = 1; aFL = 0 eF(1) = eF(3) = .2; eF(2) = .3; eF(4) = 0; eF(5) = eF(6) = .1

  14. Pseudocounts Solution for small training sets: Add pseudocounts Akl = # times kl transition occurs in  + rkl Ek(b) = # times state k in  emits b in x + rk(b) rkl, rk(b) are pseudocounts representing our prior belief Larger pseudocounts  Strong priof belief Small pseudocounts ( < 1): just to avoid 0 probabilities

  15. 2. When the right answer is unknown We don’t know the true Akl, Ek(b) Idea: • We estimate our “best guess” on what Akl, Ek(b) are • Or, we start with random / uniform values • We update the parameters of the model, based on our guess • We repeat

  16. 2. When the right answer is unknown Starting with our best guess of a model M, parameters : Given x = x1…xN for which the true  = 1…N is unknown, We can get to a provably more likely parameter set  i.e.,  that increases the probability P(x | ) Principle: EXPECTATION MAXIMIZATION • Estimate Akl, Ek(b) in the training data • Update  according to Akl, Ek(b) • Repeat 1 & 2, until convergence

  17. Estimating new parameters To estimate Akl: (assume “|CURRENT”, in all formulas below) At each position i of sequence x, find probability transition kl is used: P(i = k, i+1 = l | x) = [1/P(x)]  P(i = k, i+1 = l, x1…xN) = Q/P(x) where Q = P(x1…xi, i = k, i+1 = l, xi+1…xN) = = P(i+1 = l, xi+1…xN | i = k) P(x1…xi, i = k) = = P(i+1 = l, xi+1xi+2…xN | i = k) fk(i) = = P(xi+2…xN | i+1 = l) P(xi+1 | i+1 = l) P(i+1 = l | i = k) fk(i) = = bl(i+1) el(xi+1) akl fk(i) fk(i) akl el(xi+1) bl(i+1) So: P(i = k, i+1 = l | x, ) = –––––––––––––––––– P(x | CURRENT)

  18. Estimating new parameters • So, Akl is the E[# times transition kl, given current ] fk(i) akl el(xi+1) bl(i+1) Akl = i P(i = k, i+1 = l | x, ) = i ––––––––––––––––– P(x | ) • Similarly, Ek(b) = [1/P(x | )]{i | xi = b} fk(i) bk(i) fk(i) bl(i+1) akl k l xi+2………xN x1………xi-1 el(xi+1) xi xi+1

  19. The Baum-Welch Algorithm Initialization: Pick the best-guess for model parameters (or arbitrary) Iteration: • Forward • Backward • Calculate Akl, Ek(b), given CURRENT • Calculate new model parameters NEW : akl, ek(b) • Calculate new log-likelihood P(x | NEW) GUARANTEED TO BE HIGHER BY EXPECTATION-MAXIMIZATION Until P(x | ) does not change much

  20. The Baum-Welch Algorithm Time Complexity: # iterations  O(K2N) • Guaranteed to increase the log likelihood P(x | ) • Not guaranteed to find globally best parameters Converges to local optimum, depending on initial conditions • Too many parameters / too large model: Overtraining

  21. Alternative: Viterbi Training Initialization: Same Iteration: • Perform Viterbi, to find * • Calculate Akl, Ek(b) according to * + pseudocounts • Calculate the new parameters akl, ek(b) Until convergence Notes: • Not guaranteed to increase P(x | ) • Guaranteed to increase P(x | , *) • In general, worse performance than Baum-Welch

  22. Conditional Random Fields A brief description of a relatively new kind of graphical model

  23. 1 1 1 1 … 2 2 2 2 … … … … … K K K K … Let’s look at an HMM again 1 Why are HMMs convenient to use? • Because we can do dynamic programming with them! • “Best” state sequence for 1…i interacts with “best” sequence for i+1…N using K2 arrows Vl(i+1) = el(i+1) maxk Vk(i) akl = maxk( Vk(i) + [ e(l, i+1) + a(k, l) ] ) (where e(.,.) and a(.,.) are logs) • Total likelihood of all state sequences for 1…i+1 can be calculated from total likelihood for 1…i by only summing up K2 arrows 2 2 K x1 x2 x3 xN

  24. 1 1 1 1 … 2 2 2 2 … … … … … K K K K … Let’s look at an HMM again 1 • Some shortcomings of HMMs • Can’t model state duration • Solution: explicit duration models (Semi-Markov HMMs) • Unfortunately, state i cannot “look” at any letter other than xi! • Strong independence assumption: P(i | x1…xi-1, 1…i-1) = P(i | i-1) 2 2 K x1 x2 x3 xN

  25. 1 1 1 1 … 2 2 2 2 … … … … … K K K K … Let’s look at an HMM again 1 • Another way to put this, features used in objective function P(x, ): • akl, ek(b), where b   • At position i: all K2akl features, and all K el(xi) features play a role • OK forget probabilistic interpretation for a moment • “Given that prev. state is k, current state is l, how much is current score?” • Vl(i) = Vk(i – 1) + (a(k, l) + e(l, i)) = Vk(i – 1) + g(k, l, xi) • Let’s generalize g!!! Vk(i – 1) + g(k, l, x, i) 2 2 K x1 x2 x3 xN

  26. “Features” that depend on many pos. in x i-1 i • What do we put in g(k, l, x, i)? • The “higher” g(k, l, x, i), the more we like going from k to l at position i • Richer models using this additional power • Examples • Casino player looks at previous 100 pos’ns; if > 50 6s, he likes to go to Fair g(Loaded, Fair, x, i) += 1[xi-100, …, xi-1 has > 50 6s]  wDON’T_GET_CAUGHT • Genes are close to CpG islands; for any state k, g(k, exon, x, i) += 1[xi-1000, …, xi+1000 has > 1/16 CpG]  wCG_RICH_REGION x7 x8 x9 x10 x1 x2 x3 x4 x5 x6

  27. “Features” that depend on many pos. in x x7 x8 x9 x10 x1 x2 x3 x4 x5 x6 Conditional Random Fields—Features • Define a set of features that you think are important • All features should be functions of current state, previous state, x, and position i • Example: • Old features: transition kl, emission b from state k • Plus new features: prev 100 letters have 50 6s • Number the features 1…n: f1(k, l, x, i), …, fn(k, l, x, i) • features are indicator true/false variables • Find appropriate weights w1,…, wn for when each feature is true • weights are the parameters of the model • Let’s assume for now each feature has a weight wj • Then, g(k, l, x, i) = j=1…nfj(k, l, x, i)  wj

  28. “Features” that depend on many pos. in x x7 x8 x9 x10 x1 x2 x3 x4 x5 x6 Define Vk(i): Optimal score of “parsing” x1…xi and ending in state k Then, assuming Vk(i) is optimal for every k at position i, it follows that Vl(i+1) = maxk [Vk(i) + g(k, l, x, i+1)] Why? Even though at pos’n i+1 we “look” at arbitrary positions in x, we are only “affected” by the choice of ending state k Therefore, Viterbi algorithm again finds optimal (highest scoring) parse for x1…xN

  29. 1 2 3 4 5 6 … x1 x2 x3 x4 x5 x6 1 2 3 4 5 6 … x1 x2 x3 x4 x5 x6 “Features” that depend on many pos. in x • Score of a parse depends on all of x at each position • Can still do Viterbi because state i only “looks” at prev. state i-1 and the constant sequence x HMM CRF

  30. How many parameters are there, in general? • Arbitrarily many parameters! • For example, let fj(k, l, x, i) depend on xi-5, xi-4, …, xi+5 • Then, we would have up to K  |  |11 parameters! • Advantage: powerful, expressive model • Example: “if there are more than 50 6’s in the last 100 rolls, but in the surrounding 18 rolls there are at most 3 6’s, this is evidence we are in Fair state” • Interpretation: casino player is afraid to be caught, so switches to Fair when he sees too many 6’s • Example: “if there are any CG-rich regions in the vicinity (window of 2000 pos) then favor predicting lots of genes in this region” • Question: how do we train these parameters?

  31. Conditional Training • Hidden Markov Model training: • Given training sequence x, “true” parse  • Maximize P(x, ) • Disadvantage: • P(x, ) = P( | x)P(x) Quantity we care about so as to get a good parse Quantity we don’t care so much about because x is always given

  32. Conditional Training P(x, ) = P( | x)P(x) P( | x) = P(x, ) / P(x) Recall F(j, x, ) = # times feature fj occurs in (x, ) = i=1…N fj(k, l, x, i) ; count fj in x,  In HMMs, let’s denote by wj the weight of jth feature: wj = log(akl) or log(ek(b)) Then, HMM: P(x, ) =exp[j=1…n wj  F(j, x, )] CRF: Score(x, ) =exp[j=1…n wj  F(j, x, )]

  33. Conditional Training In HMMs, P( | x) = P(x, ) / P(x) P(x, ) =exp[j=1…n wjF(j, x, )] P(x) = exp[j=1…n wjF(j, x, )]=: Z Then, in CRF we can do the same to normalize Score(x, ) into a prob. PCRF( | x) = exp[j=1…n wjF(j, x, )]/ Z QUESTION: Why is this a probability???

  34. Conditional Training • We need to be given a set of sequences x and “true” parses  • Calculate Z by a sum-of-paths algorithm similar to HMM • We can then easily calculate P( | x) • Calculate partial derivative of P( | x) w.r.t. each parameter wj (not covered—akin to forward/backward) • Update each parameter with gradient descent! • Continue until convergence to optimal set of weights P( | x) = exp[j=1…n wjF(j, x, )]/ Z is convex!!!

  35. Conditional Random Fields—Summary • Ability to incorporate complicated non-local feature sets • Do away with some independence assumptions of HMMs • Parsing is still equally efficient • Conditional training • Train parameters that are best for parsing, not modeling • Need labeled examples—sequences x and “true” parses  (Can train on unlabeled sequences, however it is unreasonable to train too many parameters this way) • Training is significantly slower—many iterations of forward/backward

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