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Classification ( SVMs / Kernel method)

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Classification (SVMs / Kernel method)

Bafna/Ideker

LP: linear constraints, linear objective function

LP can be solved in polynomial time.

In QP, the objective function contains a quadratic form.

For +ve semindefinite Q, the QP can be solved in polynomial time

Bafna/Ideker

- Suppose we find a separating hyperplane (, 0) s.t.
- For all +ve points x
- Tx-0>=1

- For all +ve points x
- Tx-0 <= -1

- For all +ve points x
- What is the margin of separation?

Tx- 0=1

Tx- 0=0

Tx- 0=-1

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- Solutions with a wider margin are better.

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- In general, data is not linearly separable
- What if we also wanted to minimize misclassified points
- Recall that, each sample xi in our training set has the label yi{-1,1}
- For each point i, yi(Txi-0) should be positive
- Define i >= max {0, 1- yi(Txi-0) }
- If i is correctly classified ( yi(Txi-0) >= 1), and i = 0
- If i is incorrectly classified, or close to the boundaries i > 0
- We must minimize ii

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- Maximimize margin while minimizing misclassification
- Solved using non-linear optimization techniques
- The problem can be reformulated to exclusively using cross products of variables, which allows us to employ the kernel method.
- This gives a lot of power to the method.

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- Goal
- S.t.
- We minimize

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- For fixed >= 0, >= 0, we minimize the lagrangian

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- Substituting (1)

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- Substituting (2,3), we have the minimization problem

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- Under these conditions, the problem is a quadratic programming problem and can be solved using known techniques
- Quiz: When we have solved this QP, how do we classify a point x?

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- The SVM formulation can be solved using QP on dot-products.
- As these are wide-margin classifiers, they provide a more robust solution.
- However, the true power of SVMs approach from using ‘the kernel method’, which allows us to go to higher dimensional (and non-linear spaces)

Bafna/Ideker

- Let X be the set of objects
- Ex: X =the set of samples in micro-arrays.
- Each object xX is a vector of gene expression values

- k: X X -> R is a positivesemidefinitekernel if
- k is symmetric.
- k is +vesemidefinite

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- Quiz: Suppose the objects x are all real vectors (as in gene expression)
- Define
- Is kL a kernel? It is symmetric, but is is +ve semi-definite?

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- Recall X as a matrix, such that each column is a sample
- X=[x1 x2 …]

- By definition, the linear kernel kL=XTX
- For any c

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- Any object can be represented by a feature vector in real space.

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- Note that the feature mapping could actually be non-linear.
- On the flip side, Every kernel can be represented as a dot-product in a high dimensional space.
- Sometimes the kernel space is easier to define than the mapping

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- If an algorithm for vectorial data is expressed exclusively in the form of dot-products, it can be changed to an algorithm on an arbitrary kernel
- Simply replace the dot-product by the kernel

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- Consider a kernel k defined on a mapping
- k(x,x’) = (x)T (x’)

- It could be that is very difficult to compute explicitly, but k is easy to compute
- Suppose we define a distance function between two objects as
- How do we compute this distance?

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- Recall that SVM based classification is described as

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- Applying the kernel trick
- We can try kernels that are biologically relevant

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- Consider a string s = s1, s2,…
- Define an index set I as a subset of indices
- s[I] is the substring limited to those indices
- l(I) = span
- W(I) = cl(I) c<1
- Weight decreases as span increases

- For any string u of length k

l(I)

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- Map every string to a ||n dimensional space, indexed by all strings u of length upto n
- The mapping is expensive, but given two strings s,t,the dot-product kernel k(s,t) = (s)T(t) can be computed in O(n |s| |t|) time

s

u

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- SVM are a generic scheme for classifying data with wide margins and low misclassifications
- For data that is not easily represented as vectors, the kernel trick provides a standard recipe for classification
- Define a meaningful kernel, and solve using SVM

- Many standard kernels are available (linear, poly., RBF, string)

Bafna/Ideker

- We started out by treating the classification problem as one of separating points in high dimensional space
- Obvious for gene expression data, but applicable to any kind of data
- Question of separability, linear separation
- Algorithms for classification
- Perceptron
- Lin. Discriminant
- Max Likelihood
- Linear Programming
- SVMs
- Kernel methods & SVM

Bafna/Ideker

- Recall that we considered 3 problems:
- Group together samples in an unsupervised fashion (clustering)
- Classify based on a training data (often by learning a hyperplane that separates).
- Selection of marker genes that are diagnostic for the class. All other genes can be discarded, leading to lower dimensionality.

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- Many genes have highly correlated expression profiles.
- By discarding some of the genes, we can greatly reduce the dimensionality of the problem.
- There are other, more principled ways to do such dimensionality reduction.

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- With a high enough dimensionality, all points can be linearly separated.
- Recall that a point xi is misclassified if
- it is +ve, but Txi-0<=0
- it is -ve, but Txi+0 > 0

- In the first case choose i s.t.
- Txi-0+i >= 0

- By adding a dimension for each misclassified point, we create a higher dimension hyperplane that perfectly separates all of the points!

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- We get the intrinsic dimensionality of a data-set.

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- Consider the expression values of 2 genes over 6 samples.
- Clearly, the expression of the two genes is highly correlated.
- Projecting all the genes on a single line could explain most of the data.
- This is a generalization of “discarding the gene”.

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- Consider the mean of all points m, and a vector emanating from the mean
- Algebraically, this projection on means that all samples x can be represented by a single value T(x-m)

m

x

x-m

=

T

T(x-m)

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M

- Consider a set of 2 (k) orthonormal vectors 1, 2…
- Once projected, each samplemeans that all samples x can be represented by 2 (k) dimensional vector
- 1T(x-m), 2T(x-m)

2

1

m

x

1T(x-m)

x-m

1T

=

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M

- The generic scheme allows us to project an m dimensional surface into a k dimensional one.
- How do we select the k ‘best’ dimensions?
- The strategy used by PCA is one that maximizes the variance of the projected points around the mean

Bafna/Ideker

- Suppose all of the data were to be reduced by projecting to a single line from the mean.
- How do we select the line ?

m

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- Let each point xk map to x’k=m+ak. We want to minimize the error
- Observation 1: Each point xk maps to x’k = m + T(xk-m)
- (ak= T(xk-m))

xk

x’k

m

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Differentiating w.r.t ak

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- To minimize error, we must maximize TS
- By definition, = TS implies that is an eigenvalue, and the corresponding eigenvector.
- Therefore, we must choose the eigenvector corresponding to the largest eigenvalue.

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- X = starting matrix with n columns, m rows

X

xj

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- The two leukemias need different different therapeutic regimen.
- Usually distinguished through hematopathology
- Can Gene expression be used for a more definitive test?
- 38 bonemarrow samples
- Total mRNA was hybridized against probes for 6817 genes
- Q: Are these classes separable

Bafna/Ideker

- Each gene is represented by an expression vector v(g) = (e1,e2,…,en)
- Choose an idealized expression vector as center.
- Discriminating genes will be ‘closer’ to the center (any distance measure can be used).

Discriminating gene

Bafna/Ideker

- Q: Are there genes, whose expression correlates with one of the two classes
- A: For each class, create an idealized vector c
- Compute the number of genes Nc whose expression ‘matches’ the idealized expression vector
- Is Nc significantly larger thanNc* for a random c*?

Bafna/Ideker

- Distance measure used:
- For any binary vector c, let the one entries denote class 1, and the 0 entries denote class 2
- Compute mean and std. dev. [1(g),1(g)] of expression in class 1 and also [2(g),2(g)].
- P(g,c) = [1(g)-2(g)]/ [1(g)+2(g)]
- N1(c,r) = {g | P(g,c) == r}
- High density for some r is indicative of correlation with class distinction
- Neighborhood is significant if a random center does not produce the same density.

Bafna/Ideker

- #{g |P(g,c) > 0.3} > 709 (ALL) vs 173 by chance.
- Class prediction should be possible using micro-array expression values.

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- Choose a fixed set of informative genes (based on their correlation with the class distinction).
- The predictor is uniquely defined by the sample and the subset of informative genes.

- For each informative gene g, define (wg,bg).
- wg=P(g,c) (When is this +ve?)
- bg = [1(g)+2(g)]/2

- Given a new sample X
- xg is the normalized expression value at g
- Vote of gene g =wg|xg-bg| (+ve value is a vote for class 1, and negative for class 2)

Bafna/Ideker

- PS = [Vwin-Vlose]/[Vwin+Vlose]
- Reflects the margin of victory

- A 50 gene predictor is correct 36/38 (cross-validation)
- Prediction accuracy on other samples 100% (prediction made for 29/34 samples.
- Median PS = 0.73
- Other predictors between 10 and 200 genes all worked well.

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- Do the predictive genes reveal any biology?
- Initial expectation is that most genes would be of a hematopoetic lineage.
- However, many genes encode
- Cell cycle progression genes
- Chromatin remodelling
- Transcription
- Known oncogenes
- Leukemia targets (etopside)

Bafna/Ideker

ML when the covariance matrix is a diagonal matrix with identical variance for different classes is similar to Golub’s classifier

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- The classification of different cancers is over years of hypothesis driven research.
- Suppose you were given unlabeled samples of ALL/AML. Would you be able to distinguish the two classes?

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- SOMs was applied to group the 38 samples
- Class A1 contained 24/25 ALL and 3/13 AML samples.
- How can we validate this?
- Use the labels to do supervised classification via cross-validation
- A 20 gene predictor gave 34 accurate predictions, 1 error, and 2 of 3 uncertains

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