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Pushpak Bhattacharyya CSE Dept., IIT Bombay 5 th and 6 th Nov , 2012

CS460/626 : Natural Language Processing/Speech, NLP and the Web Lecture 31-32: Expectation Maximisation. Pushpak Bhattacharyya CSE Dept., IIT Bombay 5 th and 6 th Nov , 2012. Some Useful mathematical concepts. Convex/ concave functions Jensen’s inequality

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Pushpak Bhattacharyya CSE Dept., IIT Bombay 5 th and 6 th Nov , 2012

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  1. CS460/626 : Natural Language Processing/Speech, NLP and the WebLecture 31-32:Expectation Maximisation Pushpak BhattacharyyaCSE Dept., IIT Bombay 5th and 6th Nov, 2012

  2. Some Useful mathematical concepts • Convex/ concave functions • Jensen’s inequality • Kullback–Leibler distance/divergence

  3. Criteria for convexity • A function f(x) is said to be convex in the interval [a,b] iff

  4. Jensen’s inequality • For any convex function f(x) Where and

  5. Proof of Jensen´s inequality • Method:- By induction on N • Base case:-

  6. Another base case • N = 2

  7. Hypothesis

  8. Induction Step

  9. Proof

  10. Continued... • Examine each μi

  11. Continued... • Therefore,

  12. KL -divergence • We will do the discrete form of probability distribution. • Given two probability distribution P,Q on the random variable • X : x1,x2,x3...xN • P:p1=p(x1 ), p2=p(x2), ... pn=p(xn) • Q:q1=q(x1 ), q2=q(x2), ... qn=q(xn)

  13. KLD definition

  14. Proof: KLD>=0

  15. Proof cntd. • Apply Jensen’s inequality

  16. Convexity of –log x

  17. Interesting problem • Try to prove:-

  18. 2nd definition of convexity • Theorem:

  19. Lemma 1 a s z t b

  20. Mean Value Theorem

  21. Alternative form of z Add –λz to both sides

  22. Alternative form of convexity Add –λf(z) to both sides

  23. Proof: second derivative >=0 implies convexity (1/2) We have that (1) (2)

  24. Second derivative >=0 implies convexity (2/2) (2) Is equivalent to For some s and t, where Now since f’’(x) >=0 Combining this with (1), the result is proved

  25. Why all this • In EM, we maximize the expectation of log likelihood of the data • Log is a concave function • We have to take iterative steps to get to the maximum • There are two unknown values: Z (unobserved data) and θ(parameters) • From θ, get new value of Z (E-step) • From Z, get new value of θ(M-step)

  26. How to change θ • How to choose the next θ? • Take argmaxθ(LL(X,Z:θ) - LL(X,Z:θn)) Where, X: observed data Z: unobserved data Θ: parameter LL(X,Z:θn): log likelihood of complete data with parameter value at θn This is in lieu of, for example, gradient ascent Θ θn At every step LL(.) will Increase, ultimately reaching local/global maximum

  27. Why expectation of log likelihood? (1/2) • P(X:θ) may not be a convenient mathematical expression • Deal with P(X,Z:θ), marginalized over Z • Log(ΣZP(X,Z:θ)) is mathematically processed with multiplying by P(Z|X: θn) which for each Z is between 0 and 1 and sums to 1 • Then Jensen inequality will give Log(ΣZP(X,Z:θ)) >= Log(ΣZP(Z|X: θn)P(X,Z:θ)/P(Z|X: θn)) = ΣZP(Z|X: θn)Log(P(X,Z:θ)/P(Z|X: θn))

  28. Why expectation of log likelihood? (2/2) LL(X:θ) – LL(X:θn) =Log(ΣZP(X,Z:θ)) - Log(P(X:θn)) >= Log(ΣZP(Z|X: θn)P(X,Z:θ)/P(Z|X: θn)) - Log(P(X:θn)) = ΣZP(Z|X: θn)Log(P(X,Z:θ)/(P(Z|X: θn).P(X:θn)) since ΣZP(Z|X: θn)=1 = ΣZP(Z|X: θn)Log((P(X,Z:θ)/(P(X,Z:θn)) So, argmaxθ(LL(X:θ) – LL(X:θn)) =ΣZP(Z|X: θn)Log(P(X,Z:θ) = EZ(Log(P(X,Z:θ)), where EZ(.) is the expectation of log likelihood of complete data wrtZ

  29. Why expectation of Z? • If the log likelihood is a linear function of Z, then the expectation can be carries inside of the log likelihood and E(Z) is computed • The above is true when the hidden variables form a mixture of distributions (e..g, in tosses of two coins), and • Each distribution is an exponential distribution like multinomial/normal/poisson

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