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 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

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

Example: CpG Islands

• In CpG islands,

CG is more frequent

Other pairs (AA, AG, AT…) have different frequencies

Question: Detect CpG islands computationally

A model of CpG Islands – (1) Architecture

A+ C+ G+ T+

A- C- G- T-

CpG Island

Not CpG Island

A model of CpG Islands – (2) TransitionsHow do we estimate parameters of the model?

Emission probabilities: 1/0

1. Transition probabilities within CpG islands

Established from known CpG islands (Training Set)

2. 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

+ A C G TA .180 .274 .426 .120

C .171 .368 .274 .188

G .161 .339 .375 .125

T .079 .355 .384 .182

- A C G TA .300 .205 .285 .210

C .233 .298 .078 .302

G .248 .246 .298 .208

T .177 .239 .292 .292

= 1

= 1

= 1

= 1

= 1

= 1

= 1

= 1

Log Likehoods—Telling “CpG Island” from “Non-CpG Island”

A C G T

A -0.740 +0.419 +0.580 -0.803

C -0.913 +0.302 +1.812 -0.685

G -0.624 +0.461 +0.331 -0.730

T -1.169 +0.573 +0.393 -0.679

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

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

X Y

1-p

1-q

p q

Length distribution of region X:

P[lX = 1] = 1-pP[lX = 2] = p(1-p)…P[lX= k] = pk-1(1-p)

E[lX] = 1/(1-p)Geometric distribution, with mean

1/(1-p)

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

Learning

Re-estimate the parameters of the model based on training data

Two learning scenarios

1. 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

2. 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|)

1. When the right answer is known

Given x = x1…xN

for which the true = 1…N is known,

Define:

Akl = # times kl 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 c Ek(c)

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 unreasonable0 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 = 0eF(1) = eF(3) = .2; eF(2) = .3; eF(4) = 0; eF(5) = eF(6) = .1

Pseudocounts

Solution for small training sets:

Add pseudocounts

Akl = # times kl 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

Pseudocounts

Example: dishonest casino

We will observe player for one day, 600 rolls

Reasonable pseudocounts:

r0F = r0L = rF0 = rL0 = 1;rFL = rLF = rFF = rLL = 1;rF(1) = rF(2) = … = rF(6) = 20 (strong belief fair is fair)rL(1) = rL(2) = … = rL(6) = 5 (wait and see for loaded)

Above #s are arbitrary – assigning priors is an art

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

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

1. Estimate Akl, Ek(b) in the training data2. Update according to Akl, Ek(b)3. Repeat 1 & 2, until convergence

Estimating new parameters

To estimate Akl: (assume “| CURRENT”, in all formulas below)

At each position i of sequence x, find probability transition kl 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)

Estimating new parameters

• So, Akl is the E[# times transition kl, 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)

k l

xi+1

akl

el(xi+1)

bl(i+1)fk(i)

x1………xi-1 xi+2………xN

xi

The Baum-Welch Algorithm

Initialization:Pick the best-guess for model parameters

(or arbitrary)

Iteration:1. Forward2. Backward3. Calculate Akl, Ek(b), given CURRENT

4. Calculate new model parameters NEW : akl, ek(b)5. Calculate new log-likelihood P(x | NEW)

GUARANTEED TO BE HIGHER BY EXPECTATION-MAXIMIZATION

Until P(x | ) does not change much

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

Alternative: Viterbi Training

Initialization: Same

Iteration:1. Perform Viterbi, to find *

2. Calculate Akl, Ek(b) according to * + pseudocounts3. 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